Kyungpook Mathematical Journal 2018; 58(4): 599-613
On a Generalization of the Pentagonal Number Theorem
Ho-Hon Leung
Department of Mathematical Sciences, United Arab Emirates University, Al Ain, 15551, United Arab Emirates
e-mail : hohon.leung@uaeu.ac.ae
Received: April 3, 2018; Revised: August 20, 2018; Accepted: August 27, 2018; Published online: December 23, 2018.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

We study a generalization of the classical Pentagonal Number Theorem and its applications. We derive new identities for certain infinite series, recurrence relations and convolution sums for certain restricted partitions and divisor sums. We also derive new identities for Bell polynomials.

Keywords: integer partition, divisor sum, Bell polynomials, polygonal numbers, pentagonal number theorem.
1. Introduction

The Pentagonal Number Theorem is one of Euler’s most profound discoveries. It is the following identity:

$∏n=1∞(1-qn)=∑n=-∞∞(-1)nqn(3n-1)/2$

where n(3n − 1)/2 is called the nth pentagonal numbers. The pentagonal numbers represent the number of distinct points which may be arranged to form superimposed regular pentagons such that the number of points on the sides of each respective pentagonal is the same. Bell’s article [2] is an excellent reference about the Pentagonal Number Theorem and its applications from the historical perspective. Andrews’s article [1] is devoted to a modern exposition of Euler’s original proof of the theorem.

Let ℕ = {1, 2, …} and ℕ0 = {0, 1, 2, …}. The function p(n) is the number of integer partitions of n. The function σ(n) is the sum of all divisors of n. Euler applied the Pentagonal Number Theorem to study various properties of integer partitions and divisor sums. In particular, there is a convolution formula that connects the functions p(n) and σ(n):

$np(n)=∑k=0∞σ(k)p(n-k).$

Apart from this, there are recurrence relations for p(n) and σ(n) which are intimately related to the Pentagonal Number Theorem:

$p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+…,σ(n)=σ(n-1)+σ(n-2)-σ(n-5)-σ(n-7)+....$

The readers are invited to read the survey article by Osler et. al. [4] for a readable account of the connections between functions p(n) and σ(n).

One may wonder if there are recurrence relations and convolution formulas for (restricted) integer partitions and divisor sums in terms of other polygonal numbers (e.g. triangular numbers, heptagonal numbers …). The goal of this article is to give positive answers to this question based on a generalization of the Pentagonal Number Theorem.

Unless mentioned otherwise, throughout the paper, all equations in the variable q which involve infinite sums and infinite products hold true if |q| < 1.

2. Main Results

### 2.1. Main Theorem and Some Corollaries

Let g > 3 and n ∈ ℤ. We denote the generalized nthg-gonal number [6, p.40] by

$Pg,n=n((g-2)n-(g-4))2.$

For n ∈ ℕ, Pg,n represents the number of points which may be arranged to form regular g-gons such that the number of points on the sides of each respective g-gon is the same as n.

Let (a; q)n be the q-Pochhammer symbol for n ≥ 1. That is,

$(a;q)n:=∏k=0n-1(1-aqk)=(1-a)(1-aq)(1-aq2)…(1-aqn-1).$

Considered as a formal power series in q, the definition of q-Pochhammer symbol can be extended to an infinite product. That is,

$(a;q)∞:=∏k=0∞(1-aqk).$

We note that (q; q) is the Euler’s function.

We state the following theorem.

Theorem 2.1.1

Let g ≥ 5.

$(q;qg-2)∞(qg-3;qg-2)∞(qg-2;qg-2)∞=∑n=-∞∞(-1)nqPg,n.$
Proof

We replace q by q(g−2)/2 and z by q−(g−4)/2 in the Jacobi’s Triple Product Identity,

$∑n=-∞∞qn2zn=∏n=1∞(1-q2n)(1+q2n-1z)(1+q2n-1z-1)$

to get the desired result.

### Remark 2.1.2

If g = 5, we get Euler’s Pentagonal Number Theorem by Theorem 2.1.1.

If n ∈ ℕ, then

$(-n)((g-2)(-n)-(g-4))2=n((g-2)n+(g-4))2.$

Let g > 3 and n ∈ ℕ. Let Qg,n be the following numbers:

$Qg,n=n((g-2)n+(g-4))2.$

We note that Qg,n = Pg,−n. Theorem 2.1.1 can be restated as

$(q;qg-2)∞(qg-3;qg-2)∞(qg-2;qg-2)∞=1+∑n=1∞(-1)n(qPg,n+qQg,n).$

The nth triangular number Δn is

$Δn=n(n+1)2=P3,n.$

We note that

$Δ2n-1=P6,n, Δ2n=Q6,n.$

The following corollary is clear by (2.1) when g = 6.

