Figurate numbers and sums of powers of integers

Abstract

Recently, Marko and Litvinov conjectured that, for all positive integers $n$ and $p$, the $p$th power of $n$ admits the representation $n^p = \sum_{l=0}^{p-1} (-1)^{l} c_{p,l} F_{n}^{p-l}$, where $F_{n}^{p-l}$ is the $n$th hyper-tetrahedron number of dimension $p-l$ and $c_{p,l}$ denotes the number of $(p-l)$-dimensional facets of the $p$-dimensional simplex $x_{\sigma_1} \geq x_{\sigma_2} \geq \cdots \geq x_{\sigma_p}$ (where $\sigma$ is a permutation of $\{ 1, 2, \ldots, p \}$) formed by cutting the $p$-dimensional cube $0 \leq x_1, x_2, \ldots, x_p \leq n-1$. In this note we show that this conjecture is true for every natural number $p$ if, and only if, $c_{p,l} = (p-l)! S(p, p-l)$, where $S(p,p-l)$ are the Stirling numbers of the second kind. Furthermore, we provide several equivalent formulas expressing the sum of powers $\sum_{i=1}^{n} i^p$, $p =1,2,\ldots$, as a linear combination of figurate numbers.

Full text

arXiv:2001.03208v5 [math.NT] 18 Apr 2021
Figurate Numbers and Sums of Powers
of Integers
Jos´e Luis Cereceda
Collado Villalba, 28400 (Madrid), Spain
jl.cereceda@movistar.es
Abstract
Recently, Marko and Litvinov (ML) conjectured that, for all positive integers n
and p, the p-th power of nadmits the representation np=Pp−1
ℓ=0 (−1)lcp,ℓFp−ℓ
n, where
Fp−ℓ
nis the n-th hyper-tetrahedron number of dimension p−ℓand cp,ℓ denotes the
number of (p−ℓ)-dimensional facets formed by cutting the p-dimensional cube 0 ≤
x1, x2,...,xp≤n−1. In this paper we show that the ML conjecture is true for every
natural number p. Our proof relies on the fact that the validity of the ML conjecture
necessarily implies that cp,ℓ = (p−ℓ)!S(p, p −ℓ), where S(p, p −ℓ) are the Stirling
numbers of the second kind. Furthermore, we provide a number of equivalent formulas
expressing the sum of powers Pn
i=1 ipas a linear combination of figurate numbers.
1 Introduction
Let Fk
n=n+k−1
kbe the n-th hyper-tetrahedron number of dimension k. In particular,
F2
nis the n-th triangular number Tn=1
2n(n+ 1) and F3
nis the n-th tetrahedral number
T en=Pn
i=1 Ti=1
6n(n+ 1)(n+ 2) (see, e.g., [4, Chapter 2]). Recently, Marko and Litvinov
conjectured (see [12, Conjecture 16]) that, for all positive integers nand p, the p-th power
of ncan be put as
np=
p−1
X
ℓ=0
(−1)ℓcp,ℓFp−ℓ
n,(1)
for certain positive integer coefficients cp,0, cp,1,...,cp,p−1(note the corrected factor (−1)ℓin
(1) instead of the original one (−1)p−ℓappearing in [12]). Specifically, cp,ℓ is the number of
(p−ℓ)-dimensional simplices defined by 0 ≤x1, x2,...,xp≤n−1 in conjunction with the
conditions
xσ1L1xσ2L2. . . Lp−1xσp,
where exactly ℓsymbols Liare “=”, the remaining p−ℓ−1 symbols Liare “≥”, and where σ
is a permutation of {1,2, . . . , p}. As indicated by Marko and Litvinov [12, p. 18], every such
simplex is then a (p−ℓ)-dimensional facet of the p-dimensional simplex xσ1≥xσ2≥ · · · ≥ xσp
that is formed by cutting the original p-dimensional cube 0 ≤x1, x2,...,xp≤n−1.
An alternative characterization of the coefficients cp,ℓ can be made in terms of m-tuples
(k1, k2,...,km) of nonnegative integers with content ℓ=Pm
i=1 kiand support s(the latter
1

