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Journal of Applied Mathematics and Physics, 2021, 9, 2038-2113
https://www.scirp.org/journal/jamp
ISSN Online: 2327-4379
ISSN Print: 2327-4352
DOI:
10.4236/jamp.2021.98132 Aug. 31, 2021 2038
Journal of Applied Mathematics and Physics
Pythagoreans Figurative Numbers: The
Beginning of Number Theory and Summation of
Series
Ravi P. Agarwal
Department of Mathematics, Texas A & M University-Kingsville, Kingsville, Texas, USA
Abstract
In this article we shall examine several different types
of figurative numbers
which have been studied extensively over the period of 2500 years, and cur-
rently scattered on hundreds of websites. We shall discuss their computation
through simple recurrence relations, patterns and properties, and mutual re-
lationships which have led to curious results in the field of elementary num-
ber theory. Further,
for each type of figurative numbers we shall show that
the addition of first finite numbers and infinite addition of their inverses of-
ten require new/strange techniques. We sincerely hope that besides experts,
students and teachers of mathematics will also be benefited with this article.
Keywords
Figurative Numbers, Patterns and Properties, Relations, Sums of Finite and
Infinite Series, History
1. Introduction
Pythagoras of Samos (around 582-481 BC, Greece) and his several followers, es-
pecially, Hypsicles of Alexandria (around 190-120 BC, Greece), Plutarch of
Chaeronea (around 46-120, Greece), Nicomachus of Gerasa (around 60-120,
Jordan-Israel), and Theon of Smyrna (70-135, Greece) portrayed natural num-
bers in orderly geometrical configuration of points/dots/pebbles and labeled
them as
figurative
numbers
. From these arrangements, they deduced some asto-
nishing number-theoretic results. This was indeed the beginning of the number
theory, and an attempt to relate geometry with arithmetic. Nicomachus in his
book, see [1], originally written about 100 A.D., collected earlier works of Py-
thagoreans on natural numbers, and presented cubic figurative numbers (solid
How to cite this paper:
Agarwal, R.P.
(20
21)
Pythagoreans Figurative Numbers:
The Beginning of Number Theory and
Summ
ation of Series.
Journal of Applied
Math
ematics and Physics
,
9
, 2038-2113.
https://doi.org/10.4236/jamp.2021.98132
Received:
July 13, 2021
Accepted:
August 28, 2021
Published:
August 31, 2021
Copyright © 20
21 by author(s) and
Scientific
Research Publishing Inc.
This work is
licensed under the Creative
Commons Attribution International
License (CC BY
4.0).
http://creativecommons.org/licenses/by/4.0/
Open Access
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2039
Journal of Applied Mathematics and Physics
numbers). Thus, figurate numbers had been studied by the ancient Greeks for
polygonal numbers, pyramidal numbers, and cubes. The connection between
regular geometric figures and the corresponding sequences of figurative num-
bers was profoundly significant in Plato’s science, after Plato of Athens (around
427-347 BC, Greece), for example in his work
Timaeus
. The study of figurative
numbers was further advanced by Diophantus of Alexandria (about 250, Greece).
His main interest was in figurate numbers based on the Platonic solids (tetrahe-
dron, cube, octahedron, dodecahedron, and icosahedron), which he documented
in
De
solidorum
elementis
. However, this treatise was lost, and rediscovered on-
ly in 1860. Dicuilus (flourished 825, Ireland) wrote
Astronomical
Treatise
in
Latin about 814-816, which contains a chapter on triangular and square num-
bers, see Ross and Knott [2]. After Diophantus’s work, several prominent ma-
thematicians took interest in figurative numbers. The long list includes: Leonar-
do of Pisa/Fibonacci (around 1170-1250, Italy), Michael Stifel (1486-1567, Ger-
many), Gerolamo Cardano (1501-1576, Italy), Johann Faulhaber (1580-1635,
German), Claude Gaspard Bachet de Meziriac (1581-1638, France), René Des-
cartes (1596-1650, France), Pierre de Fermat (1601-1665, France), John Pell
(1611-1685, England). In 1665, Blaise Pascal (1623-1662, France) wrote the
Traité
du
triangle
arithmétique
,
avec
quelques
autres
petits
traitez
sur
la
mesme
matiére
which contains some details of figurate numbers. Work of Leonhard
Euler (1707-1783, Switzerland) and Joseph Louis Lagrange (1736-1813, France)
on figurate numbers opened new avenues in number theory. Octahedral num-
bers were extensively examined by Friedrich Wilhelm Marpurg (1718-1795,
German) in 1774, and Georg Simon Klügel (1739-1812, Germany) in 1808. The
Pythagoreans could not have anticipated that figurative numbers would engage
after 2000 years leading scholars such as Adrien-Marie Legendre (1752-1833,
France), Karl Friedrich Gauss (1777-1855, Germany), Augustin-Louis Cauchy
(1789-1857, France), Carl Guslov Jacob Jacobi (1804-1851, Germany), and Wac-
law Franciszek Sierpiński (1882-1969, Poland). In 2011, Michel Marie Deza
(1939-2016, Russia-France) and Elena Deza (Russia) in their book [3] had given
an extensive information about figurative numbers.
In this article we shall systematically discuss most popular polygonal, centered
polygonal, three dimensional numbers (including pyramidal numbers), and four
dimensional figurative numbers. We shall begin with triangular numbers and
end this article with pentatope numbers. For each type of polygonal figurative
numbers, we shall provide definition in terms of a sequence, possible sketch, ex-
plicit formula, possible relations within the class of numbers through simple re-
currence relations, properties of these numbers, generating function, sum of first
finite numbers, sum of all their inverses, and relations with other types of poly-
gonal figurative numbers. For each other type of figurative numbers mainly we
shall furnish definition in terms of a sequence, possible sketch, explicit formula,
generating function, sum of first finite numbers, and sum of all their inverses.
The study of figurative numbers is interesting in its own sack, and often these
numbers occur in real world situations. We sincerely hope after reading this ar-
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
ticle it will be possible to find new representations, patterns, relations with other
types of popular numbers which are not discussed here, extensions, and real ap-
plications.
2. Triangular Numbers
In this arrangement rows contain
1,2,3, 4, ,n
dots (see Figure 1).
From Figure 1 it follows that each new triangular number is obtained from
the previous triangular number by adding another row containing one more dot
than the previous row added, and hence
n
t
is the sum of the first
n
positive in-
tegers,
i.e.
,
() ( )
12
1 123 1 ,
nn n
ttntnn nn
−−
= += + −+= =+++ + −+
(1)
i.e.
, the differences between successive triangular numbers produce the sequence
of natural numbers. To find the sum in (1) we shall discuss two methods which
are innovative.
Method 1. Since
( )
( ) ( )
123 1
1 2 21
n
n
t nn
t nn n
=+++ + − +
=+−+−+++
An addition of these two arrangements immediately gives
() ( )()( )
2 11 1 1
n
t n n n nn
= ++ ++ + += +
and hence
( )
2
1
111
.
2 22
n
nk
nn
t k nn
=
+
= = = +
∑
(2)
Thus, it immediately follows that
11t=
,
2
123t=+=
,
3
123 6t=++=
,
4
123410t=+++=
,
5
15t=
,
6
21t=
,
7
28t=
,
. This method was first
employed by Gauss. The story is his elementary school teacher asked the class to
add up the numbers from 1 to 100, expecting to keep them busy for a long time.
Young Gauss found the Formula (2) instantly and wrote down the correct an-
swer 5050.
Method 2. From Figure 2
Proof
without
words
of (2) is immediate, see Alsina
and Nelsen [4]. However, a needless explanation is a “stairstep” configuration
made up of one block plus two blocks plus three blocks, etc, replicated it as the
shaded section in Figure 2, and fit them together to form an
( )
1
nn×+
rectan-
gular array. Because the rectangle is made of two identical stairsteps (each
Figure 1. Triangular numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
Figure 2. Proof of (2) without words.
representing
n
t
) and the rectangle’s area is the product of base and height, that
is,
( )
1nn
+
, then the stairstep’s area must be half of the rectangle’s, and hence
(2) holds.
To prove (2) the
Principle
of
mathematical
induction
is routinely used. The
relation (1) is a special case of an arithmetic progression of the finite sequence
{ }
, 0,1, , 1
k
ak n= −
where
k
a a kd= +
, or
()
, 0
k
a a k dk
= + − ≥≥
,
i.e.
,
( ) ( ) ( ) ( )
( )
1
0
2 1.
n
k
S akd a ad a d a n d
−
=
= + =++++ +++−
∑
(3)
For this, following the Method 1, it immediately follows that
( )
[ ]
01
21 .
22
n
nn
S an d aa
−
= +− = +
(4)
Thus, the mean value of the series is
()
01
2
n
S Sn a a
−
= = +
, which is similar
as in discrete uniform distribution. For
1ad= =
, (3) reduces to (1), and (4)
becomes same as (2). From (4), it is also clear that
()() ( )
() ( )
( )
1 11
00
21 2 1
22
2 1.
2
n nm
km k k
a kd a kd a kd
nm
an d am d
nm a nm d
− −−
= = =
+= +− +
= +− − + −
−
= + +−
∑ ∑∑
(5)
Ancient Indian Sulbas (see Agarwal and Sen [5]) contain several examples of
arithmetic progression. Aryabhata (born 2765 BC) besides giving the Formula (4)
also obtained
n
in terms of
S
, namely,
( )
2
82 2
1
1
2
Sd a d a
nd
+−−
= +
(6)
He also provided elegant results for the summation of series of squares and
cubes. In Rhind Papyruses (about 1850 and 1650 BC) out of 87 problems two
problems deal with arithmetical progressions and seem to indicate that Egyptian
scriber Ahmes (around 1680-1620 BC) knew how to sum such series. For exam-
ple, Problem 40 concerns an arithmetic progression of five terms. It states: divide
R. P. Agarwal
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Journal of Applied Mathematics and Physics
100 loaves among 5 men so that the sum of the three largest shares is 7 times the
sum of the two smallest
(
( ) ( ) ( ) ( )
2 3 4 100xxdxdxdxd++++ ++ ++ =
,
( ) ( ) ( ) ( )
7 234xxd xd xd xd++ =+ ++ ++
,
10 6x=
,
55 6d=
). There is a
discussion of arithmetical progression in the works of Archimedes of Syracuse
(287-212 BC, Greece), Hypsicles, Brahmagupta (born 30 BC, India), Diophantus,
Zhang Qiujian (around 430-490, China), Bhaskara II or Bhaskaracharya (working
486, India), Alcuin of York (around 735-804, England), Dicuil, Fibonacci, Jo-
hannes de Sacrobosco (around 1195-1256, England), Levi ben Gershon (1288-1344,
France). Abraham De Moivre (1667-1754. England) predicted the day of his own
death. He found that he slept 15 minutes longer each night, and summing the
arithmetic progression, calculated that he would die on November 27, 1754, the
day that he would sleep all 24 hours. Peter Gustav Lejeune Dirichlet (1805-1859,
Germany) showed that there are infinitely many primes in the arithmetic progres-
sion
an b+
, where
a
and
b
are relatively prime. Enrico Bombieri (born 1940, Italy)
is known for the distribution of prime numbers in arithmetic progressions. Te-
rence Chi-Shen Tao (born 1975, Australia-USA) showed that there exist arbitra-
rily long arithmetic progressions of prime numbers.
The following equalities between triangular numbers can be proved rather
easily.
2
22
1
12
1 21
11
3
3
nn n
nn n
nn n
n m nm
n m n m nm
tt t
tt t
tt t
t t nm t
tt t t t
−
−
++
+
−−
+=
+=
+=
++ =
+
Instead of adding the above finite arithmetic series
{ }
k
a
, we can multiply its
terms which in terms of Gamma function
Γ
can be written as
( )
1
01 1 0
,
nn
nk
an
d
a a a a kd d a
d
−
−=
Γ+
= +=
Γ
∏
(7)
provided
a
/
d
is nonpositive.
The triangular number
n
t
solves the
handshake
problem
of counting the
number of handshakes if each person in a room with
( )
1
n+
people shakes
hands once with each person. Similarly a fully connected network of
( )
1n+
computing devices requires
n
t
connections. The triangular number
n
t
also
provides the number of games played by
( )
1
n+
teams in a
Round-Robin
Tournament
in which each team plays every other team exactly once and no
ties are allowed. Further, the triangular number
n
t
is the number of ordered
pairs
( )
,xy
, where
1xyn≤≤≤
. For an
( )
1n+
sided-polygon, the num-
ber of diagonals is
( )( )
1
1 2 2 2 , 2
nn
n n tt n
+
+− =− ≥
. From Figure 1, it fol-
lows that the number of line segments between closest pairs of dots in the
triangles is
( )
1
3 3 1 2
nn
t nn
−
= = −
, or recursively,
( )
1
31
nn
n
−
= +−
,
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
1
0=
. Thus, for example,
4
18=
. A problem of Christoff Rudolff
(1499-1545, Poland) reads: I am owed 3240
florins
. The debtor pays me 1
flo-
rin
the first day, 2 the second day, 3 the third day, and so on. How many days
it takes to pay off the debt (80 days). For the Pythagoreans the fourth trian-
gular number
4
10t=
(decade) was most significant of all: it contains in it-
self the first four integers, one, two, three, and four
123410+++ =
, it was
considered to be a symbol of “perfection”, being the sum of 1 (a point), 2 (a
line), 3 (a plane) and 4 (a solid); it is the smallest integer
n
for which there
are just as many primes between 1 and
n
as nonprimes, and it gives rise to
the tetraktys (see Figure 1 and its alternative form Figure 3). To them, the
tetraktys was the sum of the divine influences that hold the universe together,
or the sum of all the manifest laws of nature. They recognized tetraktys as
fate, the universe, the heaven, and even God. Pythagoras also called the Deity
a Tetrad or Tetracyts, meaning the “four sacred letters”. These letters origi-
nated from the four sacrad letters JHVH, in which the ancient Jews called
God our Father, the name “Jehovah”. The tetraktys was so revered by the
members of the brotherhood that they shared the following oath and their
most jealously guarded secret, “I swear by him who has transmitted to our
minds the holy tetraktys, the roots and source of ever-flowing nature”. For
Plato (Plato, meaning broad, is a nickname, his real name was Aristocles, he
died at a wedding feast) number ten was the archetypal pattern of the universe.
According to Eric Temple Bell (1883-1960, UK-USA), see [6], “Pythagoras
asked a merchant if he could count. On the merchants’s replying that he could,
Pythagoras told him to go ahead. One, two, three, four ..., he began, when Py-
thagoras shouted Stop! What you name four is really what you would call ten.
The fourth number is not four, but decade, our tetractys and inviolable oath by
which we swear”. Inadvertently, the tetractys occurs in the following: the ar-
rangement of bowling pins in ten-pin bowling, the baryon decuplet, an arch-
bishop’s coat of arms, the “Christmas Tree” formation in association football, a
Chinese checkers board, and the list continues. The number
5
15t=
gives the
number and arrangement of balls in Billiards. The 36th triangular number,
i.e.
,
is 666 (The Beast of Revelation-Christians often seems to have difficulties with
numbers). The 666th triangular number,
i.e.
,
666
t
is 222111. On triangular
numbers an interesting article is due to Fearnehough [7].
Figure 3. Alternative form of tetraktys.
R. P. Agarwal
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Journal of Applied Mathematics and Physics
No triangular number has as its last digit (unit digit) 2, 4, 7 or 9. For this, let
()
mod10nk≡
, then
( ) ( )( )
1 1 mod10nk+≡ +
; here
09k≤≤
. Thus, it fol-
lows that
()()( )
1 2 1 2 mod10
n
t nn kk
=+≡+
. This relation gives only
choices for
k
as
0,1,3,5,6
and 8.
We shall show that for an integer
1k>
,
()
mod , 1
n
t kn
≥
repeats every
k
steps if
k
is odd, and every 2
k
steps if
k
is even,
i.e.
, if
is the smallest posi-
tive integer such that for all integers
n
( )( ) ( ) ( )
11
mod ,
22
n n nn k
+ ++ +
≡
(8)
then
k=
if
k
is odd, and
2k
=
if
k
is even. For this, note that
( )( ) ( ) ( )
11 1
,
222
n n nn n
+ ++ + +
−=+
and hence if (8) holds, then
( ) ( )
1 0 mod .
2
nk
+
+≡
For
nk=
and
1n=
the above equation respectively gives
( ) ( ) ( ) ( )
11
0 0 mod and 0 mod .
22
kk
++
+≡ +≡
Combining these two relations, we find
( )
0 mod k
≡
and hence
for some positive integer .ck c=
(9)
Now if
k
is odd, then in view of
( )
12k+
is an integer, we have
( ) ( )
1 0 mod .
2
kk
nk k
+
+≡
This implies that
k≥
, because
is the smallest integer for which (8)
holds. But, then from (9) it follows that
k=
.
If
k
is even, then
1k+
is odd, and so
( ) ( )
1 2 0 modkk k+≡
/
. Thus,
k≠
,
but
( ) ( ) ( )
21
2 0 mod
2
kk
nk k
+
+≡
and so 2
k
satisfies (8). This implies that
2k≥
, which again from (9) gives
2k=
.
For example, for
( )
mod3 , 1
n
tn≥
, we have
1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0,
and for
( )
mod 4 , 1
n
tn≥
,
1, 3, 2, 2,3, 1,0, 0, 1, 3, 2, 2,3, 1,0, 0, .
Triangular numbers and binomial coefficients are related by the relation
11
.
21
n
nn
tn
++
= =
−
R. P. Agarwal
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Journal of Applied Mathematics and Physics
Thus, triangular numbers are associated with Pascal’s triangle
1
1 1
2
1 1
1 4 4 1
1 5 5 1
1 6 20 6 1
1 7 35 35 7 1
1 6 56 70 56 8 1
11
33
6
10 10
15 15
21 21
28 28
For the origin of Pascal triangle see Agarwal and Sen [5].
The only triangular numbers which are the product of three consecutive in-
tegers are 6, 120, 210, 990, 185,136, 258, 474, 216, see Guy [8].
A number is called palindromic if it is identical with its reverse,
i.e.
, reading
the same forward as well as backward. There are 28
palindromic
triangular
numbers
less than 1010, namely, 1, 3, 6, 55, 66, 171, 595, 666, 3003, 5995, 8778,
15,051, 66,066, 617,716, 828,828, 1,269,621, 1,680,861, 3,544,453, 5,073,705,
5,676,765, 6,295,926, 351,335,153, 61,477,416, 178,727,871, 1,264,114,621,
1,634,004,361, 5,289,009,825, 6,172,882,716. The largest known palindromic
triangular numbers containing only odd digits and even digits are
32850970
539593131395935t=
and
128127032
8208268228628028t=
. It is
known, see Trigg [9], that an infinity of palindromic triangular numbers exist
in several different bases, for example, three, five, and nine; however, no infi-
nite sequence of such numbers has been found in base ten.
Let
m
be a given natural number, then it is
n
-th triangular number,
i.e.
,
n
mt
=
if and only if
( )
1 18 2nm=−+ +
. This means if and only if
81m+
is a perfect square.
If
n
is a triangular number, then
9 1, 25 3nn++
and
49 6n+
are also tri-
angular numbers. This result of 1775 is due to Euler. Indeed, if
m
nt
=
, then
31
91 m
nt
+
+=
,
52
25 3
m
nt
+
+=
and
73
49 6
m
nt
+
+=
. An extension of Euler’s
result is the identity
( )
( )
2
21
2 1 , 1, 2,
mk k mk
k ttt k
++
+ += =
i.e.
,
( ) ( ) ( ) ( ) ( )
221 21 1
11
21 .
22 2
k mk k mk
mm kk
k++ +++
++
+⋅ + =
From the identity
( ) ( ) ( ) ( )
22
11
4 11
22
xx yy xy xy
++
+ += + + + −
it follows that if
n
is the sum of two triangular numbers, then
41n+
is a sum of
two squares.
Differentiating the expansion
( )
1
0
1
n
n
xx
−∞
=
−=
∑
twice, we get
R. P. Agarwal
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Journal of Applied Mathematics and Physics
( ) ( ) ( )( )
21
321
21 1,
1
nn
nn
nn x n nx
x
∞∞
−−
= =
= −=+
−
∑∑
(10)
and hence
( )
( ) ( ) ( )( )
0
3100
11
0.
22
1
n nn
n
nnn
nn nn
xx x x tx
x
∞∞∞
= = =
++
=+==
−
∑∑∑
Hence,
( )
3
1xx
−
−
is the
generating
function
of all triangular numbers. In
1995, Sloane and Plouffe [10] have shown that
21
0
1
12 e .
2!
n
xn
n
x
xx t
n
∞
+
=
++ =
∑
To find the sum of the first
n
triangular numbers, we need an expression for
2
1
n
k
k
=
∑
(a general reference for the summation of series is Davis [11]). For
this, we begin with Pascal’s identity
( )
3
32
1 3 3 1, 1kk kk k−− = −+ ≥
and hence
( ) ( ) ( )
()
()
3
3 3 33 3 3 3 2
1 11
1 0 2 1 3 2 1 3 3 1,
n nn
k kk
nn k k
= = =
− + − + − ++ −− = − +
∑ ∑∑
which in view of (2) gives
( )( )
232 32
1
11111111 1 2 1.