### Corollary 2.1.3

$(q;q4)∞(q3;q4)∞(q4;q4)∞=1-∑k=0∞(qΔ4k+1+qΔ4k+2)+∑k=0∞(qΔ4k+3+qΔ4k+4).$

### Corollary 2.1.4

$∑n=0∞qn2+n(q;q)n=(∑n=0∞p(n)qn) (1+∑n=1∞(-1)n(qP7,n+qQ7,n)).$
Proof

Based on one of the Rogers-Ramanujan identities [9, 11], we get

$∑n=0∞qn2+n(q;q)n=1(q2;q5)∞(q3;q5)∞=(q;q5)∞(q4;q5)∞(q5;q5)∞(q;q5)∞(q2;q5)∞(q3;q5)∞(q4;q5)∞(q5;q5)∞=(1(q;q)∞) ((q;q5)∞(q4;q5)∞(q5;q5)∞).$

We get the desired result based on the generating function of p(n) and the identity (2.1) when g = 7.

### Corollary 2.1.5

$1+∑n=1∞(-1)n(q5P5,n+q5Q5,n)=(∑n=0∞qn2(q;q)n) (1+∑n=1∞(-1)n(qP7,n+qQ7,n)).$
Proof

We replace q by q5 in the Pentagonal Number Theorem to get

$(q5;q5)∞=1+∑n=1∞(-1)n(q5P5,n+q5Q5,n).$

On the other hand,

$(q5;q5)∞=(1(q;q)∞(q4;q5)∞) ((q;q5)∞(q4;q5)∞(q5;q5)∞).$

We recall one of the Rogers-Ramanujan identities [9, 11],

$∑n=0∞qn2(q;q)n=1(q;q)∞(q4;q5)∞.$

By putting (2.3), (2.5) into (2.4), we get the desired result based on the identity (2.1) when g = 7.

2.2. Recurrence Relations for Some Restricted Integer Partitions

### Definition 2.2.1

The function (p2,5 + p3,5)(n) is the number of partitions of n such that each part is either congruent to 2 modulo 5 or 3 modulo 5. We extend the domain of (p2,5 + p3,5)(n) to ℤ by setting (p2,5 + p3,5)(x) = 0 if x ∉ ℕ0.

### Theorem 2.2.2

$(p2,5+p3,5)(n)=p(n)+∑k=1∞(-1)k(p(n-P7,k)+p(n-Q7,k)).$
Proof

By Corollary 2.1.4 and the generating function of (p2,5 + p3,5)(n), we get

$∑n=0∞(p2,5+p3,5)(n)qn=1(q2;q5)∞(q3;q5)∞=(∑n=0∞p(n)qn) (1+∑n=1∞(-1)n(qP7,n+qQ7,n)).$

We get the desired result by comparing coefficients of qn on both sides of the equation.

### Example 2.2.3

Let n = 8. Then we have 8 = 2 + 2 + 2 + 2 = 3 + 3 + 2. So, (p2,5 + p3,5)(8) = 3. The generalized heptagonal numbers are P7,1 = 1, Q7,1 = 4, P7,2 = 7… and hence

$p(8)-p(8-P7,1)-p(8-Q7,1)+p(8-P7,2)=p(8)-p(7)-p(4)+p(1)=22-15-5+1=3=(p2,5+p3,5)(8).$

### Definition 2.2.4

The function (p1,5 + p4,5)(n) is the number of partitions of n such that each part is either congruent to 1 modulo 5 or 4 modulo 5. We extend the domain of (p1,5 + p4,5)(n) to ℤ by setting (p1,5 + p4,5)(x) = 0 if x ∉ ℕ0.

### Theorem 2.2.5

$(p1,5+p4,5)(n)=M+∑k=1∞(-1)k+1((p1,5+p4,5)(n-P7,k)+(p1,5+p4,5)(n-Q7,k))).$

where

$M={(-1)m,n=5P5,m or 5Q5,m for some m;0,otherwise.$
Proof

The generating function of (p1,5 + p4,5)(n) is

$∑n=0∞(p1,5+p4,5) (n)qn=1(q;q5)∞(q4;q5)∞.$

Now the result is obvious by comparing coefficients of qn in (2.4) and by Corollary 2.1.5.

### Example 2.2.6

Let n = 9. The generalized heptagonal numbers are P7,1 = 1, Q7,1 = 4, P7,2 = 7…. By Theorem 2.2.5, we get

$(p1,5+p4,5)(9)=(p1,5+p4,5)(8)+(p1,5+p4,5)(5)-(p1,5+p4,5)(2).$

It can be easily verified since

$(p1,5+p4,5)(9)=4,9=1+1+1+1+1+1+1+1+1=4+1+1+1+1+1=4+4+1,(p1,5+p4,5)(8)=3,1+1+1+1+1+1+1+1=4+1+1+1+1=4+4,(p1,5+p4,5)(5)=2,1+1+1+1+1=4+1,(p1,5+p4,5)(2)=1,1+1.$

### Definition 2.2.7

The function q(n) is the number of partitions of n such that each part is distinct. We extend the domain of q(n) to ℚ by setting q(x) = 0 if x ∉ ℕ0.

We obtain two recurrence relations for q(n).