being defined as the number of indices isuch that ki>0). As shown in [12, Proposition 14],
the coefficients cp,ℓ are given by the combinatorial formula
cp,ℓ =X
(k1,k2,...,km)
p!
(k1+ 1)!(k2+ 1)! ···(km+ 1)! ,(2)
where the sum runs over all m-tuples of nonnegative integers having the content ℓand the
support s=m+ℓ+ 1 −p, and such that kj>0 implies that kj+1 = 0 for every j < m.
Example 1. As a simple example, we may use formula (2) to calculate the coefficients cp,ℓ for
p= 5. As is readily verified, for this case the list of m-tuples satisfying the above conditions
is: (0,0,0,0) for ℓ= 0; (1,0,0,0), (0,1,0,0), (0,0,1,0), and (0,0,0,1) for ℓ= 1; (2,0,0),
(0,2,0), (0,0,2), (1,0,1,0), (0,1,0,1), and (1,0,0,1) for ℓ= 2; (3,0), (0,3), (1,0,2), and
(2,0,1) for ℓ= 3; and 4 for ℓ= 4. Hence, from (2) we get c5,0= 120, c5,1= 240, c5,2= 150,
c5,3= 30, and c5,4= 1.
Furthermore, by using (2), one can also deduce that, for example, for all p≥1, cp,0=p!;
for all p≥2, cp,1=1
2(p−1)p!; for all p≥3, cp,2=1
24 p!(p−2)(3p−5); and that, for all
p≥1, cp,p−1= 1.
Consider now the sum of powers of integers Σp
n=Pn
i=1 ip. Since Fk
n=Pn
i=1 Fk−1
i,
representation (1) for npimmediately implies that
Σp
n=
p−1
X
ℓ=0
(−1)ℓcp,ℓFp+1−ℓ
n,(3)
expressing Σp
nas a linear combination of figurate numbers. In what follows, we refer to either
(1) or (3) (with the coefficients cp,ℓ given by (2)) as the ML conjecture. The crucial point
we want to remark here is that Σp
ncan, in fact, be expressed in the polynomial form (see,
e.g., [7, Equation (7.5)], [15], and [2, Section 4])
Σp
n=
p
X
i=1
(−1)p−ii!S(p, i)n+i
i+ 1,(4)
where S(p, i) are the Stirling numbers of the second kind. Correspondingly, formula (4) can
be written in terms of figurate numbers as
Σp
n=
p−1
X
ℓ=0
(−1)ℓ(p−ℓ)!S(p, p −ℓ)Fp+1−ℓ
n.(5)
For example, setting p= 5 in (5) yields
Σ5
n= 120F6
n−240F5
n+ 150F4
n−30F3
n+F2
n.
Comparing the formulas for Σp
nin (3) and (5), and noting that the polynomials representing
the figurate numbers Fk
nare linearly independent, it is clear that if formula (3) for Σp
nis true
then necessarily the coefficients cp,ℓ in (2) should be of the form cp,ℓ = (p−ℓ)!S(p, p −ℓ) for
ℓ= 0,1,...,p−1. Conversely, if cp,ℓ = (p−ℓ)!S(p, p −ℓ) for ℓ= 0,1,...,p−1, then formula
(3) for Σp
nis true by virtue of (5). This can be summarized as follows.
2

Proposition 2. For p≥1and ℓ= 0,1,...,p−1, let cp,ℓ be the coefficient defined in (2).
Then,
ML conjecture ⇔cp,ℓ = (p−ℓ)!S(p, p −ℓ).
The rest of the paper is organized as follows. In Section 2, we give an alternative deriva-
tion of Proposition 2by determining the elements of the transition matrix connecting the
bases {n, n2,...,np}and {F1
n, F 2
n,...,Fp
n}, and its inverse. In Section 3, we prove that the
ML conjecture is true. Specifically, by using the representation of the Stirling numbers of the
second kind given in equation (12), we show that the coefficients defined in (2) are indeed
identical to cp,ℓ = (p−ℓ)!S(p, p −ℓ) for all p≥1 and ℓ= 0,1,...,p−1. In Section 4, we give
several alternative representations for the coefficients cp,ℓ. Finally, in Section 5, we provide a
number of equivalent formulas expressing the sum of powers Pn
i=1 ipas a linear combination
of figurate numbers.
2 Matrix formulation
For k≥1, the figurate numbers Fk
ncan be expanded in the basis {n, n2,...,nk}as
Fk
n=n+k−1
k=n(n+ 1)(n+ 2) ···(n+k−1)
k!=1
k!
k
X
r=1
s(k, r)nr,(6)
where the numbers s(k, r) are (unsigned) Stirling numbers of the first kind. For example,
we have that
F1
n=n,
F2
n=1
2n+1
2n2,
F3
n=1
3n+1
2n2+1
6n3,
F4
n=1
4n+11
24 n2+1
4n3+1
24 n4,
F5
n=1
5n+5
12 n2+7
24 n3+1
12 n4+1
120 n5,
or, in matrix form,