32233266
n
kknnnnnnnnn n
=
=++−=++= + +
∑
(11)
Archimedes as proposition 10 in his text
On
Spirals
stated the formula
( ) ( )
( )
2 22 2
1 1 2 31 2nn n n+ ++++ = +++
(12)
from which (11) is immediate. It is believed that he obtained (12) by letting
k
the
successive values
1,2, , 1n−
in the relation
( ) ( ) ( )
22
22
2,n k nk k knk nk= +− = + −+−
and adding the resulting
1n−
equations, together with the identity
22
22nn=
, to arrive at
( )
( )
( ) ( ) ( )
2 22 2
1 21 2 2 1 1 2 2 11 .nn n n n n+ = + ++ + −+ −++−
(13)
Next, letting
1, 2, ,kn=
in the formula
( )
2
21 2 1kk k=+ ++ + −
and adding
n
equations to get
( ) ( ) ( ) ( )
22 2
1 2 1 2 21 1 2 2 11.n nn n n+++ =++++ −+ −++−
(14)
From (13) and (14), the Formula (12) follows.
Another proof of (11) is given by Fibonacci. He begins with the identity
( )( ) ( ) ( )
2
12 1 1 2 1 6 .kk k k k k k+ += − −+
and takes
1, 2, 3, ,kn=
to get the set of equations
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
( ) ( ) ( ) ( )( ) ( )
( )( ) ( ) ( )
2
2
2
2
2
123 61
235 123 62
347 235 63
1 2 1 2 12 3 6 1
12 1 1 2 1 6.
n nn n n n n
nn n n n n n
⋅⋅= ⋅
⋅⋅=⋅⋅+⋅
⋅⋅= ⋅⋅+⋅
− −= − − −+ −
+ += − −+
On adding these
n
equations and cancelling the common terms, (11) follows.
Now from (2) and (11), we have
( )( )
2
11 1
1 11
1 2.
2 26
nn n
k
kk k
t k k nn n
= = =
= + = ++
∑∑ ∑
(15)
Relation (15) is due to Aryabhata.
For an alternative proof of (15), we note that
( )
( )
( ) ( ) ( )
00
22 2 2
1
1
123 23 3
12 ,
nn
n k nk
kk
n
k
n t t tt
n n nn
nk
= =
=
+− = −
=+++++++++++++
=+++ =
∑∑
∑
and hence in view of (11), we have
( )
( ) ( ) ( )( )
( )( )
2
11
1
11
1 12 1
26
1
1 2.
6
nn
kn
kk
tnt k
nn
n nn n
nn n
= =
=+−
+
=+ −++
= ++
∑∑
From (15) it follows that
( ) ( ) ( )( )
2
1
1
3 1 1 1,
6
n
k
km t nm nm n m
= +
= − − + + +−
∑
which in particular for
4, 7mn= =
gives
2
567
64 8ttt++= =
,
i.e.
, three suc-
cessive triangular numbers whose sum is a perfect square. Similarly, we have
2
5678
10tttt+++=
.
From (15), we also have
( )( )
113 2
nkn
kt nt
== +
∑
, which means
n
t
divides
1
nk
k
t
=
∑
if
32nm= −
,
1, 2,m=
.
The reciprocal of the
( )
1n+
-th triangular number is related to the integral
()( )
11
00 1
21
dd .
12
n
n
x y xy nn t
+
−= =
++
∫∫
The sum of reciprocals of the first
n
triangular numbers is
( )
11 1
1 2 11 1
2 21 ,
1 11
nn n
kk k
k
t kk k k n
= = =
= = −=−
+ ++
∑∑ ∑
(16)
and hence
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
1
11
2 lim 1 2.
1
n
kk
tn
∞
→∞
=
= −=
+
∑
(17)
Jacob Bernoulli (1654-1705, Switzerland) in 1689 summed numerous conver-
gent series, the above is one of the examples. In the literature this procedure is
now called
telescoping
, also see Lesko [12].
Pythagoras theorem states that if
a
and
b
are the lengths of the two legs of a
right triangle and
c
is the length of the hypothenuse, then the sum of the
areas of the two squares on the legs equals the area of the square on the hy-
potenuse,
i.e.
,
222
.abc+=
(18)
A set of three positive integers
a
,
b
and
c
which satisfy (18) is called
Pythago-
rean
triple
and written as ordered triple
( )
,,abc
. A Pythagorean triangle
( )
,,abc
is said to be
primitive
if
,,abc
have no common divisor other than 1.
For the origin, patterns, extensions, astonishing directions, and open problems,
of Pythagoras theorem and his triples, see Agarwal [13] [14], and an interesting
article of Beauregard and Suryanarayan [15]. There are Pythagorean triples (not
necessarily primitive) each side of which is a triangular number, for example,
( )
( )
132 143 164
, , 8778,10296,13530ttt =
. It is not known whether infinitively many
such triples exist.
A number is called
perfect
if and only if it is equal to the sum of its positive
divisors, excluding itself. For example,
3
6t=
is perfect, because
( )
61 2 3 6++=
. The numbers
( )
7 31 127
28,496,8128 , ,tt t
are also perfect that
Pythagoreans discovered. For mystical reasons, such numbers have been
given considerable attention in the past. Especially, Pythagoreans praised the
number six eulogistically, concluding that the universe is harmonized by it
and from it comes wholeness, permanence, as well as perfect health. In fact,
Plato asserted that the creation is perfect because the number 6 is perfect.
They also realized that like squares, six equilateral triangles (see Figure 4)
meeting at a point (add up to 360˚) leave no space in tilling a floor.
Till very recently only 51 even perfect numbers of the form
( )
1
2 21
pp−
−
have been discovered. It is not known whether there are any odd perfect num-
bers, and if there exist infinitely many perfect numbers. The following result due
Figure 4. Tilling a floor.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
to Euclid of Alexandria (around 325-265 BC, Greece) and Euler states that an
even number is perfect if and only if it has the form
( )
1
2 21
pp−
−
, where
21
p
−
is a prime number (known as Pére Marin Mersenne’s, 1588-1648, France, prime
number). In 1575 it was observed that
( ) ( )
1
21
2 212212 p
p p pp
t
−
−
−= − =
,
i.e.
,
every known perfect number is also a triangular number.
Fermat numbers are defined as
2
2 1, 0
n
n
Fn=+≥
. First few Fermat’s num-
bers are 3, 5, 17, 257, 65537. We shall show that for
0n>
, Fermat number
n
F
is never a triangular number,
i.e.
, there is no integer
m
which satisfies
( )
2
2 1 12
n
mm+= +
. This means the discriminant of the equation
( )
22
22 1 0
n
mm+− +=
is not an integer. Suppose to contrary that there ex-
ists an integer
p
such that
( )
2
1 82 1
np+ +=
, but then
( )( )
23 2
2 9 33
n
p pp
+= −= + −
, which implies that there exist integers
r
and
s
such that
32
r
p+=
and
32
s
p−=
. Hence, we have
226
rs
−=
for which
the only solution is
3, 1rs= =
. This means,
23 3
2 22
n
+= ×
, or
2
22
n
=
,
which is true only for
0n=
.
We shall find all
square
triangular
numbers
,
i.e.
, all positive integers
n
and
the corresponding
m
so that
( )
2
12nn m+=
. This equation can be written
as, so called Pell’s Equation (for its origin, see Agarwal [5])
22
21ba−=
,
where
21
bn= +
and
2am=
. We note that if
( )
11
, , 1
kk
ab k
−− ≥
is an in-
teger solution of
22
21ba−=±
, then
()
,
kk
ab
defined by the recurrence re-
lations
11 11
, 2 , 1
kkkkkk
aabbabk
−− −−
=+ =+≥
(19)
satisfy
( ) ( )
( )
22
22 2 2
11 11 1 1
2 2 2 2 ,
kk kk kk k k
ba ab ab b a
−− −− − −
−= + − + =− −
and hence
22
21ba−=
. From this observation we conclude that if
( )
11
, , 1
kk
ab k
−−
≥
is an integer solution of
22
21ba−=
, then so is
( ) ( )
11 1 1 1 1
, 32,43
kk k k k k
ab a b a b
++ − − − −
=++
. Since
( )
( )
00
, 0,1ab =
is a solution of
22
21ba−=
(its fundamental solution is
( ) ( )
, 2,3ab =
), it follows that the iter-
ative scheme
11
1 10 0
32
4 3 , 0, 1
kk k
kk k
xx y
yx yx y
−−
−−
= +
=+==
(20)
gives all solutions of
22
21ba−=
. System (20) can be written as
1 10 1
1 10 1
6 , 0, 2
6 , 1, 3.
k kk
k kk
x xx x x
y yy y y
+−
+−
=−==
=−==
(21)
Now in (21) using the substitution
2 , 2 1,
k kk k
x my n
= = +
we get
1 10 1
1 1 01
6 , 0, 1
6 2, 0, 1
k kk
k kk
m mm m m
n nn n n
+−
+−
=−==
=−+ = =
(22)
Clearly, (22) generates all (infinite) solutions
()
,
kk
mn
of the equation
( )
2
12nn m
+=
. First few of these solutions are
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
1,1 , 6,8 , 35, 49 , 204,288 , 1189,1681 ,
6930,9800 , 40391,57121 , 235416,332928 .
For
1k≥
, explicit solution of the system (22) can be computed (for details
see Agarwal [16] [17]) rather easily, and appears as
( ) ( )
( ) ( )
1322 3 22
42
1322 322 2
4
kk
k
kk
k
m
n
= + −−
= + +− −
(23)
This result is originally due to Euler which he obtained in 1730. While com-
pare to the explicit solution (23) the computation of
( )
,
kk
mn
from the recur-
rence relations (22) is very simple, the following interesting relation follows
from (23) by direct substitution
22
1 21
.
kk k
mm m
−−
−=
(24)
Hence the difference between two consecutive square triangular numbers is
the square root of another square triangular number.
Now we note that the system (19) can be written as
1 10 1
1 10 1
2 , 0, 1
2 , 1, 1
n nn
n nn
a aa a a
b bb b b
+−
+−
=+==
=+==
and its (integer) solution is
( ) ( )
( ) ( )
1
12 12,
22
112 12.
2
nn
n
nn
n
a
b
= + −−
= + +−
(25)
From this, and simple calculations the following relations follow
( )
2
2 22
2 2 21 21 21
, 1 2 2 , 2 1.
k kk k k k k k k
m ab n b m n b a
++ +
= = −= = = −
It is apparent that if
( )
,
kk
mn
is a solution of
()
2
12nn m+=
, then
()()
( )
21,21 , 1
kk
pmpnp+ +≥
is a solution of
( )
2
2 12nn p m++ =
. Now, if
n
is even, we have
( ) ( )
( ) ()
22
22
12
2
2
24
13
22
21
2
21 ,
22
nn n p
nn
tt t n n
nn p
np
np
++
++
=++
++
+
+−
(26)
and, when
n
is odd,
( ) ( )
( ) ( )
22
22
12
22
13
24
22
21
21 2
22
nn n p
nn
tt t n n
nn p
np np
++
++
=++
++
+−
+
(27)
and hence the right side is a perfect square for
( )
21
k
n pn= +
. Therefore, the
product of
( )
21p+
consecutive triangular numbers is a perfect square for each
R. P. Agarwal
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Journal of Applied Mathematics and Physics
1p≥
and
1k≥
. In particular, for
2pk= =
,
2
5 40nn= =
, from (26) we
have
( ) ( ) ( ) ( ) ( ) ( )
22222 2
40 41 42 43 44
41 21 43 22 30 24435180ttttt = =
and for
3
3, 7 343pk n n= = = =
, from (27), we find
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
343 344 345 346 347 348 349
2222222
2
172 345 173 347 174 349 245
52998536784979800 .
ttttttt
=
=
Similarly, if
n
is even, we have
( ) ( )
( ) ( )
22
22
1 21
22
24
2 13
22
2
22 21 ,
22
nn n p
nn
tt t n n
nn p
np np
+ +−
++
=++
+
+−
+−
(28)
and hence the right side is a perfect square for
2
k
n pn
=
(which is always
even). Therefore, two times the product of
2p
consecutive triangular numbers
is a perfect square for each
1p≥
and
1
k≥
. In particular, for
2pk= =
,
2
4 32nn= =
, from (28) we have
( ) ( ) ( ) ( ) ( )
222 2 2
32 33 34 35
2 33 17 35 24 471240tttt = =
and for
3
3, 6 294pk n n= = = =
, we find
( ) ( ) ( ) ( ) ( ) ( )
( )
222222
294 295 296 297 298 299
2
2 295 148 297 149 299 210
121315678684200 .
tttttt =
=
From the equality
( )
()
( )
( )
( ) ()
2
41411 1
4 21
22
nn nn nn n
+ ++ +
= +
it follows that if the triangular number
n
t
is square, then
( )
41nn
t
+
is also square.
Since
1
t
is square, it follows that there are infinite number of square triangular
numbers. This clever observation was reported in 1662, see Pietenpol
et al.
[18].
From this, the first four square triangular numbers, we get are
1 8 288
,,ttt
and
332928
t
.
There are infinitely many triangular numbers that are simultaneously expres-
sible as the sum of two cubes and the difference of two cubes. For this, Bur-
ton [19] begins with the identity
( ) ( ) ( ) ( ) ( )
2 33 33
64343
27 1 9 3 9 1 9 3 9 1k kk k kk k−= − + − = + − +
and observed that if
k
is odd then this equality can be written as
( ) ( ) ( ) ( ) ( )
2 33 3 3
21 12 2 2 2 ,n abcd+ −= + = −
which is the same as
33 3 3
.
n
tabcd=+=−
For
1,3k=
and 5 this gives
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
()()()
() ( )( ) ( )
33 33
13
33 3
3
9841
33 33
210937
3465
360 121 369 122
2805 562 2820 563
t
t
t
=+=−
=+=−
= += −
In 1844, Eugéne Charles Catalan (1814-1894) conjectured that 8 and 9 are the
only numbers which differ by 1 and are both exact powers
3
82=
,
2
93=
.
This conjecture was proved by Preda Mihăilescu (Born 1955, Romania) after
one hundred and fifty-eight years, and published two years later in [20]. Thus
the only solution in natural numbers of the Diophantine equation
1
ab
xy−=
for
,1ab>
,
,0xy>
is
3x=
,
2a=
,
2y=
,
3b=
. Now
since
()
3
12nn m
+=
can be written as
( ) ( )
23
21 2 1nm+− =
, the only solu-
tion of this equation is
2 13
n+=
,
22m=
,
i.e.
,
( )
1,1
is the only cubic tri-
angular number.
In 2001, Bennett [21] proved that if
a
,
b
and
n
are positive integers with
3n≥
, then the equation
1
nn
ax by−=
, possesses at most one solution in
positive integers
x
and
y
. This result is directly applicable to show that for the
equation
( )
1 2 , 3
p
nn m p+= ≥
the only solution is
()
1,1
. For this, first we
note that integers
,2 1tt+
and
1,2 1
tt++
are coprime,
i.e.
, they do not
have any common factor except 1. We also recall that if the product of co-
prime numbers is a
p
-th power, then both are also of
p
-the power. Now let
n
be even,
i.e.
,
2
nt
=
, then the equation
( )
12
p
nn m+=
is the same as
( )
21
p
tt m+=
. Thus, it follows that
p
tx=
and
21
p
ty+=
, and hence
21
pp
yx−=
, which has only one solution, namely,
0, 1xy= =
which gives
0t=
, and hence
0n=
and so
( )
0,0
is the solution of
( )
12
p
nn m+=
,
but we are not interested in this solution. Now we assume that
n
is odd,
i.e.
,
21nt= +
, then the equation
( )
12
p
nn m+=
is the same as
( )( )
12 1
p
ttm+ +=
. Thus, we must have
1
p
tx+=
and
21
p
ty+=
, which
gives
21
pp
yx
−=−
. The only solution of this equation is
1xy= =
, and
hence again
0t=
and so
( )
0,0
is the undesirable solution of
( )
12
p
nn m+=
.
Startling
generating
function
of all square triangular numbers is recorded by
Plouffe [22] as
( )
( )
()
22 23
2
1 6 35 .
1 34 1
xx xx x
xx x
+=++ +
− −+
(29)
3. Square Numbers Sn
In this arrangement rows as well as columns contain
1,2,3,4, , n
dots, (see
Figure 5).
From Figure 5 it is clear that a square made up of
( )
1n+
dots on a side can
be divided into a smaller square of side
n
and an
L
, shaped border (a gnomon),
which has
( )
1 21n nn++= +
dots (called
( )
1n+
th
gnomonic
number
and
denoted
as
1n
g
+
), and hence
( ) ( )
22
1
1 2 1,
nn
S Sn n n
+
−=+ −= +
(30)
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
Figure 5. Square numbers.
i.e.
, the differences between successive nested squares produce the sequence of
odd numbers. From (30) it follows that
( ) () ( )
()
( )
( )
2
2 2 22 2 2 2
10 21 32 1 135 21nn n
− + − + − ++ −− =++++ −
and hence
( ) ( )
2
1
2 1 1357 2 1 .
n
n
k
k n nS
=
−=+++++ −= =
∑
(31)
An alternative proof of (31) is as follows
() ( )
135 2 3 2 1
n
S nn=++++−+−
()()()
21 23 25 31.
n
Sn n n
=−+−+−+++
An addition of these two arrangements immediately gives
2
2 2 2 2 2.
n
S nn nn= + ++ =
Figure 6 provides proof of (31) without words. Here odd integers, one block,
three blocks, five blocks, and so on, arranged in a special way. We begin with a
single block in the lower left corner; three shaded blocks surrounded it to form a
22×
square; five unshaded blocks surround these to form a
33×
square; with
the next seven shaded blocks we have a
44×
square; and so on. The diagram
makes clear that the sum of consecutive odd integers will always yield a (geome-
tric) square.
Comparing Figure 1 and Figure 5 or Figure 2 and Figure 6, it is clear that
n
-th square number is equal to the
n
-th triangular number increased by its pre-
decessor,
i.e.
,
2
1
.
n nn
Stt n
−
=+=
(32)
Indeed, we have
()
123 1
n
t nn=+++ + − +
( ) ( )
1
1 2 2 1.
n
t nn
−
=+++−+−
An addition of these two arrangements in view of (31) gives
( )
2
1
135 2 1 .
nn n
tt n n S
−
+ =+++ + − = =
Of course, directly from (1), (2), and (32), we also have
( ) ( )
( )
2
2
11
11 ,
22
nn nn n
nn n n
tt n tt S
−−
+−
+= + ==− =
or simply from (1) and (2),
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Journal of Applied Mathematics and Physics
Figure 6. Proof of (31) without words.
( )
2
11
2 1.
nn n n
t t t nnn nn S
−−
+ = += −+= =
From (32), we find the identities
( )
( ) ( )
( )
2
2 1 4 3 2 21
1
2
22
24 2
n
k nn
kt tt tt t t
n
−
=
= ++ + ++ +
= + ++
∑
and
( ) ( ) ( )
( )
21
1 3 2 5 4 21 2
1
2
222
1 3 5 2 1.
n
k nn
k
t t tt tt t t
n
+
+
=
=++++++ +
=++++ +
∑
It also follows that
( ) ( )
2
2
22 1 1
22 .
22
nn n
n n nn
t t nS
++
−= − ==
(33)
We also have equalities
( )
( )
2
94 31 32 1
32 1 ,
nn n
tt n S
++ +
−= +=
(34)
( ) ( )
11
1234
1 1,
nn
nn
SSSS S t
++
−+−++− =−
(35)
and
( ) ( )
22
2
01
4,
nn
nn
kk
t k tk
= =
+= +
∑∑
(36)
which is the same as
( ) ( )
22
22
22
2 22
0 1 01
2 2 2 or
n n nn
n nk n nk
k k kk
n nk n nk S S
++ + +
= = = =
++ = + + =
∑ ∑ ∑∑
and, in particular, for
4n=
reduces to
222222222
36 37 38 39 40 41 42 43 44 .++++=+++
The following equality is of exceptional merit
( ) ( )
111 1
,
nn nn nn
SS S S
+++ +
+= −
(37)
which, in particular, for
5n=
gives
22 2 2
5 6 31 30+= −
.
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Journal of Applied Mathematics and Physics
Relation (30) reveals that every odd integer
()
21
n+
is the difference of two
consecutive square numbers
1n
S
+
and
n
S
. Relation (32) shows that every
square integer
2
n
is a sum of two consecutive triangular numbers
n
t
and
1n
t
−
, whereas (33) displays it is the difference of 2
n
-th and two times
n
-th
triangular numbers.
From the equalities
( ) ( )
22
22 2
8,
n
t nn nn=+++
( ) ( )
22
22 2
81 1 2,
n
t n nn+= − + +
( ) ( )
22
22 2
82 1 1
n
t nn nn+= +− + ++
it follows that there are infinite triples of consecutive numbers which can be
written as the sum of two squares.
No square number has as its last digit (unit digit) 2, 3, 7 or 8.
From (10) it follows that
( )
2 12
30 10
2 d1
d1
1
nn n
n nn
xnx x nx nx x xx
x
∞ ∞∞
−
= = =
=+=+
−
−∑ ∑∑
and hence
( ) ( )
( )
( )
2
32 3
0
1
2 .
11 1
n
n
xx
xx nx
xx x
∞
=
+
−= =
−− −
∑
(38)
Therefore,
( )( )
3
11xx x
−
+−
is the
generating
function
of all square numbers.
From (38) it also follows that the generating function for all gnomonic numbers
is
( )
( ) ( )
211
1 21 .