F1
n
F2
n
F3
n
F4
n
F5
n






=







1 0 0 0 0
1
2
1
20 0 0
1
3
1
2
1
60 0
1
4
11
24
1
4
1
24 0
1
5
5
12
7
24
1
12
1
120













n
n2
n3
n4
n5






=A5






n
n2
n3
n4
n5






,
where, following [12], we call the matrices Ap(for any p≥1) Fermat matrices. By inverting
A5we get the transition matrix from the basis {F1
n, F 2
n, F 3
n, F 4
n, F 5
n}to {n, n2, n3, n4, n5},
namely






n
n2
n3
n4
n5






=






1 0 0 0 0
−1 2 0 0 0
1−6 6 0 0
−1 14 −36 24 0
1−30 150 −240 120












F1
n
F2
n
F3
n
F4
n
F5
n






.
3

Moreover, recalling that Fk
n=Pn
i=1 Fk−1
i, we can express the last matrix equation in the
equivalent way






Σ1
n
Σ2
n
Σ3
n
Σ4
n
Σ5
n






=






1 0 0 0 0
−1 2 0 0 0
1−6 6 0 0
−1 14 −36 24 0
1−30 150 −240 120












F2
n
F3
n
F4
n
F5
n
F6
n






.
Let Apbe a Fermat matrix, and let a′
k,j denote the elements of A−1
p,k, j = 1,2,...,p.
Thus, for arbitrary p≥1, we have





Σ1
n
Σ2
n
.
.
.
Σp
n





=




a′
1,1a′
1,2··· a′
1,p
a′
2,1a′
2,2··· a′
2,p
.
.
..
.
.....
.
.
a′
p,1a′
p,2··· a′
p,p










F2
n
F3
n
.
.
.
Fp+1
n





,
from which it follows that
Σp
n=
p
X
i=1
a′
p,iFi+1
n=
p−1
X
i=0
a′
p,p−iFp+1−i
n.(7)
Comparing the rightmost side of (7) with (3), and noting that the polynomials F2
n, F 3
n,...,Fp+1
n
are linearly independent, it is concluded that the ML conjecture is equivalent to having (cf.
[12])
a′
p,i = (−1)p−icp,p−i, i = 1,2. . . , p. (8)
Next, we show the following result concerning the elements a′
k,j .
Proposition 3. The elements a′
k,j of the inverse of the Fermat matrix, A−1
p, are given by
a′
k,j = (−1)k−jj!S(k, j ), k, j = 1,2,...,p. (9)
Proof. Let us denote the elements of the Fermat matrix Apas ak,j. Then, from (6) it is clear
that
ak,j =s(k, j )
k!, k, j = 1,2,...,p. (10)
Now, consider the product matrices Mp=ApA−1
pand Np=A−1
pAp, where A−1
pand Aphave
elements given in (9) and (10), respectively. It is easily verified that the matrix elements
Mk,j and Nk,j of Mpand Npare
Mk,j =j!
k!
p
X
r=1
(−1)r−js(k, r)S(r, j),
Nk,j =
p
X
r=1
(−1)k−rS(k, r)s(r, j),
4

for 1 ≤k, j ≤p. Therefore, invoking the well-known orthogonality relations (see, e.g., [5,
Thm. 6.24])
k
X
r=0
(−1)rs(k, r)S(r, j) = (−1)kδk,j ,
k
X
r=0
(−1)rS(k, r)s(r, j) = (−1)kδk,j ,
and taking into account that s(k, 0) = S(k, 0) = 0 for k≥1, and that s(k, j) = S(k, j) = 0
for k < j, we obtain Mk,j = (−1)k−j(j!/k!)δk,j and Nk,j = (−1)2kδk ,j , and thus both Mpand
Npturn out to be the identity matrix Ip.
Note that, for the case in which k=p, equation (9) reads (after renaming the index j
as i)a′
p,i = (−1)p−ii!S(p, i). Hence, from (8), we conclude that the ML conjecture is true if,
and only if, cp,p−i=i!S(p, i) or, equivalently, cp,ℓ = (p−ℓ)!S(p, p −ℓ), for ℓ= 0,1,...,p−1,
thus recovering Proposition 2.
Table 1displays the first few rows of the triangular array for the numbers cp,ℓ = (p−
ℓ)!S(p, p −ℓ), where ℓ= 0,1,...,p−1. It is worth pointing out that, starting from the well-
known triangular recurrence relation for the Stirling numbers of the second kind, namely,
S(k, j) = jS(k−1, j) + S(k−1, j −1) (with initial conditions S(0,0) = 1 and S(0, j) =
S(j, 0) = 0 for j > 0), one can derive the following recurrence relation which is fulfilled by
the numbers cp,ℓ:
cp,ℓ =




p!,if ℓ= 0;
(p−ℓ)(cp−1,ℓ +cp−1,ℓ−1),if 0 < ℓ < p −1;
1,if ℓ=p−1.
(11)
Of course, the entries in Table 1can be computed using the recursive formula (11). For
example, c9,3is determined by the values of c8,3and c8,2as follows: c9,3= 6(c8,3+c8,2) =
6(126000 + 191520) = 1905120. Moreover, the alternating sum of the entries in the p-th row
in Table 1is given by Pp−1
ℓ=0 (−1)ℓcp,ℓ = 1 for all p≥1. This quickly follows from (5) by
noting that Σp
1= 1 for all p≥1.
p\ℓ0 1 2 3 4 5 6 7 8
1 1
2 2 1
3 6 6 1
4 24 36 14 1
5 120 240 150 30 1
6 720 1800 1560 540 62 1
7 5040 15120 16800 8400 1806 126 1
8 40320 141120 191520 126000 40824 5796 254 1
9 362880 1451520 2328480 1905120 834120 186480 18150 510 1
Table 1: Triangular array for the numbers cp,ℓ up to p= 9.
5

3 Proof of the ML conjecture
Since the recursive formula given in (11) completely defines the numbers cp,ℓ = (p−ℓ)!S(p, p−
ℓ), a way to prove the ML conjecture is to show that the coefficients cp,ℓ defined in (2) satisfy
the said recurrence (11). We shall not pursue this way here. Instead, we are going to prove
the ML conjecture by using the following representation for the Stirling numbers of the
second kind (see, e.g., [14, Equation (2.27)])
S(n, k) = n!
k!X
(r1,r2,...,rk)
1
r1!r2!···rk!,(12)
where the summation extends over all positive integer solutions of the equation r1+r2+···+
rk=n. Using the above representation, we can express the numbers cp,ℓ = (p−ℓ)!S(p, p −ℓ)
as
cp,ℓ =X
(r1,r2,...,rp−ℓ)
p!
r1!r2!···rp−ℓ!,(13)
where each of the (p−ℓ)-tuples (r1, r2,...,rp−ℓ) in the summation has the content Pp−ℓ
i=1 ri=
p, with ri≥1.
Example 4. For ℓ=p−2 and p≥2, from (13) we have
cp,p−2=X
r1+r2=p
p!
r1!r2!=
p−1
X
j=1
p!
j!(p−j)! =
p−1
X
j=1 p
j= 2p−2.
Similarly, for ℓ=p−3 and p≥3, from (13) we have
cp,p−3=X
r1+r2+r3=p
p!
r1!r2!r3!=X
i+j+k=pp
i, j, k−3−3
p−1
X
t=1 p
t= 3p−3·2p+ 3,(14)
where p
i,j,kare the trinomial coefficients, with i,j, and kbeing nonnegative integers.
On the other hand, by renaming ki+ 1 as jifor i= 1,2,...,m, equation (2) can be
rewritten as
cp,ℓ =X
(j1,j2,...,jm)
p!
j1!j2!···jm!,(15)
where now the summation extends over all m-tuples of positive integers (j1, j2,...,jm) with
the content Pm
i=1 ji=ℓ+m, where m=p+t−ℓ−1 (with tbeing the number of indices i
for which ji≥2), and such that ji≥2 implies that ji+1 = 1 for every i < m.
Example 5. For ℓ=p−2 and p≥2, it can be seen that the allowed m-tuples for this case
are: (p−1,1), (1, p −1), and (a, 1, b), with a+b=pand a, b ≥2. From (15) we then have
cp,p−2= 2p+X
a+b=p
p!
a!b!= 2p+
p−2
X
j=2
p!
j!(p−j)! = 2p+
p−2
X
j=2 p
j= 2p−2.
6