1
nn
n
nn
xx n x gx
x
∞∞
= =
+= −=
−
∑∑
The sum of the first
n
square numbers is given in (11). For the exact sum of
the reciprocals of the first
n
square numbers no formula exists; however, the
problem of summing the reciprocals of all square numbers has a long history
and in the literature it is known as
the
Basel
problem
. Euler in 1748 consi-
dered
sin , 0xxx≠
which has roots at
, 1nn± π ≥
. Then, he wrote this
function in terms of infinite product
246
222
22 22 22
sin 13! 5! 7!
11 1 ,
123
x xxx
x
xxx
=−+−+
=−− −
πππ
which on equating the coefficients of
2
x
, gives
22 22 22
11 1 1 ,
6123
++
ππ
= +
π
and hence
2
222
111 1.6449340668.
6
123
π
+ + += ≈
(39)
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
The above demonstration of Euler is based on manipulations that were not
justified at the time, and it was not until 1741 that he was able to produce a truly
rigorous proof. Now in the literature for (39) several different proofs are known,
e.g., for a recent elementary, but clever demonstration, see Murty [23].
The following result provides a characterization of all Pythagorean triples,
i.e.
,
solutions of (18): Let
u
and
v
be any two positive integers, with
uv>
, then
the three numbers
22 22
, 2 , a u v b uv c u v=−= =+
(40)
form a Pythagorean triple. If in addition
u
and
v
are of opposite parity-one even
and the other odd-and they are coprime,
i.e.
, that they do not have any common
factor other than 1, then
( )
,,
abc
is a primitive Pythagorean triple. The con-
verse,
i.e.
, any Pythagorean triple is necessarily of the form (40) also holds. For
the proof and history of this result see, Agarwal [14]. From (18), (32), and (40)
the following relations hold
() ( ) ( )
22 22
111
2
, ,
.
a b c aa bb cc
uv
uv uv
SS S tt tt tt
S SS
−− −
−+
+= + ++ =+
+=
(41)
The relation (30) can be written as
( ) ( )
2
221 1nn n+ += +
. With the help of
this relation we can find Pythagorean triples
( )
,,
abc
. For this, we let
2
21nm+=
, (and hence
m
is odd), then
2
12nm= −
,
2
21 1nm+= +
. Thus, it
follows that
( ) ( )
( )
22
22
22
2
12 12
11
, i.e.,
22
, odd .
mmm
mm
m
SS S m
−+
−+
+=
+=
(42)
For
3,5,7,9,m=
Equation (42) gives solutions of (18):
334 5
5 5 12 13
7 7 24 25
9 9 40 41
mab c
Similar to (42) for
m
even we also have the relation
( )
22
2
22
2
( 4) 4 ( 4) 4
1 1 , i.e.,
44
even .
mmm
mm
m
SS S m
−+
+ −= +
+=
(43)
For
4,6,8,10,m=
Equation (43) gives solutions of (18):
4435
6 6 8 10
8 8 15 17
10 10 24 26
ma b c
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DOI:
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Journal of Applied Mathematics and Physics
In (40), letting
( )
2
2um= +
and
( )
2
1vm= +
, from (18) and (32), we get the
relations
( ) ( )( )
( )
( ) ( )
( )
2
2
2 22
2 3 2 1 2 1 2 , i.e.,m mm m m+ + + + = +++
( )( ) ( ) ( )
22
23 21 2 12
,
mmm mm
SS S
+++ + ++
+=
which is the same as
( ) ( )
2
2
23 22 1 2 1
16 2 .
m m m m mm
t t t t tt
++ + + +
+ + =++
In 1875, Francois Edouard Anatole Lucas (1842-1891, French) challenged the
mathematical community to prove that the only solution of the equation
()( )
22
1
112 1
6
n
k
k nn n m
=
= + +=
∑
with
1n>
is when
24n=
and
70m=
. In the literature this has been termed
as the cannonball problem, in fact, it can be visualized as the problem of taking a
square arrangement of cannonballs on the ground and building a square pyra-
mid out of them. It was only in 1918, George Neville Watson (1886-1965, Britain)
used elliptic functions to provide correct (filling gaps in earlier attempts) proof
of Lucas assertion. Simplified proofs of this result are available, e.g., in Ma [24]
and Anglin [25].
4. Rectangular (Oblong, Pronic, Heteromecic) Numbers Rn
In this arrangement rows contain
()
1n+
whereas columns contain
n
dots, see
Figure 7.
From Figure 7 it is clear that the ratio
()
1nn
+
of the sides of rectangles
depends on
n
. Further, we have
( ) ( )
2468 2
21234 2 1
n
n
Rn
n t nn
=++++ +
= ++++ + = = +
(44)
i.e.
, we add successive even numbers, or two times triangular numbers. It also
follows that rectangular number
1n
R
+
is made from
n
R
by adding an
L
--shaped
border (a gnomon), with
( )
21n+
dots,
i.e.
,
( )
12 1,
nn
RR n
+−= +
(45)
i.e.
, the differences between successive nested rectangular numbers produce the
sequence of even numbers.
Figure 7. Rectangular numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
Thus the odd numbers generate a limited number of forms, namely, squares,
while the even ones generate a multiplicity of rectangles which are not similar.
From this the Pythagoreans deduced the following correspondence:
odd limited and even unlimited.↔↔
We also have the relations
( )
( ) ( ) ( )
( )
12
11
1 1 22 1
23
22 2
2
n n n nn n
n n n nn nn
nn n n n n
RS t tt t
RS t tt tt n
−
−−
+− +
+= ++ = + = =
−= −+ =− =
(46)
( )
( ) ( )
( ) ( )
2
1 11
22
2
2
21
2 6 21
2 1 4 1 18 1 4 1 4
nnn nn n
n nn n
RSS tt t n
S n nn t t t
+ −+
+
++ = + + = +
= + = + += += + −
(47)
( )
2
21 9 4 31
3
n nn
S tt
+ ++
= −
11
1.
nn n
Rt t
+−
=+−
From (31) and (46) it follows that
( )
( ) ( )
( )
21 11 3 2 21 22
1
2
1
13 2 1 .
nkk nn
kt t tt t t
nn
−+
−−
=
− =+ − ++ −
=++ + − =
∑
Relation (44) shows that the product of two consecutive positive integers
n
and
( )
1n+
is the same as two times
n
-th triangular numbers. According to
historians with this relation Pythagoreans’ enthusiasm was endless. Relation
(45) reveals that every even integer 2
n
is the difference of two consecutive
rectangular numbers
n
R
and
1n
R
−
. Relation (46) displays that every positive
integer
n
is the difference of
n
-th and
()
1n−
-th triangular numbers. Rela-
tion (47) is due to Plutarch), it says an integer
n
is a triangular number if and
only if
81n+
is a perfect odd square.
Let
m
be a given natural number, then it is
n
-th rectangular number,
i.e.
,
n
mR=
if and only if
( )
1 14 2
nm=−+ +
.
From (10) it is clear that
( )
3
21xx
−
−
is the
generating
function
of all rec-
tangular numbers.
From (15)-(17) and (44) it is clear that
( )( )
1
11
1
1 2,
3
1 11
1 , 1.
1
n
k
k
n
kk
kk
R nn n
Rn R
=
∞
= =
= ++
=−=
+
∑
∑∑
(48)
There is no rectangular number which is also a perfect square, in fact, the
equation
( )
2
1nn m+=
has no solutions (the product of two consecutive in-
tegers cannot be a prefect square).
To find all
rectangular
numbers
which
are
also
triangular
numbers
, we need
to find integer solutions of the equation
( ) ( )
1 12nn mm+= +
. This equa-
tion can be written as Pell’s equation
22
21ba−=−
(its fundamental solu-
tion is
( ) ( )
, 1,1ab =
) where
21bm= +
and
21an= +
. For this, corres-
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
ponding to (22) the system is
1 1 12
1 1 12
34 16, 3, 119
34 16, 2, 84.
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −+ = =
(49)
This system genetrates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
1 12nn mm+= +
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
3,2 , 119,84 , 4059, 2870 , 137903,97512 , 4684659,3312554 .
For
1k≥
, explicit solution of the system (49) can be written as
( ) ( )
( ) ( )
41 41
41 41
121 21 2
4
221 21 22
8
kk
k
kk
k
m
n
−−
−−
= + −− −
= + +− −
Fibonacci
numbers
denoted as
n
F
are defined by the recurrence relation
1 20 1
, 0, 1
nn n
FF F F F
−−
=+==
or the closed from expression
115 15 .
22
5
nn
n
F
+−
= −
First few of these numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144. For the
origin of Fibonacci numbers, see Agarwal and Sen [5].
Lucas
numbers
denoted
by
n
L
are defined by the same recurrence relation as Fibonacci numbers except
first two numbers as
01
2, 1LL= =
or the closed from expression
15 15
.
22
nn
n
L
+−
= +
First few of these numbers are 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521,
843, 1364, 2207, 3571, 5778, 9349. Clearly, Fibonacci numbers 1, 3, 21, 55 are al-
so triangular numbers
1 2 6 10
,,,tttt
. In 1989, Luo [26] had used (47) to show that
these are the only Fibonacci numbers which are also triangular. This conjecture
was made by Verner Emil Hoggatt Jr. (1921-1980, USA) in 1971. Similarly, only
Lucas numbers which are also triangular are 1, 3, 5778,
i.e.
,
1 2 107
,,ttt
. From the
above explicit expressions the following relations can be obtained easily
11nn n
LF F
−+
= +
and
( )
11
5
nnn
FLL
−+
= +
.
5. Pentagonal Numbers Pn
The pentagonal numbers are defined by the sequence
1,5,12,22,35,51,
,
i.e.
,
beginning with 5 each number is formed from the previous one in the sequence
by adding the next number in the related sequence
( )
4,7,10, , 3 2n−
. Thus,
514= +
,
12 1 4 7 5 7=++=+
,
22 1 4 7 10 12 10=+++ = +
, and so on (see Fig-
ure 8 and Figure 9).
Thus,
n
-th pentagonal number is defined as
( ) ( )
132147 32.
nn
PP n n
−
= + − =+++ + −
(50)
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
Figure 8. Pentagonal numbers.
Figure 9. Pentagonal numbers.
Comparing (50) with (3), we have
1, 3ad= =
and hence from (4) it follows
that
()( )( )
31
3 13
11
31 .
2 32 3
nn
nn
n
Pn t
−
−
= −= =
(51)
It is interesting to note that
n
P
is the sum of
n
integers starting from
n
,
i.e.
,
( ) ( ) ( )
1 2 2 1,
n
Pnn n n=++++++ −
(52)
whose sum from (4) is the same as in (51).
Note that from (50), we have
( )
( )
( ) ( )
12 12
12
12
2 32
2 3 32 3 2
2 3.
nnn nn
nn
nn
PPP PP n
PP n n
PP
−− −−
−−
−−
= − − − +−
= − − −− + −
= −+
From (32) and (51), we also have
( )
( )
211
1 21 1
1
2
2.
n n nn
nn nn
nn
P nt tt
tt t t
−−
− −−
−
= += ++
=+=−
(53)
Relation (51) shows that pentagonal number
n
P
is the one-third of the
( )
31n−
-th triangular number, whereas relation (53) reveals that it is the sum
of
n
-th triangular number and two times of
()
1n−
-th triangular number,
and it is the difference of
( )
21n−
-th triangular number and
( )
1n−
-th tri-
angular number.
Let
m
be a given natural number, then it is
n
-th pentagonal number,
i.e.
,
n
mP=
if and only if
( )
1 1 24 6nm=++
.
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DOI:
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Journal of Applied Mathematics and Physics
As in (38), we have
( )
( ) ( )
( )
( )
3 23
0
1 21
31
22
1 11
n
n
n
x x xx
x
Px x xx
∞
=
++
= −=
− −−
∑
and hence
( )()
3
2 11
xx x
−
+−
is the
generating
function
of all pentagonal num-
bers.
From (2), (11) and (51) it is easy to find the sum of the first
n
pentagonal
numbers
( )
2
1
11.
2
n
k
k
P nn
=
= +
∑
(54)
To find the sum of the reciprocals of all pentagonal numbers, we begin with
the series
( ) ( )
3
1
2
31
k
k
fx x
kk
∞
=
=−
∑
and note that
( ) ()
11
21
1,
31
kk
k
fkk P
∞∞
= =
= =
−
∑∑
()
31
1
1
6,
31
k
k
fx x
k
∞−
=
′=−
∑
()
32 3
1
6
6 .
1
k
k
x
fx x x
∞−
=
′′ = = −
∑
Now since
( ) ( )
0 00ff
′
= =
, we have
( ) ( )
3
0
6d
1
x
t
fx xt t
t
= − −
∫
and hence
( ) ( )
( )
( )
1
3
0
11
22
00
2
6
1 1 d
1
21 1
3 d d,
112 32
t
f tt
t
ttt
tt t
= − −
+
= −
++ ++
∫
∫∫
which immediately gives
1
1 3ln 3 1.4820375018.
3
kk
P
∞
=
π
= −≈
∑
(55)
To find all
square
pentagonal
numbers
, we need to find integer solutions of
the equation
( )
2
3 12nn m−=
. This equation can be written as Pell’s equa-
tion
22
61ba−=
(its fundamental solution is
( ) ( )
, 2,5ab =
), where
61bn= −
and
2am=
. For this, corresponding to (22) the system is
1 11 2
1 1 12
98 , 1, 99
98 16, 1, 81
k kk
k kk
m mm m m
n nn n n
+−
+−
=−==
= −− = =
(56)
This system genetrates all (infinite) solutions
( )
,
kk
mn
of the equation
R. P. Agarwal
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Journal of Applied Mathematics and Physics
( )
2
3 12nn m−=
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 99,81 , 9701,7921 , 950599,776161 , 93149001,76055841 .
For
1
k≥
, explicit solution of the system (56) can be written as
( ) ( )
( ) ( )
12
21 21
21 21
15 6 12 5 6 12
46
11
5 6 12 5 6 12 .
6
26
kk
kk
kk
kk
m
n
−−
−−
+
= + −−
×
= + +− +
×
To find all
pentagonal
numbers
which
are
also
triangular
numbers
, we need
to find integer solutions of the equation
( ) ( )
3 12 12n n mm−= +
. This eq-
uation can be written as Pell’s equation
22
32ba−=−
(its fundamental so-
lution is
( ) ( )
, 3,5ab =
) where
61bn= −
and
21am= +
. For this, corres-
ponding to (22) the system is
1 1 12
1 1 12
14 6, 1, 20
14 2, 1, 12
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(57)
This system generates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
3 12 12n n mm−= +
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 20,12 , 285,165 , 3976,2296 , 55385,31977 .
For
1k≥
, explicit solution of the system (57) can be written as
()( ) ( )()
( )()()( )
21 21
21 21
13 32 3 3 32 3 6
12
11 32 3 1 32 3 2
12
kk
k
kk
k
m
n
−−
−−
= + + +− − −
= + + +− − +
To find all
pentagonal
numbers
which
are
also
rectangular
numbers
, we need
to find integer solutions of the equation
( ) ( )
3 12 1n n mm−= +
. This equa-
tion can be written as Pell’s equation
22
65ba−=−
(its fundamental solu-
tion is
() ( )
, 1,1ab =
) where
61bn= −
and
21am
= +
. For this, corres-
ponding to (22) the system is
1 1 12
1 1 12
98 48, 3, 341
98 16, 3, 279
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(58)
This system genetrates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
3 12 1n n mm−= +
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
3,3 , 341,279 , 33463,27323 , 3279081,2677359 , 321316523,262353843 .
For
1k≥
, explicit solution of the system (58) can be written as
( )( ) ( )( )
( )( ) ( )( )
21 21
21 21
16 6 5 26 6 6 5 26 12
24
16 15 26 6 15 26 2
12
kk
k
kk
k
m
n
−−
−−
= + + +− − −
= ++ +−− +
6. Hexagonal Numbers Hn
The hexagonal numbers are defined by the sequence
1,6,15,28, 45,
,
i.e.
, be-
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Journal of Applied Mathematics and Physics
ginning with 6 each number is formed from the previous one in the sequence by
adding the next number in the related sequence
( )
5,9,13,17,21, , 4 3n−
.
Thus,
615= +
,
15 1 5 9 6 9=++=+
,
28 1 5 9 13 15 13=+++ = +
, and so on
(see Figure 10).
Thus,
n
-th hexagonal number is defined as
( ) ( )
1
4 3 15913 4 3.
nn
HH n n
−
= +−=+++++−
(59)
Comparing (59) with (3), we have
1, 4
ad= =
and hence from (4) it follows
that
( ) ( )( ) ( )
2 12
4 2 2 1.
22
n
nn
n
H n nn
−
= −= = −
(60)
From (60) it is clear that
21nn
Ht
−
=
,
i.e.
, alternating triangular numbers are
hexagonal numbers.
Let
m
be a given natural number, then it is
n
-th hexagonal number,
i.e.
,
n
mH=
if and only if
( )
1 18 4nm=++
.
As in (38), we have
( )
( ) ( )
( )
( )
32 3
0
1 31
211 1
n
n
n
x x xx
x
Hx xx x
∞
=
++
= −=
−− −
∑
and hence
( ) ( )
3
3 11xx x
−
+−
is the
generating
function
of all hexagonal num-
bers.
From (2), (11), and (60) it is easy to find the sum of the first
n
hexagonal
numbers
( )( )
1
11 4 1.
6
n
k
kH nn n
=
=+−
∑
(61)
To find the sum of the reciprocals of all hexagonal numbers, as for pentagon-
al numbers we begin with the series
( ) ( )
2
1
21
n
k
fx x n n
∞
=
= −
∑
, and get
()
1
2
0
1
11
1 2 d 2 ln 2 1.3862943611.
1
kk
t
ft
Ht
∞
=
−
= = = ≈
−
∑∫
(62)
To find all
square
hexagonal
numbers
, we need to find integer solutions of
the equation
( )
2
21nn m−=
. This equation can be written as Pell’s equation
22
21ba−=
(its fundamental solution is
( ) ( )
, 2,3ab =
), where
41bn= −
and
2
am=
. For this, corresponding to (22) the system is
Figure 10. Hexagonal numbers.
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DOI:
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Journal of Applied Mathematics and Physics
1 11 2
1 1 12
34 , 1, 35
34 8, 1, 25
k kk
k kk
m mm m m
n nn n n
+−
+−
=−==
= −− = =
(63)
This system generates all (infinite) solutions
( )
,
kk
mn
of the equation
()
2
21
nn m−=
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 35,25 , 1189,841 , 40391, 28561 , 1372105,970225 .
For
1k≥
, explicit solution of the system (63) appears as
( ) ( )
( ) ( )
21 21
2121
21 21
2
21
1
3 22 3 22
42
1
322 322 2
8
kk
k kk
kk
kk
mab
na
−−
−−
−−
−
= = + −−
= = + +− +
here,
n
a
and
n
b
are as in (25).
To find all
hexagonal
numbers
which
are
also
rectangular
numbers
, we need
to find integer solutions of the equation
( ) ( )
21 1n n mm
−= +
. This equation
can be written as Pell’s equation
22
21ba−=−
(its fundamental solution is
( ) ( )
, 1,1ab =
) where
41bn= −
and
21
am= +
. For this, corresponding to
(22) the system is
1 1 12
1 1 12
34 16, 2, 84
34 8, 2, 60
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(64)
This system generates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
21 1n n mm−= +
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
2,2 , 84,60 , 2870,2030 , 97512,68952 , 3312554,2342330 .
For
1k≥
, explicit solution of the system (64) can be written as
( ) ( )
( ) ( )
41 41
41 41
221 21 22
8
121 21 2
8
kk
k
kk
k
m
n
−−
−−
= + +− −
= + −− +
To find all
hexagonal
numbers
which
are
also
pentagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
2 1 3 12
nn mm−= −
. This equa-
tion can be written as Pell’s equation
22
32ba−=−
(its fundamental solu-
tion is
( ) ( )
, 1,1
ab
=
) where
61bm= −
and
41an= −
. For this, corres-
ponding to (22) the system is
1 1 12
1 1 12
194 32, 1, 165
194 48, 1, 143
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(65)
This system generates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
2 1 3 12nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
()
1,1 , 165,143 , 31977, 27693 , 6203341,5372251 ,
1203416145,1042188953 .
For
1k≥
, explicit solution of the system (65) can be written as
( )( ) ( )( )
( )( ) ( )( )
42 42
44 44
1312 3 312 3 2
12
19 53 2 3 9 53 2 3 6
24
kk
k
kk
k
m
n
−−
−−
= − + −+ − +
= + + +− − +
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2065
Journal of Applied Mathematics and Physics
7. Generalized Pentagonal Numbers (Centered Hexagonal
Numbers, Hex Numbers) (GP)n
The generalized pentagonal numbers are defined by the sequence
1,7,19,37,61,
,
i.e.
, beginning with 7 each number is formed from the previous one in the se-
quence by adding the next number in the related sequence
()
6,12,18, ,6 1n−
. Thus,
716= +
,
19 1 6 12 7 12=++ =+
,
37 1 6 12 18 19 18=++ + = +
, and so on (see Figure 11). These numbers are also
called centered hexagonal numbers as these represent hexagons with a dot in the
center and all other dots surrounding the center dot in a hexagonal lattice. These
numbers have practical applications in materials logistics management, for ex-
ample, in packing round items into larger round containers, such as Vienna
sausages into round cans, or combining individual wire strands into a cable.
Thus,
n
-th generalized pentagonal number is defined as
( ) ( ) ( ) ( )
( )
1
611612 61
1612 1.
nn
GP GP n n
n
−
= +−=++++−
=+ ++ + −
(66)
Hence, from (2) it follows that
()( ) ( )
1 1 12
1
16 13 1
2
6 4.
n
n nnn
nn
GP n n
tt t t t
− −−
−
=+ =+−
=+=++
(67)
Incidentally,
()
2
7
GP =
occurs in uds baryon octet, whereas
( )
5
61GP =
makes a part of a Chinese checkers board.