Moreover, for ℓ=p−3 and p≥3, the allowed m-tuples for this case are: (p−2,1,1),
(1, p −2,1), (1,1, p−2), (a, 1, b, 1), (a, 1,1, b), (1, a, 1, b), and (c, 1, d, 1, e), where a+b=p−1
with a, b ≥2, and c+d+e=pwith c, d, e ≥2. Thus, from (15) we obtain
cp,p−3= 3p(p−1) + 3 X
a+b=p−1
p!
a!b!+X
c+d+e=p
p!
c!d!e!.
Now, we have
X
c+d+e=p
p!
c!d!e!=X
r1+r2+r3=p
p!
r1!r2!r3!−3p(p−1) −3X
a+b=p−1
p!
a!b!,
where r1, r2, r3≥1. Therefore, it follows that
cp,p−3=X
r1+r2+r3=p
p!
r1!r2!r3!= 3p−3·2p+ 3,
where we have used the previous result in (14). In the same manner, for ℓ=p−4 and p≥4,
starting from either (13) or (15) we find that
cp,p−4= 4p−4·3p+ 6 ·2p−4.(16)
Indeed, since j!S(p, j) = Pj
r=0(−1)rj
r(j−r)p, the validity of the ML conjecture will retro-
spectively enable us to deduce that both (13) and (15) lead to the general result
cp,p−j=
j
X
r=0
(−1)rj
r(j−r)p, j = 1,2,...,p.
In view of Proposition 2, and making use of (13) and (15), it follows that the validity of
the ML conjecture is then equivalent to the following number theoretic identity.
Proposition 6. For all p≥1and ℓ= 0,1,...,p−1, we have
X
(j1,j2,...,jm)
p!
j1!j2!···jm!=X
(r1,r2,...,rp−ℓ)
p!
r1!r2!···rp−ℓ!.(17)
Next, we sketch the proof of Proposition 6and, as a consequence, of the ML conjecture.
Proof. The first step towards proving (17) is to expand the summation in the left-hand
side of (17) over all the m-tuples (j1, j2,...,jm) of positive integers fulfilling the conditions
stipulated after equation (15). Recalling that tis the number of ji’s for which ji≥2, it is
not hard to show that, for ℓ=p−j, equation (15) can be decomposed as
cp,p−j=




j
X
t=1 j
tX
s1+s2+···+st=p+t−j
p!
s1!s2!···st!,for j= 1,2,...,p−1; (18)
p!,for j=p, (19)
7

where s1, s2,...,stare all integers ≥2. For example, for j= 4, we have
cp,p−4= 4 X
s1=p−3
p!
s1!+ 6 X
s1+s2=p−2
p!
s1!s2!+ 4 X
s1+s2+s3=p−1
p!
s1!s2!s3!
+X
s1+s2+s3+s4=p
p!
s1!s2!s3!s4!.
As an even more concrete example, we may use the last equation to calculate the coefficient
c9,5as follows
c9,5= 4 X
s1=6
9!
s1!+ 6 X
s1+s2=7
9!
s1!s2!+ 4 X
s1+s2+s3=8
9!
s1!s2!s3!+X
s1+s2+s3+s4=9
9!
s1!s2!s3!s4!
= 9! 41
6! + 6 2
2!5! +2
3!4!+ 4 3
2!2!4! +3
2!3!3!+4
2!2!2!3!= 186480.
Of course, equation (16) reproduces this result as 186480 = 49−4·39+6 ·29−4. Furthermore,
as an aside, it is to be noted that the number of individual summands involved in (18)
amounts to
N(p, j) =
j
X
t=1 j
tp−j−1
t−1=p−1
j−1.
For the example above for which p= 9 and j= 4, we have N(9,4) = 4·1+6·4+4·6+1·4 = 56.
Returning to the main argument, let us, for convenience, rewrite (18) in the form
cp,p−j=X
s1+s2+···+sj=p
p!
s1!s2!···sj!+W(p, j),(20)
where
W(p, j) =
j−1
X
t=1 j
tX
s1+s2+···+st=p+t−j
p!
s1!s2!···st!.
The second (and last) crucial point is to realize that W(p, j) is precisely the summation
W(p, j) = X
w1+w2+···+wj=p
p!
w1!w2!···wj!,
where w1, w2,...,wjare all integers ≥1 and at least one of them is equal to 1. (Note that
the case w1=w2=... =wj= 1 is excluded since, as far as equation (18) is concerned,
j≤p−1.) For example, we have that
W(p, 4) = 4 X
s1+1+1+1=p
p!
s1!1!1!1! + 6 X
s1+s2+1+1=p
p!
s1!s2!1!1! + 4 X
s1+s2+s3+1=p
p!
s1!s2!s3!1!.
8

Now, for ℓ=p−j, (13) reads as
cp,p−j=X
r1+r2+···+rj=p
p!
r1!r2!···rj!,(21)
where r1, r2,...,rjare all integers ≥1. Hence, it is clear that, by definition,
X
r1+r2+···+rj=p
p!
r1!r2!···rj!=X
s1+s2+···+sj=p
p!
s1!s2!···sj!+W(p, j),(22)
for j= 1,2,...,p−1. Therefore, noting that both (13) and (15) give cp,0=p!, and taking
into account (20), (21), and (22), it follows that the identity (17) holds for all p≥1 and
ℓ= 0,1,...,p−1, and, consequently, it is concluded that the ML conjecture is true.
We end this section with the following observation. It is a well-known, classical result
that the number of surjections T(m, n) from the set A(having melements) onto the set B
(having nelements, with m≥n) is given by n!S(m, n); see, e.g., [11]. Since cp,p−j=j!S(p, j),
this means, in particular, that the formula for cp,p−jin (18)-(19) actually gives the number of
surjections T(p, j) from a p-element set to a j-element set for each j= 1,2,...,p. Moreover,
from the recurrence relation T(p, j) = j(T(p−1, j) + T(p−1, j −1)), we immediately get
the recursive formula
cp,p−j=




1,if j= 1;
j(cp−1,p−j+cp−1,p−j−1),if 0 < j < p −1;
p!,if j=p,
which is, of course, the same as that in (11).
4 Alternative representations for the coefficients cp,ℓ
Next, we state the following alternative formula for the coefficients cp,ℓ:
cp,ℓ =




p!,for ℓ= 0;
p!(p−ℓ)
ℓ
X
k=1 p−ℓ−1
k−12ℓ−p
ℓ−k(k−1)!
(k+ℓ)! S(k+ℓ, k),for 1 ≤ℓ≤p−1. (23)
Recalling that cp,ℓ = (p−ℓ)!S(p, p −ℓ), it is easily seen that the representation (23) is a
direct consequence of the identity (see [19, Thm. 2.3])
S(p+ℓ, p) =
ℓ
X
k=0 ℓ−p
ℓ−kℓ+p
ℓ+kS(k+ℓ, k),
9

which is in turn implied by the results found in Gould’s paper [8]. We can write down the
expressions for cp,0, cp,1, . . . , cp,7obtained from (23) as follows
cp,0=p!,
cp,1=p!
2!(p−1),
cp,2=p!
3!(p−2) 3p−5
4,
cp,3=p!
4!(p−3)2p−2
2,
cp,4=p!
5!(p−4) 15p3−150p2+ 485p−502
48 ,
cp,5=p!
6!(p−5)2(p−4) 3p2−23p+ 38
16 ,
cp,6=p!
7!(p−6) 63p5−1575p4+ 15435p3−73801p2+ 171150p−152696
576 ,
cp,7=p!
8!(p−7)2(p−6) 9p4−198p3+ 1563p2−5182p+ 6008
144 .
(24)
In view of these and successive coefficients, it seems safe to conjecture that (p−ℓ)2(p−ℓ+1)
constitutes a factor of cp,ℓ for all odd ℓwith ℓ≥3. A table of coefficients related to those in
(24) is given in Equations (1.18)-(1.22) of Gould’s book [9]. It is to be noted, on the other
hand, that (23) may also be obtained by considering the generalized Bernoulli numbers or
N¨orlund polynomials B(α)
kdefined by the generating function
x
ex−1α
=
∞
X
k=0
B(α)
k
xk
k!,|x|<2π,
where the parameter αstands for any arbitrary (real or complex) number. It turns out that
B(α)
kis a polynomial in αof degree k. In particular, B(1)
k=Bkare the ordinary Bernoulli
numbers. The Stirling numbers of the second kind are related to the generalized Bernoulli
numbers by means of (see [16, Equation (14.12)])
S(n+k, n) = n+k
kB(−n)
k.
Therefore, we have that
cp,ℓ = (p−ℓ)!S(p, p −ℓ) = p!
ℓ!B(−p+ℓ)
ℓ.(25)
Moreover, B(α)
ℓis given explicitly by Equation (15) of Srivastava and Todorov’s paper [18]
B(α)
ℓ=
ℓ
X
k=0
(−1)kα+k−1
kα+ℓ
ℓ−kk+ℓ
k−1
S(k+ℓ, k).(26)
10