Since
( ) ( )
3
3
13 1 1nn n n+ −= −−
, generalized pentagonal numbers are dif-
ferences of two consecutive cubes, so that the
( )
n
GP
are the gnomon of the
cubes.
Clearly,
( ) ( ) ( )
22 11
21 1 2
nn
n
n GP n n R t
−−
− − = −= =
.
Let
m
be a given natural number, then it is
n
-th generalized pentagonal
number,
i.e.
,
( )
n
m GP=
if and only if
( )
3 12 3 6nm=+−
.
From (10) and (67), we have
( ) ( )
( )
( )
2
2
33
0
14
6
111
n
n
n
x xx
xx
GP x xxx
∞
=
++
=+=
−−−
∑
and hence
( )
( )
3
2
14 1x xx x
−
++ −
is the
generating
function
of all generalized
pentagonal numbers.
Figure 11. Generalized pentagonal numbers (centered hexagonal numbers).
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
From (15) and (67) it is easy to find the sum of the first
n
generalized penta-
gonal numbers
( ) ( ) ( )
13
1
1 11
6 6 1 1.
n nn
kk
k
k kk
GPntntnnnnn
−
−
= = =
=+ =+ =+ − +=
∑ ∑∑
(68)
Since from (32) and (46), we have
( )( )
22 3
1 11nn nn nn
tt tt tt n
− −−
−=+ − =
from (68) it follows that
( )
22 3
1
1.
n
nn
k
kGP t t n
−
=
=−=
∑
(69)
Thus the equation
2 32
c ab
= +
has an infinite number of integer solutions.
In fact, for each
1
n≥
equations
22n
cab= +
and
22n
cab= +
have infinite
number of solutions (see Agarwal [14]).
To find the sum of the reciprocals of all generalized pentagonal numbers we
need the following well-known result, e.g., see Andrews
et al.
[27], page 536,
and Efthimiou [28]
2
22
1
1 1 1e
2 .
1e
s
s
k
sks s
−
−
π
π
∞
=
π +
+=
+−
∑
(70)
Now from (70), we have
( ) ( )
( ) ( )
22
11 1
22
11
33
33
33
3
3
3
3
1 14 1
3
3 31 2 1 13
4 11 1
34
1 3 1 12
41 1e 1 1e
3 3 2 3 12
32 8
1e 1e
1e
2 1e
1e
31 e
1e
31e
kk k
k
kk
GP kk k
kk
∞∞ ∞
= = =
∞∞
= =
−2π π
2π π
2π π
π
π
−
−−
−−
−
−
−
π
π
−
= =
−+ −+
= −
++
++
= −− −
−−
+
= −+
ππ
π
+
−
−
=+
π
∑∑ ∑
∑∑
and hence
( )
1
1 tanh 1.3052841530.
3 23
kk
GP
∞
=
ππ
= ≈
∑
(71)
To find all
square
generalized
pentagonal
numbers
, we need to find integer
solutions of the equation
( )
2
13 1nn m+ −=
. This equation can be written as
Pell’s equation
22
31ba−=
(its fundamental solution is
( ) ( )
, 1, 2ab
=
),
where
2bm
=
and
21an= −
. For this, corresponding to (22) the system is
1 11 2
1 1 12
14 , 1, 13
14 6, 1, 8
k kk
k kk
m mm m m
n nn n n
+−
+−
=−==
= −− = =
(72)
This system genetrates all (infinite) solutions
( )
,
kk
mn
of the equation
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2067
Journal of Applied Mathematics and Physics
()
2
13 1
nn m
+ −=
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 13,8 , 181,105 , 2521,1456 , 35113,20273 .
For
1k≥
, explicit solution of the system (72) appears as
( ) ( )
( ) ( )
21 21
21 21
123 23
4
31
23 23
12 2
kk
k
kk
k
m
n
−−
−−
= + +−
= + −− +
To find all
generalized
pentagonal
numbers
which
are
also
triangular
num-
bers
, we need to find integer solutions of the equation
( ) ( )
13 1 12nn mm+ −= +
. This equation can be written as Pell’s equation
22
63ba−=
(its fundamental solution is
( ) ( )
, 1, 3ab =
), where
21bm= +
and
21an= −
. For this, corresponding to (22) the system is
1 1 12
1 1 12
10 4, 1, 13
10 4, 1, 6
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(73)
This system genetrates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
13 1 12nn mm+ −= +
. First few of these solutions are
() ( )( )( ) ()
1,1 , 13,6 , 133,55 , 1321,540 , 13081,5341 .
For
1k≥
, explicit solution of the system (73) can be written as
()( ) ( )()
()()()()
11
11
13 6 5 26 3 6 526 2
4
12 6 5 26 2 6 526 4
8
kk
k
kk
k
m
n
−−
−−
= + + +− − −
= + + +− − +
There is no generalized pentagonal number which is also a rectangular num-
ber, in fact, the equation
( ) ( )
13 1 1
nn mm+ −= +
has no solutions. For this,
we note that this equation can be written as Pell’s equation
22
32ba−=
,
where
21bm= +
and
21
an= −
. Now reducing this equation to
( )
mod3
gives
( )
2
2 mod3b=
, which is impossible since all squares
( )
mod3
are ei-
ther 0 or 1
( )
mod3
.
To find all
generalized
pentagonal
numbers
which
are
also
pentagonal
num-
bers
, we need to find integer solutions of the equation
( ) ( )
13 1 3 12nn m m+ −= −
. This equation can also be written as Pell’s equa-
tion
22
18 7ba−=
(its fundamental solution is
( ) ( )
, 1, 5ab =
), where
61
bm= −
and
21an= −
. For this, corresponding to (22) the system is
1 1 12
1 1 12
1154 192, 1, 889
1154 576, 1, 629
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(74)
This system genetrates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
13 1 3 12
nn m m+ −= −
. First few of these solutions are
( ) ( ) ( ) ( )
()
1,1 , 889,629 , 1025713,725289 , 1183671721,836982301 ,
1365956140129,965876849489 .
For
1k≥
, explicit solution of the system (74) can be written as
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2068
Journal of Applied Mathematics and Physics
( )( )
( )( )
1378879 267903 2 577 408 2
10404
378879 267903 2 577 408 2 1734
k
k
k
m
=−+
++ − +
( )( )
( )( )
1126293 2 178602 577 408 2
6936
126293 2 178602 577 408 2 3468 .
k
k
k
n
= −+
− + −+
To find all
generalized
pentagonal
numbers
which
are
also
hexagonal
num-
bers
, we need to find integer solutions of the equation
( ) ( )
13 1 2 1nn m m+ −= −
. This equation can also be written as Pell’s equation
22
63ba−=
(its fundamental solution is
( ) ( )
, 1, 3ab =
), where
41bm= −
and
21an= −
. For this, corresponding to (22) the system is
1 1 12
1 1 12
10 2, 1, 7
10 4, 1, 6
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(75)
This system generates all (infinite) solutions
( )
,
kk
mn
of the equation
( ) ( )
13 1 2 1nn m m+ −= −
. First few of these solutions are
( ) ( ) () ( ) ( )
1,1 , 7,6 , 67,55 , 661,540 , 6541,5341 .
For
1k≥
, explicit solution of the system (75) can be written as
()()( )()
( )( ) ( )()
11
11
13 6 5 26 3 6 526 2
8
12 6 5 26 2 6 526 4
8
kk
k
kk
k
m
n
−−
−−
= + + +− − +
= + + +− − +
8. Heptagonal Numbers (Heptagon Numbers) (HEP)n
These numbers are defined by the sequence
1,7,18,34,55,81,
,
i.e.
, beginning
with 7 each number is formed from the previous one in the sequence by adding
the next number in the related sequence
( )
6,11,16,21, , 5 4n−
. Thus,
716= +
,
18 1 6 11 7 11=++ =+
,
34 1 6 11 16 18 16=++ + = +
, and so on (see
Figure 12).
Thus,
n
-th heptagonal number is defined as
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
1
5 4 1 6 11 16 5 4
1 1 5 1 2 5 1 15.
nn
HEP HEP n n
n
−
= +−=+++++−
=++ ++× + ++ −
(76)
Figure 12. Heptagonal numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
Comparing (76) with (3), we have
1, 5ad= =
, and hence from (4) it follows
that
( ) ( ) ( ) ( )
1
1
53 14 1 4.
22
nn
n
n
HEP n n n n t t
−
= − = ++ − = +
(77)
For all integers
0k≥
it follows that
( )
41k
HEP +
and
( )
42k
HEP
+
are odd,
whereas
( )
43k
HEP +
and
()
44
k
HEP
+
are even.
From (77) the following equality holds
( ) ( )( )
1 52
5 25 1
5 1 5 20 1 .
2
nn n
n
nn
HEP t t t
−−
−−
+= + += =
Let
m
be a given natural number, then it is
n
-th heptagonal number,
i.e.
,
( )
n
m HEP=
if and only if
( )
3 9 40 10nm=++
.
From (10) and (77), we have
( )
( )
234
3
41
7 18 34
1
xx xx x x
x
+=++ + +
−
and hence
( )( )
3
4 11xx x
−
+−
is the
generating
function
of all heptagonal num-
bers.
In view of (15) and (77), we have
( ) ( ) ( )
1
1
1 5 2.
6
n
k
kHEP n n n
=
=+−
∑
(78)
The sum of reciprocals of all heptagonal numbers is (see
https://en.wikipedia.org/wiki/Heptagonal_number)
( ) ( )
1
1 1 2 15 1
25 10 5 ln 5 ln 10 2 5
15 3 3 2
15 1
ln 10 2 5
32
1.3227792531.
kk
HEP
∞
=
+
=π− + + −
−
++
≈
∑
(79)
To find all
square
heptagonal
numbers
, we need to find integer solutions of
the equation
( )
2
5 32nn m−=
. This equation can be written as Pell’s equa-
tion
22
40 9ba−=
(its fundamental solutions are
( ) ( ) ( )
, 1, 7 , 2 ,1 3ab =
and
( )
9,57
), where
10 3bn= −
and
am=
. For
( )
1,7
, corresponding to (22)
the system is
1 11 2
1 1 12
1442 , 1, 1519
1442 432, 1, 961
k kk
k kk
m mmm m
n nn n n
+−
+−
=−==
= −− = =
(80)
This system genetrates infinite number of solutions
( )
,
kk
mn
of the equation
( )
2
5 32nn m−=
. First four of these solutions are
( ) ( ) ( ) ( )
1,1 , 1519,961 , 2190397,1385329 , 3158550955,1997643025 .
For
( )
2,13
recurrence relations remain the same as in (80) with
12
77, 111035mm= =
and
12
49, 70225nn= =
. This leads to another set of in-
finite number of solutions
( )
,
kk
mn
of the equation
( )
2
5 32nn m−=
. First
four of these solutions are
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
( ) ( ) ( )
()
77,49 , 111035,70225 , 160112393,101263969 ,
230881959671,146022572641 .
For
( )
9,57
also recurrence relations remain the same as in (80) with
12
9, 12987mm= =
and
12
6, 8214nn= =
. This leads to further set of infinite
number of solutions
( )
,
kk
mn
of the equation
( )
2
5 32nn m−=
. First four of
these solutions are
( ) ( ) ( ) ( )
9,6 , 12987,8214 , 18727245,11844150 , 27004674303,17079255654 .
To find all
heptagonal
numbers
which
are
also
triangular
numbers
, we need
to find integer solutions of the equation
( ) ( )
5 32 12n n mm−= +
. This eq-
uation can be written as Pell’s equation
22
54ba−=
(its fundamental solu-
tions are
( ) ()
, 3,7
ab =
and
( )
1,3
), where
10 3bn= −
and
21
am= +
.
For
()
3,7
corresponding to (22) the system is
1 1 12
1 1 12
322 160, 1, 493
322 96, 1, 221
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(81)
This system genetrates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
5 32 12
n n mm−= +
. First four of these solutions are
( ) ( ) ( ) ( )
1,1 , 493,221 , 158905,71065 , 51167077,22882613 .
For
( )
1,3
recurrence relations remain the same as in (81) with
12
10, 3382mm= =
and
12
5, 1513nn= =
. This leads to another set of infinite
number of solutions
( )
,
kk
mn
of the equation
( ) ( )
5 32 12n n mm−= +
. First
four of these solutions are
( ) ( ) ( ) ( )
10,5 , 3382,1513 , 1089154,487085 , 350704366,156839761 .
To find all
heptagonal
numbers
which
are
also
rectangular
numbers
, we need
to find integer solutions of the equation
( ) ( )
5 32 1n n mm−= +
. This equa-
tion can be written as Pell’s equation
22
10 1ba−=−
(its fundamental solu-
tion is
( ) ( )
, 1, 3ab =
), where
10 3bn= −
and
21am= +
. For this, corres-
ponding to (22) the system is
1 1 12
1 1 12
1442 720, 18, 26676
1442 432, 12, 16872
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(82)
This system genetrates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
5 32 1n n mm−= +
. First four of these solutions are
( ) ( ) ( ) ( )
18,12 , 26676,16872 , 38467494, 24328980 , 55470100392,35082371856 .
To find all
heptagonal
numbers
which
are
also
pentagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
5 32 3 12nn mm−= −
. This equa-
tion can be written as Pell’s equation
22
15 66ba−=
(its fundamental solution
is
( ) ( )
, 1, 9
ab
= −
), where
()
3 10 3bn= −
and
61am= −
. For this, corres-
ponding to (22) the system is
1 1 12
1 1 12
62 10, 1, 54
62 18, 1, 42
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(83)
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
This system genetrates infinite number of solutions
()
,
kk
mn
of the equation
( ) ( )
5 32 3 12nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 54,42 , 3337, 2585 , 206830,160210 , 12820113,9930417 .
To find all
heptagonal
numbers
which
are
also
hexagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
5 32 2 1nn mm−= −
. This equ-
ation can be written as Pell’s equation
22
54ba−=
(its fundamental solu-
tion is
( ) ( )
, 1, 3ab = −
), where
10 3bn= −
and
41am= −
. For this, cor-
responding to (22) the system is
1 1 12
1 1 12
322 80, 1, 247
322 96, 1, 221
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(84)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
5 32 2 1nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 247,221 , 79453,71065 , 25583539,22882613,
8237820025,7368130225 .
To find all
heptagonal
numbers
which
are
also
generalized
pentagonal
num-
bers
, we need to find integer solutions of the equation
( ) ( )
5 32 13 1
n n mm
−=+ −
. This equation can be written as Pell’s equation
22
30 19ba−=
(its fundamental solution is
( ) ( )
, 1, 7ab =
), where
10 3bn= −
and
21am
= −
. For this, corresponding to (22) the system is
1 1 12
1 1 12
22 10, 1, 13
22 6, 1, 14
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(85)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
5 32 13 1
n n mm−=+ −
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 13,14 , 275,301 , 6027,6602 , 132309,144937.
9. Octagonal Numbers On
These numbers are defined by the sequence
1,8,21,40,65,96,133,176,
,
i.e.
,
beginning with 8 each number is formed from the previous one in the sequence
by adding the next number in the related sequence
( )
7,13,19,25, , 6 5n−
.
Thus,
817= +
,
21 1 7 13 8 13=++ =+
,
40 1 7 13 19 21 19=++ + = +
, and so on
(see Figure 13).
Thus,
n
-th octagonal number is defined as
Figure 13. Octagonal numbers.
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2072
Journal of Applied Mathematics and Physics
( ) ( )
( ) ( ) ( )
( )
1
6 5 1 7 13 19 6 5
1 1 6 1 2 6 1 16.
nn
OO n n
n
−
= +−=+++++−
=++ ++× + ++ −
(86)
Comparing (86) with (3), we have
1, 6ad= =
, and hence from (4) it follows
that
() ( )
1
64 32 5.
2
n nn
n
O n nn t t
−
= −= −=+
(87)
For all integers
0k≥
it follows that
21k
O+
are odd, whereas
22k
O+
are
even (in fact divisible by 4).
Let
m
be a given natural number, then it is
n
-th octagonal number,
i.e.
,
n
mO
=
if and only if
( )
1 13 3nm=++
.
From (10) and (87), we have
( )
( )
234
3
51
8 21 40
1
xx xx x x
x
+=++ + +
−
and hence
( )( )
3
5 11xx x
−
+−
is the
generating
function
of all octagonal num-
bers.
In view of (15) and (87), we have
( )( )
1
11 2 1.
2
n
k
k
O nn n
=
= +−
∑
(88)
To find the sum of the reciprocals of all octagonal numbers, following
Downey [29] we begin with the series
( ) ( )
32
1
1
32
k
k
fx x
kk
∞−
=
=−
∑
and note that
( ) ( ) ( )
( )
3
33 3
11 1
ln 1
11 1
1 , .
32
k
kk k
k
x
f fx x
kk O k x
∞∞ ∞
−
= = =
−
′
= = = = −
−
∑∑ ∑
Thus, we have
( )
( ) ( )
( )
()
()
()
( )
( )
33
3 23
00
3
22
0
3
22
0
2
22
21
ln 1 ln 1 31
d d
2
21
ln 1 1 2 21
41
21
ln 1
1
3 d lim 0
2
ln 1 ln 1 1ln 1
2
22
1 3 21 3
ln 1 tan .
4
12 34
2 12
3
xx
x
t
tx
fx t t
t xt
xt
t
x tt
t
tt
t
xx xx
xx
x
xx
++
+
+
→
−
−−
=−=−
−
−+
=−−
−++
−
−=
++
++ −
= + −−
+
+ ++ + − π
∫∫
∫
Now since
( )
2
1
11
lim ln 1 1 0
2
xxx
−
→
− −=
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
it follows that
1
13 3
ln3 1.2774090576.
4 12
kk
O
∞
=
= ≈π+
∑
(89)
To find all
square
octagonal
numbers
, we need to find integer solutions of the
equation
( )
2
32nn m−=
. This equation can be written as Pell’s equation
22
31ba−=
(its fundamental solution is
( ) ( )
, 1, 2ab =
), where
31bn= −
and
am=
. For this, corresponding to (22) the system is
1 11 2
1 1 12
14 , 1, 15
14 4, 1, 9
k kk
k kk
m mm m m
n nn n n
+−
+−
=−==
= −− = =
(90)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( )
2
32nn m−=
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 15,9 , 209,121 , 2911,1681 , 40545, 23409 .
To find all
octagonal
numbers
which
are
also
triangular
numbers
, we need to
find integer solutions of the equation
( ) ( )
3 2 12n n mm−= +
. This equation
can be written as Pell’s equation
22
6 10ba−=
(its fundamental solutions
are
( ) ( )
, 1, 4ab =
and
( )
3,8
), where
( )
43 1bn= −
and
21am= +
. For
( )
1, 4
corresponding to (22) the system is
1 1 12
1 1 12
98 48, 6, 638
98 82, 3, 261
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(91)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
() ( )
3 2 12n n mm−= +
. First few of these solutions are
( ) ( ) ( ) ( )
( )
6,3 , 638,261 , 62566,25543 , 6130878,2502921 ,
600763526,245260683 .
For
( )
3,8
recurrence relations remain the same as in (91) with
12
1, 153
mm= =
and
12
1, 63nn= =
. This leads to another set of infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
3 22 12
n n mm−=+
. First few of these solutions
are
( ) ( ) ( ) ( ) ( )
1,1 , 153,63 , 15041,6141 , 1473913,601723 , 144428481,58962681 .
There is no octagonal number which is also a rectangular number, in fact, the
equation
( ) ( )
32 1n n mm−= +
has no solutions. For this, we note that this
equation can be written as Pell’s equation
22
31ba−=
(its fundamental so-
lution is
( ) ( )
, 1, 2ab =
), where
( )
23 1bn= −
and
21am= +
. For this,
Pell’s equation all solutions can be generated by the system (corresponding to
(21))
2 1 12
2 1 12
4 , 1, 4
4 , 2, 7
k kk
k kk
a a aa a
b b bb b
++
++
=−==
=−==
(92)
Now an explicit solution of the second equation of (92) can be written as
( ) ( )
1
23 23.
2
kk
k
b
= + +−
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
Next, if
()
23 3
k
kk
st
+=+
, then
( )
23 3
k
kk
st−=−
, and hence it fol-
lows that
kk
bs=
. We note that
()
1
2 mod6s≡
and
( )
2
1 mod6s≡
. Thus, from
the second equation of (92) mathematical induction immediately gives
( )
21
2 mod6s
−
≡
and
()
2
1 mod6s≡
for all
1≥
. In conclusion
1
kk
bs= ≡
or
( )
2 mod6
. Finally, reducing the relation
( )
23 1bn= −
to
( )
mod6
gives
( )
2 mod6b≡−
. Hence, in view of
0
b>
, we conclude that
k
bs
≠
for all in-
tegers
k
, and therefore, the equation
( ) ( )
32 1n n mm−= +
has no solution.
To find all
octagonal
numbers
which
are
also
pentagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
3 2 3 12nn mm−= −
. This equ-
ation can be written as Pell’s equation
22
87ba−=−
(its fundamental solu-
tions are
( ) ( )
, 1,1ab =
and
( )
2,5
), where
61bm= −
and
31an
= −
. For
( )
1,1
corresponding to (22) the system is
1 1 12
1 1 12
1154 192, 1, 1025
1154 384, 1, 725
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(93)
This system genetrates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
3 2 3 12nn mm−= −
. First four of these solutions are
( ) ( ) ( ) ( )
1,1 , 1025,725 , 1182657,836265 , 1364784961,965048701 .
For
()
2,5
recurrence relations remain the same as in (93) with
12
11, 12507mm
= =
and
12
8, 8844nn= =
. This leads to another set of infinite
number of solutions
( )
,
kk
mn
of the equation
() ( )
3 2 3 12
nn mm−= −
. First
four of these solutions are
( ) ( ) ( ) ( )
11,8 , 12507,8844 , 14432875,10205584 , 16655525051,11777234708 .
To find all
octagonal
numbers
which
are
also
hexagonal
numbers
, we need to
find integer solutions of the equation
( ) ( )
32 21
nn mm−= −
. This equation
can be written as Pell’s equation
22
6 10ba−=
(its fundamental solutions
are
() ( )
, 1,4ab =
and
( )
3,8
), where
()
43 1bn= −
and
41am= −
. For
( )
3,8
corresponding to (22) the system is
1 1 12
1 1 12
98 24, 1, 77
98 32, 1, 63
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(94)
This system genetrates infinite number of solutions
( )
,
kk
mn
of the equation
() ( )
32 21nn mm
−= −
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 77,63 , 7521,6141 , 736957,601723 , 72214241,58962681 .
With
( ) ( )
, 1, 4ab
=
the system corresponding to (21) is
2 1 12
2 1 12
10 , 1, 13
10 , 4, 32
k kk
k kk
a a aa a
b b bb b
++
++
=−==
=−==
(95)
Now note that
( )
11 mod4a≡
and
( )
2
1 mod4a≡
. Thus, from the first equa-
tion of (95) mathemtical induction immediately gives
( ) ( ) ( )
2
10 mod4 1 mod4 1 mod 4
k
a
+
≡ −≡
for all
1k≥
. Next reducing the rela-
tion
41am= −
to
( )
mod4
gives
( )
1 mod4a≡−
. Hence, in view of
0b>
,
we conclude that
k
aa≠
for all integers
k
, and therefore, the equation
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2075
Journal of Applied Mathematics and Physics
( ) ( )
32 21nn mm
−= −
has no solution.
To find all
octagonal
numbers
which
are
also
generalized
pentagonal
numbers
,
we need to find integer solutions of the equation
()( )
3 2 13 1
n n mm
−=+ −
.
This equation can be written as Pell’s equation
22
7ba−=
, where
( )
23 1bn= −
and
()
32 1
am
= −
. For the equation
22
7ba−=
the only
meaningful integer solution is
4, 3
ba= =
and it gives
( ) ( )
, 1,1mn =
.
To find all
octagonal
numbers
which
are
also
heptagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
3 2 5 32nn mm−= −
. This equ-
ation can be written as Pell’s equation
22
30 39ba−=−
(its fundamental
solution is
( ) ( )
, 2, 9ab = −
), where
( )
3 10 3bm= −
and
( )
23 1an= −
. For
this, corresponding to (22) the system is
1 1 12
1 1 12
482 144, 1, 345
482 160, 1, 315
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(96)
This system genetrates infinite number of solutions
()
,
kk
mn
of the equation
( ) ( )
3 2 5 32nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 345,315 , 166145,151669 , 80081401,73103983 ,
38599068993,35235967977 .
10. Nonagonal Numbers Nn
These numbers are defined by the sequence
1,9,24, 46,75,111,154,
,
i.e.
, be-
ginning with 9 each number is formed from the previous one in the sequence by
adding the next number in the related sequence
( )
8,15,22, 29, , 7 6
n−
. Thus,
918= +
,
24 1 8 15 9 15=++ =+
,
46 1 8 15 22 24 22
=++ + = +
, and so on (see
Figure 14).
Thus,
n
-th nonagonal number is defined as
( ) ( )
( ) ( ) ( )
( )
1
7 6 1 8 15 22 7 6
1 1 7 1 2 7 1 17.
nn
NN n n
n
−
= +−=+++++−
=++ ++× + ++ −
(97)
Comparing (97) with (3), we have
1, 7ad= =
, and hence from (4) it follows
that
( )
1
75 6.
2
n nn
n
N n tt
−
= −=+
(98)
For all integers
0k≥
it follows that
41 42
,
kk
NN
++
are odd, whereas
43 44
,
kk
NN
++
are even.
Figure 14. Nonagonal numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
Let
m
be a given natural number, then it is
n
-th nonagonal number,
i.e.
,
n
mN=
if and only if
( )
5 25 56 14nm=++
.
From (10) and (98), we have
( )
( )
234
3
61 9 24 46
1
xx xx x x
x
+=++ + +
−
and hence
( ) ( )
3
6 11xx x
−
+−
is the
generating
function
of all nonagonal num-
bers.
In view of (15) and (98), we have
( )( )
1
11 7 4.
6
n
k
k
N nn n
=
=+−
∑
(99)
The sum of reciprocals of all nonagonal numbers is
( ) ( )
11 1
1 2 14 2
75 5755
25
5 75
25 7
1.2433209262;
kk k
k
N kk k n
γ
∞∞ ∞
= = =
= = −
−−
=− Ψ− − +
≈
∑∑ ∑
(100)
here,
()
x
Ψ
is the
digamma
function
defined as the logarithmic derivative of
the
gamma
function
( )
xΓ
,
i.e.
,
( ) ( ) ( )
x xx
′
Ψ=Γ Γ
, and
0.5772156649
γ
=
is the
Euler-Mascheroni
constant
.
To find all
square
nonagonal
numbers
, we need to find integer solutions of
the equation
( )
2
7 52nn m−=
. This equation can be written as Pell’s equa-
tion
22
14 25ba−=
(its fundamental solution are
() ( )
, 2,9ab =
and
( )
6,23
), where
14 5bn= −
and
2am=
. For
( )
2,9
corresponding to (22)
the system is
1 11 2
1 1 12
30 , 1, 33
30 10, 1, 18
k kk
k kk
m mm m m
n nn n n
+−
+−
=−==
= −− = =
(101)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( )
2
7 52nn m−=
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 33,18 , 989,529 , 29637,15842 , 888121, 474721 .
For
( )
6,23
recurrence relations remain the same as in (101) with
12
3, 91mm= =
and
12
2, 49nn= =
. This leads to another set of infinite number
of solutions
( )
,
kk
mn
of the equation
( )
2
7 52nn m−=
. First few of these so-
lutions are
() ( ) ( ) ( ) ( )
3,2 , 91,49 , 2727,1458 , 81719,43681 , 2448843,1308962 .
To find all
nonagonal
numbers
which
are
also
triangular
numbers
, we need
to find integer solutions of the equation
( ) ( )
7 52 12n n mm−= +
. This
equation can be written as Pell’s equation
22
7 18ba−=
(its fundamental
solutions are
( ) ( ) ( )
, 1,5 , 3,9ab =
and
( )
7,19
), where
14 5bn= −
and
21am= +
. For
( )
3,9
corresponding to (22) the system is
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
1 1 12
1 1 12
16 7, 1, 25
16 5, 1, 10
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(102)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
7 52 12n n mm−= +
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 25,10 , 406,154 , 6478,2449 , 1032249,39025.
For
( )
1,5
and
()
7,19
there are no integer solutions.
There is no nonagonal number which is also a rectangular number, in fact,
the equation
()( )
7 52 1
n n mm−= +
has no solutions.
To find all
nonagonal
numbers
which
are
also
pentagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
7 52 3 12nn mm−= −
. This
equation can be written as Pell’s equation
22
21 204ba−=
(its fundamental
solutions are
( ) ( )
, 5,27ab =
and
( )
125,573
), where
( )
3 14 5bn
= −
and
61am= −
. For
( )
5,27
corresponding to (22) the system is
1 1 12
1 1 12
12098 2016, 1, 10981
12098 4320, 1, 7189
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
+ −− = =
(103)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
7 52 3 12nn mm−= −
. First four of these solutions are
( ) ( ) ( )
( )
1,1 , 10981,7189 , 132846121,86968201 ,
1607172358861,1052141284189 .
For
( )
125,573
recurrence relations remain the same as in (103) with
12
21, 252081mm= =
and
12
14, 165026nn= =
. This leads to another set of in-
finite number of solutions
()
,
kk
mn
of the equation
( ) ( )
3 2 3 12nn mm−= −
.
First four of these solutions are
( ) ( ) ( )
( )
21,14 , 252081,165026 , 3049673901,1996480214,
36894954600201,24153417459626 .
To find all
nonagonal
numbers
which
are
also
hexagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
7 52 2 1nn mm−= −
. This equ-
ation can be written as Pell’s equation
22
7 18ba−=
(its fundamental solu-
tions are
() ( ) ( )
, 1,5 , 3,9
ab =
and
( )
7,19
), where
14 5bn= −
and
41am= −
. For
()
3,9
corresponding to (20) the system is
1
1 11
8 21
8 3 , 9, 3
k kk
k kk
b ba
a a bb a
+
+
= +
=+==
This system gives first four integer solutions
( )
,
kk
mn
of the equation
( ) ( )
7 52 2 1nn mm−= −
rather easily, which appear as
( ) ( ) ( ) ( )
1,1 , 13,10 , 51625,39025 , 822757,621946
. Now the following system ge-
nerates infinite number of solutions
( )
21 21
, , 2
kk
mn k
++
≥
21 21 23 1 3
21 21 23 1 3
64514 16128, 1, 51625
64514 23040, 1, 39025
k kk
k kk
m m m mm
n n n nn
+ −−
+ −−
= −− ==
= −− ==
(104)
Similarly, the following system generates infinite number of solutions
( )
22
, , 2
kk
mn k≥
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
22 2 22 2 4
22 2 22 2 4
64514 16128, 13, 822757
64514 23040, 10, 621946
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(105)
The first eight solutions
( )
,
kk
mn
are
()( ) ( ) ()
() ( )
()
()
1,1 , 13,10 , 51625,39025 , 822757,621946 ,
3330519121,2517635809 , 53079328957,40124201194 ,
214865110504441,162422756519761 ,
3424359827493013,2588572715184730 .
With
( ) ( )
, 1,5ab =
and
()
7,19
there are no integer solutions of the re-
quired equation.
To find all
nonagonal
numbers
which
are
also
generalized
pentagonal
num-
bers
, we need to find integer solutions of the equation
( ) ( )
7 52 13 1n n mm−=+ −
. This equation can be written as Pell’s equation
22
42 39ba−=
(its fundamental solutions are
( ) ( )
, 1, 9ab =
and
( )
5,33
),
where
14 5bn= −
and
21am= −
. For
( )
1,9
, corresponding to (22) the
system is
1 1 12
1 1 12
674 336, 1, 403
674 240, 1, 373
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(106)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
5 32 13 1n n mm−=+ −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 403,373 , 271285,251161 , 182845351,169281901 ,
123237494953,114095749873 .
For
( )
5,33
recurrence relations remain the same as in (106) with
12
66, 44148mm
= =
and
12
61, 40873nn= =
. This leads to another set of infi-
nite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
7 52 13 1n n mm
−=+ −
.
First four of these solutions are
( ) ( ) ( ) ( )
66,61 , 44148,40873 , 29755350,27548101 , 20055061416,18567378961 .
To find all
nonagonal
numbers
which
are
also
heptagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
7 52 5 32nn mm−= −
. This
equation can be written as Pell’s equation
22
35 310ba−=
(its fundamental
solution is
( ) ( )
, 7,45ab =
), where
( )
5 14 5bn= −
and
10 3am= −
. For this,
corresponding to (22) the system is
1 1 12
1 1 12
142 42, 1, 104
142 50, 1, 88
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(107)
This system generates
infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
7 52 5 32nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 104,88 , 14725,12445 , 2090804,1767052 , 296879401,250908889 .
To find all
nonagonal
numbers
which
are
also
octagonal
numbers
, we need to
find integer solutions of the equation
( ) ( )
7 52 3 2nn mm−= −
. This equa-
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2079
Journal of Applied Mathematics and Physics
tion can be written as Pell’s equation
22
42 57ba−=
(its fundamental solu-
tion is
( ) ( )
, 4,27ab =
), where
( )
3 14 5
bn= −
and
( )
23 1am
= −
. For this,
corresponding to (22) the system is
1 1 12
1 1 12
674 224, 1, 459
674 240, 1, 425
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(108)
This system generates infinite number of solutions
()
,
kk
mn
of the equation
( ) ( )
7 52 3 2nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 459,425 , 309141,286209 , 208360351,192904201 ,
140434567209,130017145025 .
11. Decagonal Numbers Dn
These numbers are defined by the sequence
1,10,27,52,85,126,175,
,
i.e.
, be-
ginning with 10 each number is formed from the previous one in the sequence
by adding the next number in the related sequence
( )
9,17,25,33, , 8 7n−
.
Thus,
10 1 9= +
,
27 1 9 17 10 17=++ = +
,
52 1 9 17 25 27 25=++ + = +
and so
on (see Figure 15).
Hence,
n
-th decagonal number is defined as
( ) ( )
( ) ( ) ( )
( )
1
8 7 1 9 17 25 8 7
1 1 8 1 2 8 1 18.
nn
DD n n
n
−
= +−=+++++−
=++ ++×+ ++ −
(109)
Comparing (109) with (3), we have
1, 8ad= =
, and hence from (4) it follows
that
( ) ( )
1
86 43 7.
2
n nn
n
D n nn t t
−
= −= −=+
(110)
For all integers
0k≥
it follows that
21k
D+
are odd, whereas
2k
D
are even.
Let
m
be a given natural number, then it is
n
-th decagonal number,
i.e.
,
n
mD=
if and only if
( )
3 9 16 8nm=++
.
From (10) and (110), we have
( )
( )
234
3
71
10 27 52
1
xx xx x x
x
+=++++
−
and hence
( )( )
3
7 11xx x
−
+−
is the
generating
function
of all decagonal num-
bers.
In view of (15) and (110), we have
Figure 15. Decagonal numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
( )( )
1
1
1 8 5.
6
n
k
k
D nn n
=
= +−
∑
(111)
To find the sum of the reciprocals of all decagonal numbers, as in (89) we be-
gin with the series
( )
( )
43
1
1
43
k
k
fx x
kk
∞−
=
=−
∑
and following the same steps Downey [29] obtained
( ) ( )
11
11
1 ln 2 1.2167459562.
43 6
kk
k
fkk D
∞∞
= =
= = = +
−
π≈
∑∑
(112)
To find all
square
decagonal
numbers
, we need to find integer solutions of
the equation
( )
2
43nn m−=
. This equation can be written as Pell’s equation
22
9ba−=
, where
83bn= −
and
4am=
. For the equation
22
9ba−=
the only meaningful integer solution is
5, 4ba
= =
and it gives
( ) ()
, 1,1mn =
.
To find all
decagonal
numbers
which
are
also
triangular
numbers
, we need to
find integer solutions of the equation
( ) ( )
4 3 12n n mm−= +
. This equation
can be written as Pell’s equation
22
27ba−=
(its fundamental solutions are
( ) ( )
, 1, 3ab =
and
( )
3,5
), where
83bn= −
and
21
am
= +
. For
( )
3,5
corresponding to (22) the system is
1 1 12
1 1 12
34 16, 1, 55
34 12, 1, 20
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(113)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
4 3 12n n mm−= +
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 55,20 , 1885,667 , 64051,22646 , 2175865,769285 .
For
( )
1, 3
recurrence relations remain the same as in (113) with
12
4, 154mm= =
and
12
2, 55nn= =
. This leads to another set of infinite num-
ber of solutions
()
,
kk
mn
of the equation
( ) ( )
4 3 12n n mm−= +
. First few of
these solutions are
( ) ( ) ( ) ( ) ( )
4,2 , 154,55 , 5248,1856 , 178294,63037 , 6056764,2141390 .
There is no decagonal number which is also a rectangular number, in fact,
the equation
( ) ( )
43 1n n mm−= +
has no solutions.
To find all
decagonal
numbers
which
are
also
pentagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
4 3 3 12nn mm−= −
. This equ-
ation can be written as Pell’s equation
22
6 75ba−=
(its fundamental solu-
tion is
( ) ( )
, 5,15ab =
), where
( )
38 3bn= −
and
61am= −
. For
( )
5,15
corresponding to (22) the system is
1 1 12
1 1 12
98 16, 1, 91
98 36, 1, 56
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(114)
This system generates infinite number of solutions
()
,
kk
mn
of the equation
( ) ( )
4 3 3 12nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 91,56 , 8901,5451 , 872191,534106 , 85465801,52336901 .
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
To find all
decagonal
numbers
which
are
also
hexagonal
numbers
, we need to
find integer solutions of the equation
( ) ( )
43 21
nn mm
−= −
. This equation
can be written as Pell’s equation
22
27ba−=
(its fundamental solution is
( ) ( )
, 3,5ab =
), where
83bn= −
and
41am= −
. For this, corresponding to
(22) the system is
1 1 12
1 1 12
34 8, 1, 28
34 12, 1, 20
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(115)
This system generates infinite number of solutions
()
,
kk
mn
of the equation
( ) ( )
43 21nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 28,20 , 943,667 , 32026,22646 , 1087933,769285 .
To find all
decagonal
numbers
which
are
also
generalized
pentagonal
num-
bers
, we need to find integer solutions of the equation
( ) ( )
4 3 13 1n n mm−=+ −
. This equation can be written as Pell’s equation
22
12 13ba−=
(its fundamental solution is
( ) ( )
, 1, 5ab =
), where
83bn
= −
and
21am= −
. For this, corresponding to (22) the system is
1 1 12
1 1 12
194 96, 1, 119
194 72, 1, 103
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(116)
This system generates infinite number of solutions
()
,
kk
mn
of the equation
( ) ( )
4 3 13 1n n mm−=+ −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 119,103 , 22989,19909 , 4459651,3862171 ,
865149209,749241193 .
To find all
decagonal
numbers
which
are
also
heptagonal
numbers
, we need
to find integer solutions of the equation
( ) ( )
4 3 5 32nn mm−= −
. This equ-
ation can be written as Pell’s equation
22
10 540ba
−=
(its fundamental so-
lution is
( ) ( )
, 14,50ab =
), where
( )
10 8 3bn= −
and
( )
2 10 3am= −
. For
this, corresponding to (22) the system is
1 1 12
1 1 12
1442 432, 1, 1075
1442 540, 1, 850
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(117)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
() ( )
4 3 5 32
nn mm−= −
. First few of these solutions are
( ) (
) ( ) ( )
( )
1,1 , 1075,850 , 1549717,1225159 , 2234690407,1766677888 ,
3222422016745,2547548288797 .
To find all
decagonal
numbers
which
are
also
octagonal
numbers
, we need to
find integer solutions of the equation
( ) ( )
43 3 2
nn mm−= −
. This equation
can be written as Pell’s equation
22
3 33ba−=
(its fundamental solution is
( ) ( )
, 8,15
ab =
), where
( )
38 3bn= −
and
( )
43 1am= −
. For this, corres-
ponding to (22) the system is
1 1 12
1 1 12
194 64, 1, 135
194 72, 1, 117
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(118)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
() ( )
43 3 2nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 135,117 , 26125,22625 , 5068051,4389061 ,
983175705,851455137 .
To find all
decagonal
numbers
which
are
also
nonagonal
numbers
, we need
to find integer solutions of the equation
()( )
4 3 7 52
nn mm−= −
. This equ-
ation can be written as Pell’s equation
22
14 91ba−=
(its fundamental solu-
tion is
( ) ( )
, 9,35ab =
), where
( )
78 3bn= −
and
14 5am= −
. For this,
corresponding to (22) the system is
1 1 12
1 1 12
898 320, 1, 589
898 336, 1, 551
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(119)
This system generates infinite number of solutions
()
,
kk
mn
of the equation
( ) ( )
4 3 7 52nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 589,551 , 528601, 494461 , 474682789,444025091 ,
426264615601,398734036921 .
12. Tetrakaidecagonal Numbers (TET)n
These numbers are defined by the sequence
1,14,39,76,125,
,
i.e.
, beginning
with 14 each number is formed from the previous one in the sequence by adding
the next number in the related sequence
( )
13,25,37, 49, , 12 11
n−
. Thus,
14 1 13= +
,
39 1 13 25 14 25
=++ =+
,
76 1 13 25 37 39 37=++ + = +
, and so on
(see Figure 16).
Hence,
n
-th tetrakaidecagonal number is defined as
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
1
12 11 1 13 25 37 12 11
1 1 12 1 2 12 1 1 12 .
nn
TET TET n n
n
−
= + − =+++++ −
=++ ++× + ++ −
(120)
Comparing (120) with (3), we have
1, 12ad= =
, and hence from (4) it fol-
lows that
( ) ( ) ( )
1
12 10 6 5 11 .
2
nn
n
n
TET n n n t t
−
= − = −=+
(121)
For all integers
0k≥
it follows that
( )
21k
TET +
are odd, whereas
( )
2k
TET
are even.
Let
m
be a given natural number, then it is
n
-th tetrakaidecagonal number,
i.e.
,
( )
n
m TET=
if and only if
( )
5 25 24 12
nm=++
.
Figure 16. Tetrakaidecagonal numbers.
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DOI:
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Journal of Applied Mathematics and Physics
From (10) and (121), we have
( )
( )
234
3
11 1 14 39 76
1
xx xx x x
x
+=+++ +
−
and hence
( )( )
3
11 1 1xx x
−
+−
is the
generating
function
of all tetrakaidecagonal
numbers.
In view of (15) and (121), we have
( ) ( )( )
1
1
1 4 3.