Hence, taking equation (26) with α→ −p+ℓ, and substituting the resulting B(−p+ℓ)
ℓinto
(25), we find that
cp,ℓ =p!
ℓ
X
k=0
(−1)k−p+ℓ+k−1
k2ℓ−p
ℓ−kk!
(k+ℓ)!S(k+ℓ, k).(27)
Finally, noting that p−ℓ
k= (−1)k−p+ℓ+k−1
k, we conclude that the representation in (27) is
indeed the same as that in (23).
Furthermore, from the relation S(p, p−ℓ) = p
ℓB(−p+ℓ)
ℓ, it is readily seen that the Stirling
numbers of the second kind of the type S(p, p −ℓ) are polynomials in pof degree 2ℓ. These
can be expressed, in particular, in the form
S(p, p −ℓ) = 






1,for ℓ= 0;
ℓ−1
X
j=0
aℓ,j p
2ℓ−j,for ℓ= 1,2,...,p−1, (28)
where the coefficients aℓ,j , which were introduced by Jordan and Ward [1] (see also [21] and
references therein), satisfy the recurrence
aℓ,j = (2ℓ−j−1)aℓ−1,j + (ℓ−j)aℓ−1,j−1,
and are given explicitly by Equation (1.13) of Carlitz’s paper [1]
aℓ,j =1
(ℓ−j)!
ℓ−j
X
k=0
(−1)kℓ−j
kk
X
r=0
r!k
r2ℓ−j
r(ℓ−j−k)2ℓ−j−r.(29)
From (28) we can therefore express the coefficients cp,ℓ as
cp,ℓ =






p!,for ℓ= 0;
(p−ℓ)!
ℓ−1
X
j=0
aℓ,j p
2ℓ−j,for ℓ= 1,2,...,p−1, (30)
with the aℓ,j ’s being given by (29). Table 2shows the first few rows of the triangular array
for the Jordan coefficients. Let us note, in particular, that aℓ,0= (2ℓ−1)!!.
On the other hand, for fixed ℓ, the exponential generating function of the Stirling numbers
of the second kind S(p, ℓ) is given by (see, e.g., [14, Equation (2.19)])
(ex−1)ℓ
ℓ!=
∞
X
p=0
S(p, ℓ)xp
p!,
from which one can deduce that
cp,ℓ = (p−ℓ)!S(p, p −ℓ) = f(p)
p,ℓ (0),(31)
11

ℓ\j0 1 2 3 4 5 6 7
1 1
2 3 1
3 15 10 1
4 105 105 25 1
5 945 1260 490 56 1
6 10395 17325 9450 1918 119 1
7 135135 270270 190575 56980 6825 246 1
8 2027025 4729725 4099095 1636635 302995 22935 501 1
Table 2: Triangular array for the Jordan coefficients aℓ,j up to ℓ= 8.
where f(p)
p,ℓ (0) = dpfp,ℓ(x)/dxp|x=0 is the p-th derivative of fp,ℓ(x) = (ex−1)p−ℓwith respect
to xevaluated in the limit when x→0. As a simple example, for ℓ=p−4, we have
fp,p−4(x) = (ex−1)4=e4x−4e3x+ 6e2x−4ex+ 1.
It is easily seen that
dpfp,p−4(x)
dxp= 4pe4x−4·3pe3x+ 6 ·2pe2x−4ex,
and then, from (31), it follows that cp,p−4= 4p−4·3p+ 6 ·2p−4, thus recovering (16).
We end this section by quoting two other possible representations for cp,ℓ, namely
cp,ℓ = (p−ℓ)!
ℓ
X
k=0 ℓ−p
ℓ+kp+ℓ
ℓ−ks(k+ℓ, k),(32)
and
cp,ℓ =






p!,for ℓ= 0;
(p−ℓ)!
ℓ−1
X
j=0 ℓ
jp+ℓ−1−j
2ℓ,for ℓ= 1,2,...,p−1, (33)
where s(k+ℓ, k) are the (unsigned) Stirling numbers of the first kind, and ℓ
jare the second-
order Eulerian numbers indexed so that 0 ≤j≤ℓ−1 (see [10, Table 270 and Equation
(6.43)]).
5 Power sums as a linear combination of figurate num-
bers
Starting with (3) and using the representations for cp,ℓ given in (27), (30), (32), and (33),
we successively obtain the following formulas expressing Σp
nas a linear combination of the
12