2
n
k
k
TET n n n
=
=+−
∑
(122)
To find the sum of the reciprocals of all tetrakaidecagonal numbers, as in (89)
we begin with the series
()
( )
65
1
1
65
k
k
fx x
kk
∞−
=
=−
∑
and following the same steps Downey [29] obtained
( ) ( ) ( )
( )
11
1 11
1 4 ln 2 3 ln 3 3
6 5 10
1.1509823681.
kk
k
fk k TET
∞∞
= =
= = = +
−
≈
π+
∑∑
(123)
To find all
square
tetrakaidecagonal
numbers
, we need to find integer solu-
tions of the equation
( )
2
65nn m−=
. This equation can be written as Pell’s
equation
22
6 25ba−=
(its fundamental solution are
( ) ( )
, 2,7ab =
and
( )
4,11
), where
12 5bn= −
and
2
am=
. For (2,7) corresponding to (22)
the system is
1 11 2
1 1 12
98 , 1, 119
98 40, 1, 49
k kk
k kk
m mm m m
n nn n n
+−
+−
=−==
= −− = =
(124)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( )
2
65nn m−=
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 119,49 , 11661,4761 , 1142659,466489 , 111968921,45711121 .
For
( )
4,11
recurrence relations remain the same as in (124) with
12
21, 2059mm= =
and
12
2, 841nn= =
. This leads to another set of infinite
number of solutions
( )
,
kk
mn
of the equation
( )
2
65nn m−=
. First few of
these solutions are
() ( ) ( ) ( )
()
21,9 , 2059,841 , 201761,82369 , 19770519,8071281 ,
1937309101,790903129 .
To find all
tetrakaidecagonal
numbers
which
are
also
triangular
numbers
, we
need to find integer solutions of the equation
( ) ( )
6 5 12n n mm−= +
. This
equation can be written as Pell’s equation
22
3 22ba−=
(its fundamental
solutions are
( ) ( )
, 1, 5ab =
and
( )
3,7
), where
12 5bn= −
and
21am= +
. For
( )
3,7
corresponding to (22) the system is
1 1 12
1 1 12
194 96, 1, 341
194 80, 1, 99
k kk
k kk
m mm m m
n nn n n
+−
+−
= −+ = =
= −− = =
(125)
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DOI:
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Journal of Applied Mathematics and Physics
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
() ( )
6 5 12n n mm
−= +
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 341,99 , 66249,19125 , 12852061,3710071 ,
2493233681,719734569 .
For
( )
1,5
recurrence relations remain the same as in (125) with
12
50, 9798mm= =
and
12
15, 2829nn= =
. This leads to another set of infinite
number of solutions
( )
,
kk
mn
of the equation
( )
2
65nn m−=
. First few of
these solutions are
( ) () ( )
( ) ( )
50,15 , 9798,2829 , 1900858,548731 ,
368756750,106450905 , 71536908738,20650926759 .
There is no tetrakaidecagonal number which is also a rectangular number, in
fact, the equation
( ) ( )
65 1n n mm−= +
has no solutions.
To find all
tetrakaidecagonal
numbers
which
are
also
pentagonal
numbers
,
we need to find integer solutions of the equation
( ) ( )
6 5 3 12nn mm−= −
.
This equation can be written as Pell’s equation
22
24ba−=
, where
12 5bn= −
and
61am= −
. For the equation
22
24ba−=
the only meaningful integer
solution is
7, 5ba
= =
and it gives
() ( )
, 1,1mn =
.
To find all
tetrakaidecagonal
numbers
which
are
also
hexagonal
numbers
, we
need to find integer solutions of the equation
( ) ( )
65 21nn mm−= −
. This
equation can be written as Pell’s equation
22
3 22ba−=
(its fundamental
solutions are
( ) ( )
, 1, 5ab =
and
( )
3,7
), where
12 5bn= −
and
41am= −
.
For
( )
3,7
corresponding to (22) the system is
1 1 12
1 1 12
194 48, 1, 171
194 80, 1, 99
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(126)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
() ( )
65 21nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 171,99 , 33125,19125 , 6426031,3710071 ,
1246616841,719734569 .
With
() ( )
, 1, 5ab =
there are no integer solutions of the required equation.
To find all
tetrakaidecagonal
numbers
which
are
also
generalized
pentagonal
numbers
, we need to find integer solutions of the equation
( ) ( )
6 5 13 1n n mm−=+ −
. This equation can be written as Pell’s equation
22
18 31ba−=
(its fundamental solutions are
( ) ( )
, 1, 7ab =
and
( )
11,47
),
where
12 5bn= −
and
21am= −
. For
( )
1, 7
, corresponding to (22) the
system is
1 1 12
1 1 12
1154 576, 1, 765
1154 480, 1, 541
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(127)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
6 5 13 1n n mm−=+ −
. First few of these solutions are
( ) (
) ( ) ( )
( )
1,1 , 765,541 , 882233, 623833 , 1018095541,719902261 ,
1174881371505,830766584881 .
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
For
()
11,47
recurrence relations remain the same as in (127) with
12
188, 216376mm= =
and
12
133, 151001nn= =
. This leads to another set of
infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
6 5 13 1
n n mm
−=+ −
.
First four of these solutions are
( ) ( ) ( )
( )
188,133 , 216376,153001 , 249697140,176562541,
288150282608,203753018833 .
To find all
tetrakaidecagonal
numbers
which
are
also
heptagonal
numbers
,
we need to find integer solutions of the equation
() ( )
6 5 5 32nn mm−= −
.
This equation can be written as Pell’s equation
22
15 490ba−=
(its funda-
mental solutions are
( ) ( )
, 3,25
ab =
and
( )
7,35
), where
( )
5 12 5
bn= −
and
10 3am= −
. For (7,35), corresponding to (22) the system is
1 1 12
1 1 12
3842 1152, 1, 3081
3842 1600, 1, 1989
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(128)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
()( )
6 5 5 32
nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 3081,1989 , 11836049,7640137 , 45474096025,29353402765 ,
174711465090849,112775765781393 .
With
( ) ( )
, 3,25ab =
there are no integer solutions of the required equation.
To find all
tetrakaidecagonal
numbers
which
are
also
octagonal
numbers
, we
need to find integer solutions of the equation
( ) ( )
65 3 2nn mm−= −
. This
equation can be written as Pell’s equation
22
2 17
ba−=
(its fundamental
solutions are
( ) ( )
, 2,5ab =
and
( )
4,7
), where
12 5bn= −
and
62am= −
.
For (4,7), corresponding to (22) the system is
1 1 12
1 1 12
1154 384, 1, 861
1154 480, 1, 609
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(129)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
65 3 2nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( )
( )
1,1 , 861,609 , 993209,702305 , 1146161941,810458881 ,
1322669886321,935268845889 .
With
( ) ( )
, 2,5ab =
there are no integer solutions of the required equation.
To find all
tetrakaidecagonal
numbers
which
are
also
nonagonal
numbers
, we
need to find integer solutions of the equation
( ) ( )
6 5 7 52nn mm−= −
. This
equation can be written as Pell’s equation
22
21 700ba−=
(its fundamental
solutions are
( ) ( ) ( )
, 2,28 , 5,35ab =
and
( )
9,49
), where
( )
7 12 5bn= −
and
14 5am= −
. For (9,49), corresponding to (22) the system is
1 1 12
1 1 12
12098 4320, 1, 8509
12098 5040, 1, 6499
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(130)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
6 5 7 52nn mm−= −
. First four of these solutions are
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
() ( ) ( ) ( )
1,1 , 8509, 6499 , 102937561,78619861 , 1245338600149,951143066839 .
With
()( )
, 2,28ab =
and
( )
5,35
there are no integer solutions of the re-
quired equation.
To find all
tetrakaidecagonal
numbers
which
are
also
decagonal
numbers
, we
need to find integer solutions of the equation
( ) ( )
65 4 3nn mm−= −
. This
equation can be written as Pell’s equation
22
6 46ba−=
(its fundamental
solutions are
( ) ( )
, 3,10ab =
and
( )
5,14
), where
( )
2 12 5bn= −
and
83
am= −
. For (5,14), corresponding to (22) the system is
1 1 12
1 1 12
98 36, 1, 66
98 40, 1, 54
k kk
k kk
m mm m m
n nn n n
+−
+−
= −− = =
= −− = =
(131)
This system generates infinite number of solutions
( )
,
kk
mn
of the equation
( ) ( )
65 4 3nn mm−= −
. First few of these solutions are
( ) ( ) ( ) ( ) ( )
1,1 , 66,54 , 6431,5251 , 630136,514504 , 61746861,50416101 .
With
( ) ( )
, 3,10ab =
there are no integer solutions of the required equation.
13. Centered Triangular Numbers (ct)n
These numbers are defined by the sequence
1,4,10,19,31,46,64,85,109,
,
i.e.
,
beginning with 4 each number is formed from the previous one in the sequence
by adding the next number in the related sequence
()
3,6,9,12, ,3 1n−
. Thus,
413
= +
,
10 1 3 6 4 6=++= +
,
19 1 3 6 9 10 9=+++= +
, and so on (see Figure
17).
Hence,
n
-th centered triangular number is defined as
( ) ( ) ( ) ( )
( )
( )
1
3 1 1369 3 3
1 31 2 3 1 .
nn
ct ct n n
n
−
= + − =++++ + −
=+ +++ + −
(132)
Thus, from (2) it follows that
( ) ( )
1 12
1
13 13 .
2n nn n
n
nn
ct t t t t
− −−
−
=+ =+ =++
(133)
Let
m
be a given natural number, then it is
n
-th centered triangular number,
i.e.
,
( )
n
m ct=
if and only if
( )
( )
3 9 24 1 6nm=++ −
.
From (10) and (133), we have
Figure 17. Centered triangular numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
( )
( )
2
234
3
14 10 19
1
xx x xx x x
x
++ =++ + +
−
and hence
( )
( )
3
211xx x x−
++ −
is the
generating
function
of all centered tri-
angular numbers.
In view of (15) and (133), we have
( )
( )
2
1
11.
2
n
k
kct n n
=
= +
∑
(134)
To find the sum of the reciprocals of all centered triangular numbers we fol-
low as in (71), to obtain
( )
1
12 5
tanh 1.5670651313.
23
15
kk
ct
∞
=
= ≈
ππ
∑
(135)
14. Centered Square Numbers (cS)n
These numbers are defined by the sequence
1,5,13,25,41,61,85,113,
,
i.e.
,
beginning with 5 each number is formed from the previous one in the sequence
by adding the next number in the related sequence
( )
4,8,12,16, ,4 1n−
. Thus,
514
= +
,
13 1 4 8 5 8=++=+
,
25 1 4 8 12 13 12=+++ = +
, and so on (see Fig-
ure 18).
Hence,
n
-th centered square number is defined as
( ) ( ) ( ) ( )
( )
( )
1
4 1 1 4 8 12 4 4
14123 1.
nn
cS cS n n
n
−
= + − =+++ + + −
=+ +++ + −
(136)
Thus, from (2) it follows that
( ) ( ) ( )
2
22
1
1 12
1
14 14 12 2 1
22.
n
n
nn n n n
nn
cS t n n n n
SS t t t
−
− −−
−
=+ =+ =+ − = +−
=+=+ +
(137)
Let
m
be a given natural number, then it is
n
-th centered square number,
i.e.
,
( )
n
m cS=
if and only if
()
1 2 12nm=+−
.
From (10) and (137), we have
Figure 18. Centered square numbers.
R. P. Agarwal
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Journal of Applied Mathematics and Physics
( )
( )
2
23 4
3
15 13 25
1
xx xx x x
x
+=++ + +
−
and hence
( ) ( )
23
11xx x
−
+−
is the
generating
function
of all centered square
numbers.
In view of (15) and (137), we have
()
()
2
1
12 1.
3
n
k
k
cS n n
=
= +
∑
(138)
To find the sum of the reciprocals of all centered triangular numbers we fol-
low as in (71), to obtain
( )
1
1tanh 1.44065952.
22
kk
cS
∞
=
= ≈
ππ
∑
(139)
15. Centered Pentagonal Numbers (cP)n
These numbers are defined by the sequence
1,6,16,31,51,76,106,141,181,
,
i.e.
, beginning with 6 each number is formed from the previous one in the se-
quence by adding the next number in the related sequence
()
5,10,15,20, ,5 1n−
. Thus,
615
= +
,
16 1 5 10 6 10=++ =+
,
31 1 5 10 15 16 15=++ + = +
, and so on (see Figure 19).
Hence,
n
-th centered pentagonal number is defined as
( ) ( ) ( ) ( )
( )
( )
1
5 1 1 5 10 15 5 5
1 51 2 3 1 .
nn
cP cP n n
n
−
= + − =++ + + + −
=+ +++ + −
(140)
Thus, from (2) it follows that
( ) ( )
1 12
1
15 15 3 .
2n nn n
n
nn
cP t t t t
− −−
−
=+ =+=++
(141)
Let
m
be a given natural number, then it is
n
-th centered pentagonal number,
i.e.
,
( )
n
m cP=
if and only if
( )
( )
5 25 40 1 10nm=++ −
.
From (10) and (141), we have
( )
( )
2
234
3
31 6 16 31
1
xx x xx x x
x
++=++ + +
−
Figure 19. Centered pentagonal numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
and hence
( )
( )
3
23 11xx x x−
++ −
is the
generating
function
of all centered
pentagonal numbers.
In view of (15) and (141), we have
( )
()
2
1
15 1.
6
n
k
k
cP n n
=
= +
∑
(142)
To find the sum of the reciprocals of all centered pentagonal numbers we
follow as in (71), to obtain
( )
1
12 3
tanh 1.36061317.
25
15
kk
cP
∞
=
ππ
= ≈
∑
(143)
16. Centered Heptagonal Numbers (cHEP)n
These numbers are defined by the sequence
1,8,22, 43,71,106,148,197,253,
,
i.e.
, beginning with 8 each number is formed from the previous one in the se-
quence by adding the next number in the related sequence
( )
7,14,21, 28, ,7 1n−
. Thus,
817= +
,
22 1 7 14 8 14=++ =+
,
43 1 7 14 21 22 21=++ + = +
, and so on (see Figure 20).
Hence,
n
-th centered heptagonal number is defined as
( ) ( ) ( ) ( )
( )
( )
1
7 1 1 7 14 21 7 7
1 71 2 3 1 .
nn
cHEP cHEP n n
n
−
= + − =++ + + + −
=+ +++ + −
(144)
Thus, from (2) it follows that
( ) ( )
1 12
1
17 17 5 .
2
n nn n
n
nn
cHEP t t t t
− −−
−
=+ =+ =++
(145)
Let
m
be a given natural number, then it is
n
-th centered heptagonal number,
i.e.
,
( )
n
m cHEP=
if and only if
( )
()
7 49 56 1 14nm
=++ −
.
From (10) and (145), we have
( )
( )
2
234
3
51
8 22 43
1
xx x xx x x
x
++
=++ + +
−
and hence
( )
( )
3
25 11xx x x−
++ −
is the
generating
function
of all centered
heptagonal numbers.
Figure 20. Centered heptagonal numbers.
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Journal of Applied Mathematics and Physics
In view of (15) and (145), we have
( )
( )
2
1
1
7 1.
6
n
k
kcHEP n n
=
= −
∑
(146)
To find the sum of the reciprocals of all centered heptagonal numbers we
follow as in (71), to obtain
( )
1
12
tanh 1.264723172.
7 27
kk
cHEP
∞
=
π
= ≈
π
∑
(147)
17. Centered Octagonal Numbers (cO)n
These numbers are defined by the sequence
1,9,25,49,81,121,169,225, 289,361,
,
i.e.
, beginning with 9 each number is
formed from the previous one in the sequence by adding the next number in the
related sequence
( )
8,16,24,32, ,8 1n−
. Thus,
918= +
,
25 1 8 14 9 16=++ =+
,
49 1 8 16 24 25 24=++ + = +
, and so on (see Figure 21).
Hence,
n
-th centered octagonal number is defined as
( ) ( ) ( ) ( )
( )
( )
1
8 1 1 8 16 24 8 8
1 81 2 3 1 .
nn
cO cO n n
n
−
= + − =++ + + + −
=+ +++ + −
(148)
Thus, from (2) it follows that
( ) ( ) ( )
2
1 21 1 2
1
18 18 2 1 6 .
2
n n nnn
n
nn
cO t n S t t t
− − −−
−
=+ =+ = −= =+ +
(149)
Hence, the centered octagonal numbers are the same as the odd square num-
bers.
Let
m
be a given natural number, then it is
n
-th centered octagonal number,
i.e.
,
( )
n
m cO=
if and only if
( )
12nm= +
.
From (10) and (149), we have
( )
( )
2
234
3
61 9 25 49
1
xx x xx x x
x
++
=++ + +
−
and hence
( )
( )
3
26 11xx x x−
++ −
is the
generating
function
of all centered oc-
tagonal numbers.
Figure 21. Centered octagonal numbers.
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Journal of Applied Mathematics and Physics
In view of (15) and (149), we have
( )
( )
2
1
14 1.
3
n
k
kcO n n
=
= −
∑
(150)
To find the sum of the reciprocals of all centered octagonal numbers we use
(39), to obtain
( ) ( )
2
22 2
1 1 11
1 1 11 1 1.2337005501.
48
21
k k kk
k
cO kk
k
∞ ∞ ∞∞
= = = =
π
= =−=≈
−
∑ ∑ ∑∑
(151)
18. Centered Nonagonal Numbers (cN)n
These numbers are defined by the sequence
1,10,28,55,91,136,190, 253,325,
,
i.e.
, beginning with 10 each number is formed from the previous one in the se-
quence by adding the next number in the related sequence
( )
9,18,27,36, ,9 1n−
. Thus,
10 1 9= +
,
28 1 9 18 10 18=++ = +
,
55 1 9 18 27 28 27
=++ + = +
, and so on (see Figure 22).
Hence,
n
-th centered nonagonal number is defined as
( ) ( ) ( ) ( )
( )
( )
1
9 1 1 9 18 27 9 1
1 91 2 3 1 .
nn
cN cN n n
n
−
= +−=+++++−
=+ +++ + −
(152)
Thus, from (2) it follows that
( ) ( ) ( )( )
1
32 1 2
1 3 23 1
19 19
22
7.
n
n
n nnn
nn n n
cN t
t ttt
−
− −−
− −−
=+ =+=
==++
(153)
In 1850, Frederick Pollock (1783-1870, England) conjectured that every nat-
ural number is the sum of at most eleven centered nonagonal numbers,
which has not been proved.
Let
m
be a given natural number, then it is
n
-th centered nonagonal number,
i.e.
,
( )
n
m cN=
if and only if
()
()
9 81 72 1 18nm
=++ −
.
From (10) and (153), we have
( )
( )
2
234
3
71 10 28 55
1
xx x xx x x
x
++
=++++
−
Figure 22. Centered nonagonal numbers.
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Journal of Applied Mathematics and Physics
and hence
( )
( )
3
27 11xx x x−
++ −
is the
generating
function
of all centered
nonagonal numbers.
In view of (15) and (153), we have
( )
( )
2
1
13 1.
2
n
k
k
cN n n
=
= −
∑
(154)
To find the sum of the reciprocals of all centered heptagonal numbers we
follow as in (71), to obtain
( )
1
1 2 23
tan 1.2091995762.
369
kk
cN
∞
=
= = ≈
ππ π
∑
(155)
19. Centered Decagonal Numbers (cD)n
These numbers are defined by the sequence
1,11,31,61,101,151,211,281,
,
i.e.
, beginning with 11 each number is formed from the previous one in the se-
quence by adding the next number in the related sequence
( )
10,20,30, 40, ,10 1n−
. Thus,
11 1 10= +
,
31 1 10 20 11 20=++ =+
,
61 1 10 20 30 31 30=++ + =+
, and so on (see Figure 23).
Hence,
n
-th centered decagonal number is defined as
( ) ( ) ( ) ( )
( )
( )
1
10 1 1 10 20 30 10 1
1 10 1 2 3 1 .
nn
cD cD n n
n
−
= +−=+++++−
=+ +++ + −
(156)
Thus, from (2) it follows that
( ) ( )
2
1 12
1
1 10 1 10 5 5 1 8 .
2
n nn n
n
nn
cD t n n t t t
− −−
−
=+ =+ = − += + +
(157)
For each
( )
n
cD
the last digit is 1.
Let
m
be a given natural number, then it is
n
-th centered decagonal number,
i.e.
,
()
n
m cD=
if and only if
()
( )
5 25 20 1 10nm=++ −
.
From (10) and (157), we have
( )
( )
2
234
3
81 11 31 61
1
xx x xx x x
x
++=++++
−
Figure 23. Centered decagonal numbers.
R. P. Agarwal
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Journal of Applied Mathematics and Physics
and hence
( )
( )
3
28 11xx x x−
++ −
is the
generating
function
of all centered de-
cagonal numbers.
In view of (15) and (157), we have
( )
( )
2
1
15 2.
3
n
k
kcD n n
=
= −
∑
(158)
To find the sum of the reciprocals of all centered heptagonal numbers we
follow as in (71), to obtain
( )
1
1tan 1.189356247.
5 25
kk
cD
∞
=
= ≈
ππ
∑
(159)
20. Star Numbers (ST)n
These numbers are defined by the sequence
1,13,37,73,121,181,253,337,
,
i.e.
, beginning with 13 each number is formed from the previous one in the se-
quence by adding the next number in the related sequence
()
12,24,36, 48, ,12 1n−
. Thus,
13 1 12= +
,
37 1 12 24 13 24=++ = +
,
73 1 12 24 36 37 36=++ + = +
, and so on (see Figure 24).
Hence,
n
-th star number is defined as
( ) ( ) ( ) ( )
( )
()
1
12 1 1 12 24 36 12 1
1 12 1 2 3 1 .
nn
ST ST n n
n
−
= +−=+++++−
=+ +++ + −
(160)
Thus, from (2) it follows that
( ) ( )
2
1 12
1
1 12 1 12 6 6 1 10 .
2
n n nn
n
nn
ST t n n t t t
− −−
−
=+ =+ = − += + +
(161)
All star numbers are odd. The star number
()
77
35113ST =
is unique, since
its prime factors 13, 37, 73 are also consecutive star numbers. There are infi-
nite number of star numbers which are also triangular numbers, also square
numbers.
Let
m
be a given natural number, then it is
n
-th star number,
i.e.
,
( )
n
m ST=
if and only if
( )
( )
3 96 1 6nm=++ −
.
Figure 24. Star number (
ST
)3.
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DOI:
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Journal of Applied Mathematics and Physics
From (10) and (161), we have
( )
( )
2
234
3
10 1 13 37 73
1
xx x xx x x
x
++
=++ + +
−
and hence
( )
( )
3
2
10 1 1xx x x
−
++−
is the
generating
function
of all star num-
bers.
In view of (15) and (161), we have
( )
( )
2
1
2 1.
n
k
k
ST n n
=
= −
∑
(162)
To find the sum of the reciprocals of all star numbers we follow as in (71), to
obtain
( )
1
1tan 1.15917332.
23 23
kk
ST
∞
=
ππ
= ≈
∑
(163)
21. Centered Tetrakaidecagonal Numbers (cTET)n
These numbers are defined by the sequence
1,15,43,85,141,
,
i.e.
, beginning
with 15 each number is formed from the previous one in the sequence by adding
the next number in the related sequence
( )
14,28,42,56, ,14 1n−
. Thus,
15 1 14= +
,
43 1 14 28 15 28
=++ = +
,
85 1 14 28 42 43 42=+++ = +
, and so on
(see Figure 25).
Hence,
n
-th centered tetrakaidecagonal number is defined as
( ) ( ) ( ) ( )
( )
( )
1
14 1 1 14 28 42 14 1
1 14 1 2 3 1 .
nn
cTET cTET n n
n
−
= +−=+++++−
=+ +++ + −
(164)
Thus, from (2) it follows that
( ) ( )
2
1 12
1
1 14 1 14 7 7 1 12 .
2
n n nn
n
nn
cTET t n n t t t
− −−
−
=+ =+ = − += + +
(165)
Each
( )
n
cTET
is odd.
Let
m
be a given natural number, then it is
n
-th centered tetrakaidecagonal
number,
i.e.
,
( )
n
m cTET
=
if and only if
( )
( )
7 49 28 1 14
nm=++ −
.
From (10) and (165), we have
Figure 25. Centered tetrakaidecagonal numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
( )
( )
2
234
3
12 1 15 43 85
1
xx x xx x x
x
++
=++++
−
and hence
( )
( )
3
2
12 1 1xx x x
−
++−
is the
generating
function
of all centered te-
trakaidecagonal numbers.
In view of (15) and (165), we have
( )
( )
2
1
1
7 4.
3
n
k
k
cTET n n
=
= −
∑
(166)
To find the sum of the reciprocals of all centered tetrakaidecagonal numbers
we follow as in (71), to obtain
( )
1
13
tan 1.1372969963.
27
21
kk
cTET
∞
=
= ≈
ππ
∑
(167)
22. Cubic Numbers Cn
A cubic number can be written as a product of three equal factors of natural
numbers. Thus,
3
1,8,27,64, ,n
are first few cubic numbers (see Figure 26).
Last digit of a number is the same as the last digit of its cube, except that 2
becomes 8 (and 8 becomes 2) and 3 becomes 7 (and 7 becomes 3).
Nicomachus considered the following infinite triangle of odd numbers. It is
clear that the sum of the numbers in the
n
th row is
n
3.
3
3
3
3
3
3
3
3
3
1 1
3 5 2
7 9 11 3
13 15 17 19 4
21 23 25 27 29 5
31 33 35 37 39 41 6
43 45 47 49 51 53 55 7
57 59 61 63 65 67 69 71 8
73 75 77 79 81 83 85 87 89 9
91 93 95 97 99 101 103 105 107 10 3
9 10
⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
Figure 26. Cubic numbers.
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DOI:
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Journal of Applied Mathematics and Physics
In the literature often the above representation is referred to as Pascal’s trian-
gle. Now noting that numbers in each row are odd, so the general term in view
of (31) can be written as
( ) ( ) () ( )()
() ( )
322 3
11 13 15 1 21
1 135 2 1 .
kk kk kk kk k
k kk k k k k k
−++ −+ + −+ + + −+ −
=× − + +++ + − = − + =
Taking successively
1, 2, 3, ,kn=
in the above relation, adding these
n
eq-
uations, and observing that
( ) ( ) ( )
1 2 1 2 121nn n nn−+ −= + −
, we find
( )
333 3
1
1357911 2 1 1 2 3 ,
2
nn n
+
+++++++ −=++++
i.e.
, the number of terms in the left side are
( )
12nn+
. Now again from (31) we
find that the left hand side is the same as
( )
2
12nn+
, and hence it follows
that
( )
2
333 3 4322
1111
123 ,
2 424
n
nn
n n n nt
+
++++ = = + + =
(168)
i.e.
, a perfect square number. This identity is sometimes called Nicomachus’s
theorem.
From (69), the relation (168) follows immediately. In fact, we have
( )
3 22 2
1
11
.
nn
kk n
kk
k tt t
−
= =
= −=
∑∑
The
generating
function
for all cubic numbers is
( )
( )
2
23
4
41 8 27 .
1
xx x xx x
x
++
=++ +
−
From the relations
()
22
1 21
k kk+−=+
and
( ) ( )
22
1 14kk k+ −− =
it fol-
lows that all odd and multiple of 4 integers can be expressed as the difference of
two squares. However,
42k+
cannot be expressed as the difference of two
squares. Indeed, if
( )( )
22 42a b abab k− =+ −= +
, then letting
, x a by a b=+=−
gives
( ) ( )
2, 2a xy b xy=+=−
, which implies that
both
x
and
y
must be of the same parity. If both
x
and
y
are odd (even) then
xy
is odd (multiple of 4), hence in either case we have a contradiction. Now
since cube of any odd (even) integer is odd (multiple of 4), we can conclude
that every cube is a difference of two squares. Clearly, in the conclusion cube
can be replaced by any power greater than 3. For example,
3 22
3 14 13= −
,
3 22
4 17 15= −
,
7 22
5 39063 39062= −
.
There are infinite number of
square
cubic
numbers
, in fact,
( ) ( )
32
23
, 1, 2,k kk= =
.
The well known
Riemann
zeta
function
after George Friedrich Bernhard
Riemann (1826-1866, Germany) is defined as
( )
11
1 , .
23
ss
s s it
ζσ
=+++ =+
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
In 1979, Roger Apéry (1916-1994, Greek-French) published an unexpected
proof of the irrationality of
( )
3
ζ
. In the literature
( )
3
ζ
is known as Apéry
constant. In July 2020, Seungmin Kim (Korea) has computed the value of
()
3
ζ
to one trillion two hundred billion and one hundred decimal places.
Related with cubic numbers there are
centered
cubic
numbers
( ) ( ) ( )
( )
3
32
1 21 1
n
cC n n n n n= + − = − −+
. Thus,
()
n
cC
is the count of
number of points in a body-centered cubic pattern within a cube that has
1n+
points along each of its edges. First few centered cubic numbers are 1,
9, 35, 91, 189, 341. Clearly, no centered cubic number is prime. Further, the
only centered cube number that is also a square number is 9. The
generating
function
for all centered cube numbers is
( )
( )
32
234
4
5 51
9 35 91 .
1
xx x x xx x x
x
+ ++
=++ + +
−
From (168) it follows that
()
()
22
1
1
1.
2
n
k
k
cC n n
=
= +
∑
(169)
Further, from (39) and (70), we find
( )
( )
22
22
11 1
2
1 2 11
21
1
1 coth 1.1365200388.
3
kk k
k
cC kk
kk
∞∞ ∞
= = =
= = −
+
+
=ππ+− ≈π
∑∑ ∑
(170)
23. Tetrahedral Numbers (Triangular Pyramidal Numbers)
Tn
These numbers count the number of dots in pyramids built up of triangular
numbers. If the base is the triangle of side
n
, then the pyramid is formed by
placing similarly situated triangles upon it, each of which has one less in its sides
than that which precedes it (see Figure 27).
In general, the
n
th tetrahedral number
Tn
is given in terms of the sum of the
first
n
triangular numbers,
i.e.
,
1 123
,
nn n n
TT t ttt t
−
= + =+++ +
which in view of (15) is the same as
Figure 27. Tetrahedral numbers.
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
( )( )
12 2.
63
nn
nn n n
Tt
++ +
= =
(171)
Thus, first few tetrahedral numbers are 1, 4, 10, 20, 35, 56, 84, 120, 165
.
In 1850, Pollock conjectured that every natural number is the sum of at most
five tetrahedral numbers, which has not been proved. Tetrahedral numbers
are even, except for
41
, 0,1,2,
n
Tn
+
=
, which are odd, see Conway and Guy
[30]. The only numbers which are simultaneously square and tetrahedral are
12
1, 4TT= =
, and
48
19600T=
, see Meyl [31].
The
generating
function
for all tetrahedral numbers is
( )
23 4
4
4 10 20 .
1
xxx x x
x=++ + +
−
From (2), (11), and (168) it follows that
( ) ( )
32 32
11
11
3 2 6 11 6 .
6 24
nn
k
kk
T k k k nn n n
= =
= + + = + ++
∑∑
(172)
To find the sum of the reciprocals of all tetrahedral numbers we follow as in
(16) and (17), to obtain
( )( )
1 11
6 11 1 1
lim 3 3
12 1 1 2
1 11 3
lim 3 1 3 .
1 2 22
nn
n
k kk
n
kk k k k k k
nn
∞
→∞
= = =
→∞
= −− −
++ + + +
= − −− =
++
∑ ∑∑
(173)
As in (173), we find
( )
( )( ) ( ) ( )
( ) ( )
( )
111
11 1
11
23
61 11 1 1
31 31
12 1 12
11
1
31 2 3 2
2
1
3 1 2 ln 2 2 3 2 ln 2 1
2
15
12ln 2 0.8177661667.
2
kkk
kk k
kk
kk
kk k k k k k
kk
−
∞∞ ∞
−−
= = =
−−
∞∞
= =
−
=− − −− −
++ + + +
−−
=+ −−
= + −− − +
= −≈
∑∑ ∑
∑∑
(174)
The numbers
( )
2
32 1
4 23 27 10 6
nn
T T nn n
−−
− = −+
are called
truncated
te-
trahedral
numbers
and denoted as
( )
n
TT
. These numbers are assembled by
removing the
( )
1n−
th tetrahedral number from each of the four corners
from the
( )
32n−
th tetrahedral number. First few of these numbers are 1, 16,
68, 180, 375, 676, 1106
.
The
generating
function
for all truncated tetrahedral
numbers is
( )
( )
2
23 4
4
10 12 1 16 68 180 .
1
xx x xx x x
x
++
=+++ +
−
24. Square Pyramidal Numbers (SP)n
These numbers count the number of dots in pyramids built up of square num-
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
bers. First few square pyramidal numbers are 1, 5, 14, 30, 55, 91, 140
.
In general,
the
n
th square pyramidal number
( )
n
SP
is given in terms of the sum of the
first
n
square numbers,
i.e.
,
()()
222 2
1
12 ,
nn
SP SP n n
−
= + =+ ++
which in view of (11) and (171) is the same as
( ) ( ) ( ) ( )( ) ( )
22
121 22221 11
1.
6 46 4 6
nn
n
nn n n n n
SP T n t
++ + +
= = = = +
×
(175)
In 1918, George Neville Watson (1886-1965, England) proved that besides 1,
there is only one other number that is both a square and a pyramid number,
4900, (as conjectured by Lucas in 1875), the 70th square number and the 24th
square pyramidal number,
i.e.
,
( )
70 24
S SP=
.
The
generating
function
for all square pyramidal numbers is
()
( )
234
4
15 14 30 .
1
xx xx x x
x
+=++ + +
−
From (2), (11), and (168) it follows that
( )
( )
( ) ( )
2
32
11
11
2 3 1 2.
6 12
nn
k
kk
SP k k k n n n
= =
= + += + +
∑∑
(176)
To find the sum of the reciprocals of all square pyramidal numbers, we note
that
( )( ) ( )
( )
( )
( )
( )
( )
11
121 21 2
0
1
1
122
00
2
12
2
0
12
0
6 11 2
12
121 2 2 121
12 2 d
12 1 d
1
12 1 d
1
1
12 1 d .
1
nn
kk
nkk k
k
nk
k
n
n
kk k k k k
x x xx
xx xx
x
xx x
x
xx xx
x
= =
−+
=
−
=
= +−
++ + +
= +−
= −
−
= − −
−
= −
+
∑∑
∑∫
∑
∫
∫
∫
Now since
( )
11
22
00
11
d d 0 as
1 21
nn
xx
xx xx n
xn
−< = → →∞
++
∫∫
it follows that
( ) ( )( )
( )
()
1
0
11
1
0
1
16
lim 12 d
12 1 1
2
12 2 d 6 3 4 ln 2
1
1.364467667.
n
n
kk
k
xx
x
SP k k k x
xx
x
∞
→∞
= =
−
= =
++ +
= −− = −
+
≈
∑∑ ∫
∫
(177)
In 2006, Fearnehough [32] used Madhava of Sangamagramma (1340-1425,
India) series
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2100
Journal of Applied Mathematics and Physics
1111 1
1,
4 3 5 7 9 11
=−+−+− +
π
known in the literature after James Gregory (1638-1675, England) and Gottfried
Wilhelm von Leibniz (1646-1716, Germany) in the above partial fractions, to
obtain
( )
( )( )
( )
1
1
61
12 1
114 114 114 114
6123 235 347 459
1111
61 4 3579
614 1 6 3.
4
k
k
kk k
−
∞
=
−
++
= +−−+−++−−+−+
= + −+− +−
= + −= −
ππ
∑
(178)
The sum of two consecutive square pyramidal numbers,
i.e.
,
( ) ( )
( )
2
1
2 1 3, 1
nn
SP SP n n n
−
+ =+≥
are called
octahedral
numbers
, and
denoted as
( )
n
OH
. The first few octahedral numbers are 1, 6, 19, 44, 85, 146,
231, 344
.
These numbers represent the number of spheres in an octahedral
formed from close-packed spheres. Descartes initiated the study of octahe-
dral numbers around 1630. In 1850, Pollock conjectured that every positive
integer is the sum of at most 7 octahedral numbers, which for finitely many
numbers have been proved by Brady [33]. The difference between two con-
secutive octahedral numbers is a centered square number,
i.e.
,
( ) ( ) ( ) ( )
2
2
1
1
nn n
OH OH n n cS
−
− =+− =
. The
generating
function
for all oc-
tahedral numbers is
( )
( )
2
23 4
4
16 19 44 .
1
xx xx x x
x
+=++ + +
−
From (2) and (168) it follows that
( ) ( )
( )
2
1
11 1.
6
n
k
k
OH n n n n
=
= + ++
∑
(179)
Further, as in (100), we have
( )
( ) ( )
1
13 1 1
2 2 2 2 2 1.2781850979.
22 2
kk
ii
OH
γ
∞
=
= +Ψ − +Ψ + ≈
∑
(180)
The sum of two consecutive octahedral numbers,
i.e.
,
( ) ( ) ( )
( )
2
1
2 12 2 33
nn
OH OH n n n
−
+ = − −+
is called
centered
octahedral
number
or
Haüy
octahedral
numbers
(named after René Just Haüy, 1743-1822,
France) and denoted as
( )
n
cOH
. The first few centered octahedral numbers
are 1, 7, 25, 63, 129, 231, 377
.
The
generating
function
for all centered octahe-
dral numbers is
( )
( )
3
234
4
17 25 63 .
1
xx xx x x
x
+=++ + +
−
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
As earlier, we have
( )
( )
22
1
12.
3
n
k
k
cOH n n
=
= +
∑
(181)
From the definitions of
( )
n
cC
and
( )
n
SP
it follows that
( ) ( ) ( )
( )
2
1
6 2 12 2 1
nn
cC SP n n n
−
+ = − −+
. These numbers are called
Haüy
rhombic
dodecahedral
numbers
. First few of these numbers are 1, 15, 65, 175,
369, 671
.
These numbers are constructed as a centered cube with a square
pyramid appended to each face. The
generating
function
for all Haüy rhom-
bic dodecahedral numbers is
( )
( )
( )
2
23 4
4
1 10 1 15 65 175 .
1
xx x x xx x x
x
+ ++
=+++ +
−
Haüy also gave construction of another set of numbers involving cubes and
odd square numbers, namely,
( ) ( )
( ) ( )( )( )
( )
( )
32
22
3
2
2 1 61 3 2 3
2 1 2 12 12 3
2 1 8 14 7 .
nn
n nnn
n nn
− + +++ −
=−+−−−
=− −+
First few of these numbers are 1, 33, 185, 553, 1233
.
These numbers are called
Haüy
rhombic
dodecahedron
numbers
. The
generating
function
for all of these
numbers is
( )
( )
23
234
4
1 29 59 13 33 185 553 .
1
x xx xxx x x
x
++ − =++ + +
−
From the definitions of
( )
n
OH
and
()
n
SP
it follows that
( ) ( )
32
32 1
6 16 33 24 6
nn
OH SP n n n
−−
− = − +−
. These numbers are called
truncated
octahedral
numbers
. First few of these numbers are 1, 38, 201, 586,
1289, 2406
.
These numbers are obtained by truncating all six vertices of oc-
tahedron. The
generating
function
for all truncated octahedral numbers is
( )
( )
32
234
4
6 55 34 1 38 201 586 .
1
xx x x xx x x
x
+ ++
=++ + +
−
25. Pentagonal Pyramidal Numbers (PP)n
These numbers count the number of dots in pyramids built up of pentagonal
numbers. First few pentagonal pyramidal numbers are 1, 6, 18, 40, 75, 126, 196,
288
.
In general, the
n
th pentagonal pyramidal number
( )
n
PP
is given in terms
of the sum of the first
n
pentagonal numbers,
i.e.
,
( ) ( ) ( ) ( )
1
3 1 1 5 12 22 35 3 1 ,
22
nn
nn
PP PP n n
−
= +−=++++++−
which in view of (54) is the same as
( ) ( )
2
11.
2
n
n
PP n n nt= +=
(182)
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2102
Journal of Applied Mathematics and Physics
The
generating
function
for all pentagonal pyramidal numbers is
( )
( )
23 4
4
21 6 18 40 .
1
xx xx x x
x
+=++ + +
−
From (11) and (168) it follows that
( )
( )
( )
( )
32 2
11
11
1 3 7 2.
2 24
nn
k
kk
PP k k n n n n
= =
= + = + ++
∑∑
(183)
To find the sum of the reciprocals of all pentagonal pyramidal numbers, from
(16), (17), and (39), we have
( ) ( )
2
22
1 1 11
1 2 1 11
2 2 2.
13
1
k k kk
k
PP k k
kk k
∞ ∞ ∞∞
= = = =
= = − −=−
+
π
+
∑ ∑ ∑∑
(184)
Similar to (174) it follows that
( )
( )
( )
( )
11
2
2
11
1 21 2 4ln2 0.8723453446.
6
1
kk
kk
k
PP kk
−−
∞∞
= =
−−
= = +− ≈
+
π
∑∑
(185)
26. Hexagonal Pyramidal Numbers (HP)n
These numbers count the number of dots in pyramids built up of hexagonal
numbers. First few pentagonal pyramidal numbers are 1, 7, 22, 50, 95, 161, 252,
372
.
In general, the
n
th hexagonal pyramidal number
( )
n
HP
is given in terms
of the sum of the first
n
hexagonal numbers,
i.e.
,
()() ( ) ( )
1
2 1 1 6 15 28 45 2 1 ,
nn
HP HP n n n n
−
= + −=++++++ −
which in view of (61) is the same as
( ) ( )( ) ( )
11
141 41.
63
n
n
HP n n n n t= + −= −
(186)
The
generating
function
for all hexagonal pyramidal numbers is
( )
( )
234
4
31
7 22 50 .
1
xx xx x x
x
+=++ + +
−
From (11) and (168) it follows that
()
()
()( )
32 2
11
11
4 3 1 2.
66
nn
k
kk
HP k k k n n n
= =
= + −= + +
∑∑
(187)
To find the sum of the reciprocals of all hexagonal pyramidal numbers, we
follow as in (177) and (184), to obtain
()( ) ()
()
11
1 1 1 16
65 1 54 1
612ln 2 2 1 1.2414970314.
5
kk
k
HP k k k
∞∞
= =
= −+ +
+−
−≈π
= −
∑∑
(188)
Similarly, we have
( )
( ) ( ) ( ) ( )
11
11
11 1 16
61 5 1 54 1
63
1 4 1,1, 6ln 2 0.8892970462.
54
kk
kk
k
HP k k k
−
∞∞
−
= =
−= − −+ +
+−
= +Φ− − ≈
∑∑
(189)
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DOI:
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Journal of Applied Mathematics and Physics
In (189), the function
Φ
is the Dirichlet beta function.