figurate numbers F2
n, F 3
n, . . . , F p+1
n:
Σp
n=p!
p−1
X
ℓ=0
(−1)ℓFp+1−ℓ
n
ℓ
X
k=0 p−ℓ
k2ℓ−p
ℓ−kk!
(k+ℓ)!S(k+ℓ, k),
Σp
n=p!Fp+1
n+
p−1
X
j=1
(−1)j(p−j)!Fp+1−j
n
j−1
X
k=0
aj,kp
2j−k,
Σp
n=
p−1
X
ℓ=0
(−1)ℓ(p−ℓ)!Fp+1−ℓ
n
ℓ
X
k=0 ℓ−p
ℓ+kp+ℓ
ℓ−ks(k+ℓ, k),
Σp
n=p!Fp+1
n+
p−1
X
j=1
(−1)j(p−j)!Fp+1−j
n
j−1
X
k=0 j
kp+j−1−k
2j.
(34)
All four linear combinations in (34) can be rewritten in the form Pp−1
ℓ=0 fp,ℓFp+1−ℓ
n, where the
coefficient fp,ℓ is equal to (−1)ℓ(p−ℓ)!S(p, p −ℓ) for each ℓ= 0,1,...,p−1, in accordance
with (5).
In addition to the formulas in (5) or (34), there are several other alternative formulas
expressing Σp
nin terms of the figurate numbers Fk
n=n+k−1
k. For example, the following
two well-known polynomial formulas for Σp
n(see, e.g., [2,17,20]):
Σp
n=
p
X
j=1
j!S(p, j)n+ 1
j+ 1,(35)
and
Σp
n=
p
X
j=1 p
jn+j
p+ 1,(36)
can equivalently be written in terms of Fk
nas
Σp
n=
p
X
j=1
j!S(p, j)Fj+1
n−j+1,(37)
and
Σp
n=
p
X
j=1 p
jFp+1
n+j−p,(38)
respectively, where p
jare the ordinary Eulerian numbers, with the initial value p
1= 1 for
all p≥1. Note that (38) only involves figurate numbers of dimension p+ 1. Along with the
above two formulas in (35) and (36), we may quote another, not so well-known formula for
Σp
nwhich is a variant of that in (35), namely (see, e.g., [21, Equation (9)] and [3])
Σp
n=
p+1
X
j=1
(j−1)!S(p+ 1, j)n
j=
p+1
X
j=1
(j−1)!S(p+ 1, j)Fj
n−j+1.(39)
13

As an example, for p= 8, from (5), (37), (38), and (39), we obtain the equivalent
polynomial representations
Σ8
n= 40320F9
n−141120F8
n+ 191520F7
n
−126000F6
n+ 40824F5
n−5796F4
n+ 254F3
n−F2
n
= 40320F9
n−7+ 141120F8
n−6+ 191520F7
n−5
+ 126000F6
n−4+ 40824F5
n−3+ 5796F4
n−2+ 254F3
n−1+F2
n
=F9
n+ 247F9
n−1+ 4293F9
n−2+ 15619F9
n−3
+ 15619F9
n−4+ 4293F9
n−5+ 247F9
n−6+F9
n−7
= 40320F9
n−8+ 181440F8
n−7+ 332640F7
n−6+ 317520F6
n−5
+ 166824F5
n−4+ 46620F4
n−3+ 6050F3
n−2+ 255F2
n−1+F1
n.
For completeness, let us finally mention that, as is well known, the power sums Σp
ncan be
expressed as polynomials in the triangular numbers Tn(the so-called Faulhaber polynomials
[6]) as follows
Σ2k
n= Σ2
nhbk,0+bk,1Tn+bk,2Tn2+···+bk,k−1Tnk−1i,
Σ2k+1
n=Tn2hck,0+ck,1Tn+ck,2Tn2+···+ck,k−1Tnk−1i,
where bk,j and ck,j are non-zero rational coefficients for j= 0,1,...,k −1 and k≥1. In
particular, Σ3
n=Tn2.
Note added
After the completion of this work, a much shorter and direct proof of the ML conjecture was
devised by Professor Frantiˇsek Marko himself [13]. I believe, however, that the proof of the
ML conjecture presented here still retains its interest in its own right.
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14

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15