27. Generalized Pentagonal Pyramidal Numbers (GPP)n
These numbers count the number of dots in pyramids built up of generalized pen-
tagonal numbers. First few generalized pentagonal pyramidal numbers are 1, 8, 27,
64, 125, 216, 343, 512
.
In general, the
n
th generalized pentagonal pyramidal num-
ber
( )
n
GPP
is given in terms of the sum of the first
n
generalized pentagonal
numbers,
i.e.
,
( ) ( ) ( ) ( )
1
1 3 1 1 7 19 37 1 3 1 ,
nn
GPP GPP n n n n
−
= ++ − =++ + + ++ −
which in view of (68) is the same as
( )
3
.
n
GPP n=
(190)
Thus, generalized pentagonal pyramidal numbers are the same as cubic num-
bers.
28. Heptagonal Pyramidal Numbers (HEPP)n
These numbers count the number of dots in pyramids built up of heptagonal
numbers. First few heptagonal pyramidal numbers are 1, 8, 26, 60, 115, 196, 308,
456
.
In general, the
n
th heptagonal pyramidal number
( )
n
HEPP
is given in
terms of the sum of the first
n
heptagonal numbers,
i.e.
,
( ) ( ) ( ) ( )
15 3 1 7 18 34 5 3 ,
22
nn
nn
HEPP HEPP n n
−
= +−=+++++−
which in view of (78) is the same as
( ) ( )( ) ( )
11
152 52.
63
n
n
HEPP n n n t n= + −= −
(191)
The
generating
function
for all heptagonal pyramidal numbers is
( )
( )
234
4
41 8 26 60 .
1
xx xx x x
x
+=++ + +
−
From (2), (11), and (168) it follows that
( ) ( ) ( )( )
1
11 2 5 1.
24
n
k
k
HEPP n n n n
=
= ++ −
∑
(192)
The sum of reciprocals of all heptagonal pyramidal numbers appears as
( )
( )
( )
1
1 15 4 2 5 5 5
2ln 5 1 5 1 ln
14 5 5 2
55
5 1 ln 2
1.2072933193.
kk
HEPP
∞
=
−
= −− − + +
+
−−
π
≈
∑
(193)
Similarly, as in (189), we have
( )
( )
1
1
133
2 5 1,1, 9ln 2 0.9023419344.
75
k
kk
HEPP
−
∞
=
−
= +Φ− − ≈
∑
(194)
R. P. Agarwal
DOI:
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Journal of Applied Mathematics and Physics
29. Octagonal Pyramidal Numbers (OP)n
These numbers count the number of dots in pyramids built up of octagonal
numbers. First few octagonal pyramidal numbers are 1, 9, 30, 70, 135, 231, 364,
540
.
In general, the
n
th octagonal pyramidal number
( )
n
OP
is given in terms
of the sum of the first
n
octagonal numbers,
i.e.
,
( ) () ( )( )
1
3 2 1 8 21 40 3 2 ,
nn
OP OP n n n n
−
= +−=+++++−
which in view of (88) is the same as
( ) ( )( ) ( )
1121 21.
2
n
n
OP n n n t n= + −= −
(195)
The
generating
function
for all octagonal pyramidal numbers is
( )
()
234
4
51 9 30 70 .
1
xx xx x x
x
+=++ + +
−
From (2), (11), and (168) it follows that
( ) ( )( )( )
1
11 2 3 1.
12
n
k
k
OP n n n n
=
= ++ −
∑
(196)
The sum of reciprocals of all heptagonal pyramidal numbers appears as
( ) ()
1
12
4ln 2 1 1.1817258148.
3
kk
OP
∞
=
= −≈
∑
(197)
Similarly, we find
( )
( ) ( )
1
1
121 4ln 2 0.9126692876.
3
k
kk
OP
−
∞
=
−= +− ≈π
∑
(198)
30. Nonagonal Pyramidal Numbers (NP)n
These numbers count the number of dots in pyramids built up of nonagonal
numbers. First few nonagonal pyramidal numbers are 1, 10, 34, 80, 155, 266, 420,
624
.
In general, the
n
th nonagonal pyramidal number
( )
n
NP
is given in terms
of the sum of the first
n
nonagonal numbers,
i.e.
,
( ) ( ) ( )( )
17 5 1 9 24 46 7 5 ,
22
nn
nn
NP NP n n
−
= + −=+++++ −
which in view of (99) is the same as
( ) ( )( ) ( )
11
174 74.
63
n
n
NP n n n t n= + −= −
(199)
The
generating
function
for all nonagonal pyramidal numbers is
( )
(
)
234
4
61 10 34 80 .
1
xx xx x x
x
+=++++
−
From (2), (11), and (168) it follows that
( ) ( )( )( )
1
11 2 7 3.
24
n
k
k
NP n n n n
=
= ++ −
∑
(200)
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The sum of reciprocals of all heptagonal pyramidal numbers appears as
( )
1
13 4
33 28 28 1.6184840638.
88 7
kk
NP
γ
∞
=
= − Ψ− − ≈
∑
(201)
Similarly, we find
()
()
1
1
133
4 7 1,1, 15ln 2 0.9210386965.
22 7
k
kk
NP
−
∞
=
−
= +Φ− − ≈
∑
(202)
31. Decagonal Pyramidal Numbers (DP)n
These numbers count the number of dots in pyramids built up of decagonal
numbers. First few decagonal pyramidal numbers are 1, 11, 38, 90, 175, 301, 476,
708
.
In general, the
n-
th decagonal pyramidal number (
DP
)
n
is given in terms of
the sum of the first
n
decagonal numbers,
i.e.
,
( ) ( ) ( ) ( )
1
4 3 1 10 27 52 4 3 ,
nn
DP DP n n n n
−
= +−=+++++−
which in view of (111) is the same as
( ) ( )( ) ( )
11
185 85.
63
n
n
DP n n n t n= + −= −
(203)
The
generating
function
for all decagonal pyramidal numbers is
( )
( )
234
4
71 11 38 90 .
1
xx xx x x
x
+=++ + +
−
From (2), (11), and (168) it follows that
( ) ( )()( )
1
11 2 2 1.
6
n
k
k
DP n n n n
=
= ++ −
∑
(204)
The sum of reciprocals of all decagonal pyramidal numbers appears as
( )
1
16 5
39 40 40 1.1459323453.
325 8
kk
DP
γ
∞
=
= − Ψ− − ≈
∑
(205)
Similarly, we find
( )
( )
1
1
163
5 8 1,1, 18ln 2 0.9279541642.
65 8
k
kk
DP
−
∞
=
−
= +Φ− − ≈
∑
(206)
32. Tetrakaidecagonal Pyramidal Numbers (TETP)n
These numbers count the number of dots in pyramids built up of tetrakaideca-
gonal numbers. First few tetrakaidecagonal pyramidal numbers are 1, 15, 54, 130,
255, 441, 700, 1044, 1485
.
In general, the
n-
th tetrakaidecagonal pyramidal
number
( )
n
TETP
is given in terms of the sum of the first
n
tetrakaidecagonal
numbers,
i.e.
,
( ) ( ) ( ) ( )
16 5 1 14 39 76 6 5 ,
nn
TETP TETP n n n n
−
= +−=+++++−
which in view of (122) is the same as
( ) ( ) ( ) ( )
1143 43.
2
n
n
TETP n n n t n= + −= −
(207)
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The
generating
function
for all tetrakaidecagonal pyramidal numbers is
( )
()
23 4
4
11 1 15 54 130 .
1
xx xx x x
x
+=+++ +
−
From (2), (11), and (168) it follows that
()( )()()
1
11 2 3 2.
6
n
k
k
TETP n n n n
=
= ++ −
∑
(208)
The sum of reciprocals of all tetrakaidecagonal pyramidal numbers appears
as
( ) ( )
1
12
2 12ln2 3 1.1048525213.
21
kk
TETP
∞
=
+ −≈π=
∑
(209)
Similarly, we find
( )
( )
1
1
121
3 4 1,1, 10ln 2 0.9466758087.
21 4
k
kk
TETP
−
∞
=
−
= +Φ− − ≈
∑
(210)
33. Stella Octangula Numbers (SO)n
The word octangula for eight-pointed star was given by Johannes Kepler
(1571-1630, Germany) in 1609. Stella octangula numbers count the number of
dots in pyramids built up of star numbers. These numbers also arise in a para-
metric family of instances to the crossed ladders problem in which the lengths
and heights of the ladders and the height of their crossing point are all integers.
The ratio between the heights of the two ladders is a stella octangula number.
First few stella octangula numbers are 1, 14, 51, 124, 245, 426, 679, 1016, 1449
.
In
general, the
n
th stella octangula number
( )
n
SO
is given in terms of the sum of
the first
n
star numbers,
i.e.
,
( ) ( ) ( ) ( )
1
1 6 1 1 13 37 73 1 6 1 ,
nn
SO SO n n n n
−
= ++ − =++ ++ ++ −
which in view of (162) is the same as
( )
( )
( )
21
2 1 8.
n
nn
SO n n OH T
−
= −= +
(211)
The only known square stella octangula numbers are 1 and
( )
2169
9653449 3107 SO= =
, see Conway and Guy [30].
The
generating
function
for all stella octangula numbers is
( )
( )
2
23 4
4
10 1 14 51 1124 .
1
xx x xx x x
x
++
=+++ +
−
From (2) and (168) it follows that
( ) ( )
( )
2
1
11 1.
2
n
k
k
SO n n n n
=
= + +−
∑
(212)
The sum of reciprocals of all stella octangula numbers appears as
( )
1
11 1 1
2 1.1114472084.
222
kk
SO
γ
∞
=
=− +Ψ − +Ψ ≈
∑
(213)
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Journal of Applied Mathematics and Physics
Similarly, we find
()
( )
1
1
111 1
1,1, 1 1,1, 1 2 ln 2
222
0.942739143439.
k
kk
SO
−
∞
=
−
= Φ− + +Φ− − −
≈
∑
(214)
34. Biquadratic Numbers (BC)n
A biquadratic number can be written as a product of four equal factors of natu-
ral numbers. Thus, 1, 16, 81, 256, 625, 1296, 2401, 4096 are first few biquadratic
numbers.
Last digit of a biquadratic number can only be 0 (in fact 0000), 1, 5 (in fact
0625), or 6.
The
n
-th biquadratic number is the sum is the sum of the first
n
Haüy rhombic
dodecahedral numbers. Indeed, from (2), (11), and (168) it follows that
( )
()
24
1
2 12 2 1 .
n
k
k kk n
=
− − +=
∑
Fermat’s Last Theorem confirms that a fourth power cannot be the sum of
two other fourth powers, for details see Agarwal [14]. In 1770, Edward War-
ing (1736-1798, England) conjectured (known as Waring’s problem) that
every positive integer can be expressed as the sum of at most 19 fourth pow-
ers, and every integer larger than 13,792 can be expressed as the sum of at
most 16 fourth powers. In 1769, Euler conjectured that a fourth power can-
not be written as the sum of three fourth powers, but in 1988 [34], Noam
David Elkies (born 1966, USA) disproved Euler’s conjecture. Out of infinite
number of possible counterexamples the following of Elkies is notable
4 4 44
20615673 18796760 15365639 2682440 .=++
The
generating
function
for all biquadratic numbers is
( )
()
32
23 4
5
11 11 1 16 81 256 .
1
xx x x xx x x
x
+ ++
=+++ +
−
From (2), (11), (168), and the identity
( ) ( )
( )
55
5 4 32
1 11
1 1 1 5 10 10 5 1
n nn
k kk
n k k k k kk
= = =
+−= +− = + + ++
∑ ∑∑
it follows that
5432 44 32
1
5 25 31
5 10 10 5 5 10
23 6
n
k
nnnnn kn nn n
=
++++= ++ ++
∑
and hence
( )( )
( )
4 2 543
1
1 111 1
12 13 3 1 .
30 5 2 3 30
n
k
k nn n n n n n n n
=
= + + + −= + + −
∑
(215)
The following identity due to Abu-Ali al-Hassan ibn al-Hasan ibn al-Haitham
(965-1039, Iraq) combines the sum of numbers raised to the power of four
with different sums of numbers raised to the power of three
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( )
43 3
1 1 11
1.
n n nk
k k kj
k kn j
= = = =
= +−
∑ ∑ ∑∑
(216)
From (11), (15), (168), and (215) it follows that
( ) ( )
2 32 2
11
11
3 12 13 2 3 6 1 .
30 10
nn
kn k
kk
t tn n n t n n
= =
= + + += ++
∑∑
(217)
To find the sum of reciprocals of all biquadratic numbers, we shall use the
derivation of (39). First in the two expansions of
( )
sin xx
, we compare the
coefficients of
4
x
, to get
4
22
, IN
1.
5! pq
pq pq
∈
≠
π=∑
(218)
Now squaring (39), to obtain
4
4 22
1 , IN
11
2,
36
k pq
pq
k pq
∞
= ∈
≠
+π
=
∑∑
which in view of (218) gives
4 44
4
1
12.
36 5! 90
k
k
∞
=
= =
ππ
−π
∑
(219)
From (219) it immediately follows that
( )
14
4
1
17.
720
k
k
k
−
∞
=
=π
−
∑
(220)
35. Pentatope Numbers (PTOP)n
The fifth cell of any row of Pascal’s triangle starting with the 5-term row 1 4 6 4 1,
either from left to right or from right to left are defined as pentatope numbers.
First few these numbers are 1, 5, 15, 35, 70, 126, 210, 330, 495
.
Thus, the
n
-th
pentatope number is defined as
( ) ( )( )( ) ( )
2
31 11
1 2 3 3.
424 6 4
nn n
n
n
PTOP n n n n t t n T
+
+
= = + + += = +
(221)
These numbers can be represented as regular discrete geometric patterns, see
Deza [3]. In biochemistry, the pentatope numbers represent the number of
possible arrangements of
n
different polypeptide subunits in a tetrameric (tetra-
hedral) protein.
Two of every three pentatope numbers are also pentagonal numbers. In fact,
the following relations hold
( ) ( )
22
32 31
(3 ) 2 (3 ) 2
and .
nn
nn nn
PTOP P PTOP P
−−
−+
= =
(222)
The
generating
function
for all pentatope numbers is
( )
234
5
5 15 35 .
1
xxx x x
x=++ + +
−
From (2), (11), (168), and (215) it follows that
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Journal of Applied Mathematics and Physics
( ) ( )( )( )( ) ( )( )
1
11
1234 4 .
120 5
n
kn
k
PTOP n n n n n n PTOP
=
= + + + += +
∑
(223)
As in (17), we have
( ) ( )( ) ( )( )( )
11
1 8 84
.
12 1233
kk
k
PTOP kkk kk k
∞∞
= =
=−=
++ ++ +
∑∑
(224)
We also have
( )
( )
1
1
164
32ln 2 0.8473764446.
3
n
kk
PTOP
−
∞
=
−= −≈
∑
(225)
36. Partitions by Polygonal Numbers
Recall that
general
polygonal
number
can be written as
( ) ( )
2 42
r
n
p nr n r= − −−
, where
r
n
p
is the
n
th
r
-gonal number. For exam-
ple, for
3
r=
it gives triangular number, and for
4r=
gives a square number.
Fermat in 1638 claimed that every positive integer is expressible as at most
k
k
-gonal numbers (Fermat’s Polygonal Number Theorem). Fermat claimed to
have a proof of this result, however his proof has never been found. In 1750, Eu-
ler conjectured that every odd integer can be written as a sum of four squares in
such a way that
222 2
nabcd=+++
and
1abcd+++ =
. In 1770. Lagrange
proved that every positive integer can be represented as a sum of four squares,
known as
four-square
theorem
. For example, the number
22 2 2222 2
1638 4 6 25 31 1 1 6 40=++ + =+++
has several different patricians,
whereas for the number
2222
1536 0 16 16 32=+++
this is the only partition. In
1797-8, Legendre extended the theorem in with his
three-square
theorem
, by
proving that a positive integer can be expressed as the sum of three squares if
and only if it is not of the form
( )
48 7
k
m+
for integers
k
and
m
. Later, in 1834,
Jacobi gave a formula for the number of ways that a given positive integer
n
can
be represented as the sum of four squares. In 1796, Gauss proved the difficult
triangular case (every positive integer is the sum of three or fewer triangular
numbers, which is equivalent to the statement that every number of the form
83m+
is a sum of three odd squares, see Duke [35]), commemorating the occa-
sion by writing in his diary the line EΓPHKA! num
=∆+∆+∆
, and published
a proof in his book
Disquisitiones
Arithmeticae
of 1798. For this reason, Gauss’s
result is sometimes known as the Eureka theorem. For example,
16 6 10= +
,
25 1 3 21
=++
,
39 3 15 21=++
,
150 6 66 78=++
. The full polygonal number
theorem was resolved finally in 1813 by Cauchy. In 1872, Henri Léon Lebesgue
(1875-1941, France) proved that every positive integer is the sum of a square
number (possibly 02) and two triangular numbers, and every positive integer is
the sum of two square numbers and a triangular number. For further details, see
Grosswald [36], Ewell [37] [38], and Guy [39].
37. Conclusions
Triangular numbers which are believed to have been introduced by Pythagoras
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Journal of Applied Mathematics and Physics
himself play a dominant role in all types of figurative numbers we have ad-
dressed in this article. In fact, Equation (1) says natural number
n
is the differ-
ence of
n
t
and
1n
t
−
, whereas Gauss’s Eureka theorem stipulates that
n
can be
written as the sum of three triangular numbers. Equation (32) shows that square
number
n
S
is the sum of
n
t
and
1n
t
−
. Equation (44) shows that square num-
ber
n
R
is 2 times of
n
t
. Relation (51) says pentagonal number
n
P
is
( )
31
13 n
t−
, whereas (53) gives
1 21 1
2
nn n n n
Pt t t t
− −−
=+=−
. Equation (60) informs
that hexagonal number
n
H
is the same as
21n
t
−
. Relation (67) says generalized
pentagonal number
( )
n
GP
is the same as
1 1 12
64
n nnn
tt t t t
− −−
+=++
. Equation
(77) informs that heptagonal number
()
n
HEP
is the same as
1
4
nn
tt
−
+
. Relation
(87) declares that octagonal number
n
O
is equal to
1
5
nn
tt
−
+
. Equation (98) im-
plies that nonagonal number
n
N
is equal to
1
6
nn
tt
−
+
. Relation (110) says de-
cagonal number
n
D
is the same as
1
7
nn
tt
−
+
. Equation (121) informs that tetra-
kaidecagonal number
( )
n
TET
is the same as
1
11
nn
tt
−
+
. Relation (133) shows
that centered triangular number
( )
n
ct
is the same as
12nn n
tt t
−−
++
, whereas re-
lation (137) confirms that centered square number
()
n
cS
is equal to
12
2
nnn
ttt
−−
++
. Equation (141) says centered pentagonal number
( )
n
cP
is equal
to
12
3
nn n
ttt
−−
++
, whereas Equation (145) tells centered heptagonal number
( )
n
cHEP
is the same as
12
5
nn n
ttt
−−
++
. Relation (149) informs that centered
octagonal number
()
n
cO
is the same as
12
6
nnn
ttt
−−
++
, whereas (153) shows
centered nonagonal number
( )
n
cN
is the same as
12
7
nnn
ttt
−−
++
, and relation
(157) tells centered decagonal number
()
n
cD
is the same as
12
8
nn n
ttt
−−
++
.
Equations (161) shows that star number
()
n
ST
is the same as
12
10
n nn
t tt
−−
++
,
whereas Equation (165) shows that centered tetrakaidecagonal number
( )
n
cTET
is the same as
12
12
n nn
t tt
−−
++
. Relation (168) shows that the sum of
the first
n
cubic numbers is the same as
2
n
t
. Equation (171) shows that tetrahe-
dral number
n
T
is the same as
( )( )
13 2 n
nt+
. Relation (175) says square py-
ramidal number
()
n
SP
is the same as
( )( )
2
16 1
n
nt+
. From the definition of
octahedral numbers and (175) it follows that
() ( ) ( )
( )
22 2
16 1
nn
n
OH nt n t
−
= ++
. From the relation (182) it follows that penta-
gonal pyramidal number
( )
n
PP
is the same as
n
nt
. Equation (186) says hex-
agonal pyramidal number
( )
n
HP
is the same as
( )( )
13 4 1
n
nt−
. From Equation
(191) it follows that heptagonal pyramidal number
( )
n
HEPP
is the same as
()( )
13 5 2
n
nt−
. Equation (195) suggests that octagonal pyramidal number
( )
n
OP
is the same as
( )
21
n
nt−
. Relation (199) tells nonagonal pyramidal num-
ber
( )
n
NP
is the same as
( )( )
13 7 4
n
nt−
. Equation (203) informs that decagon-
al pyramidal number
( )
n
DP
is the same as
( )( )
13 8 5
n
nt−
, whereas relation
(207) indicates that tetrakaidecagonal pyramidal number
( )
n
TETP
is the same as
( )
43
n
nt−
. Thus, in conclusion almost all figurative numbers we have studied
are directly related with triangular numbers.
Acknowledgements
The author is grateful to Professor Chan Heng Huat, National University of
R. P. Agarwal
DOI:
10.4236/jamp.2021.98132 2111
Journal of Applied Mathematics and Physics
Singapore, for his generous guidance and encouragement in the preparation of
this article. He is also thankful to Professor Chao Wang and referees for their
comments.
Authors Contribution
Not applicable.
Funding
This work received no external funding.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
There is no conflict of interest.
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