### Theorem 2.2.8

$q(n)=K+∑k=1∞(-1)k+1 (q(n-2P5,k)+q(n-2Q5,k))$

where

$K={1,n=Δm for some m;0,otherwise.$
Proof

By an identity due to Guass ([6, p.40]),

$∑n=0∞qΔn=(q2;q2)∞(q;q2)∞=((q2;q2)∞))·(1(q;q2)∞)=((q2;q2)∞)·(∑n=1∞q(n)qn)$

where the last equality is due to Euler’s Theorem. We replace q by q2 in the Pentagonal Number Theorem to get

$(q2;q2)∞=1+∑n=1∞(-1)n(q2P5,n+q2Q5,n).$

By putting (2.7) into (2.6) and comparing coefficients of qn on both sides of the equation, we get the result as desired.

### Theorem 2.2.9

$q(n)=L+∑k=1∞(-1)k+1 (q(2n-P5,k2)+q(2n-Q5,k2))$

where

$L={-1,2n=Δ4k+1,Δ4k+2 for some k;1,2n=Δ4k+3,Δ4k+4 for some k;0,otherwise.$
Proof$(q;q4)∞(q3;q4)∞(q4;q4)∞=((q;q)∞) (1(q2;q4)∞).$

We replace q by q2 in Euler’s Theorem to get

$(1(q2;q4)∞)=∏k=1∞(1+q2k)=∑n=0∞q(n)q2n.$

By putting (2.9) into (2.8), applying Corollary 2.1.3 to the left hand side of (2.8) and applying the Pentagonal Number Theorem to (q; q), we get

$1-∑k=0∞(qΔ4k+1+qΔ4k+2)+∑k=0∞(qΔ4k+3+qΔ4k+4)=(1+∑n=1∞(-1)n(qP5,n+qQ5,n))·(∑n=1∞q(n)q2n).$

Comparing the coefficients of qn on both sides of (2.10), we get the result as desired.

### Example 2.2.10

Let n = 15. We note that 15 = Δ5. The generalized pentagonal numbers are

$P5,1=1, Q5,1=2, P5,2=5, Q5,2=7,P5,3=12, Q5,3=15, P5,4=22, Q5,4=26,....$

By Theorem 2.2.8,

$q(15)=1+(q(15-2P5,1)+q(15-2Q5,1))-(q(15-2P5,2)+q(15-2Q5,2))=1+q(13)+q(11)-q(5)-q(1)=1+18+12-3-1=27.$

Alternatively, by Theorem 2.2.9 and the fact that 2(15) = 30 ≠ Δm for any m ∈ ℕ.

$q(15)=0+q (30-22)+q (30-122)-q (30-222)-q (30-262)=q(14)+q(9)-q(4)-q(2)=22+8-2-1=27.$

### Definition 2.2.11

Let n, r ∈ ℕ0. Let m ∈ ℕ. The function pr,m(n) is defined to be the number of partitions of n such that each part is congruent to r modulo m. We extend the domain of pr,m(n) to ℤ by setting pr,m(x) = 0 if x ∉ ℕ0.

### Definition 2.2.12

Let n ∈ ℤ and m ∈ ℕ. The restricted integer partition $pm′(n)$ is defined by

$pm′(n):=pm-1,m(n)+p0,m(n)+p1,m(n)$

where $pm′(0):=1$ and $pm′(n)=0$ for n ∉ ℕ0.

In particular, if m = 3, then $pm′(n)=p3′(n)=p(n)$. The generating function of $pm′(n)$ is as follows:

$∑n=0∞pm′(n)qn=1(q;qm)∞(qm-1;qm)∞(qm;qm)∞.$

It is convenient to introduce the number eg,n given by

$eg,n={1,n=0;(-1)k,if n=Pg,k or Qg,k;0,otherwise.$

Theorem 2.1.1 can be rewritten as

$(q;qg-2)∞(qg-3;qg-2)∞(qg-2;qg-2)∞=∑n=0∞eg,nqn.$

We obtain the following recurrence relation for the restricted partition $pm′(n)$ if m ≥ 3.

### Theorem 2.2.13

Let m ≥ 3 and n ∈ ℕ.

$pm′(n)=∑k=1∞(-1)k+1 (pm′(n-Pm+2,k)+pm′(n-Qm+2,k)).$
Proof

By (2.11) and (2.12),

$(∑n=0∞pm′(n)qn) (∑n=0∞em+2,nqn)=1$

By comparing coefficients of qn for n ≥ 1 on both sides of the equation, we get

$pm′(n)+∑k=1∞(-1)k (pm′(n-Pm+2,k)+pm′(n-Qm+2,k))=0.$

### Remark 2.2.14

In the case m = 3, Theorem 2.2.13 is reduced to the well-known recurrence relation for the integer partition p(n).

### Definition 2.2.15

The function $pe,m′(n)$ is the number of partitions of n such that each partition has an even number of distinct parts, and each part is congruent to 0 modulo m, 1 modulo m or m − 1 modulo m.

### Definition 2.2.16

The function $po,m′(n)$ is the number of partitions of n such that each partition has an odd number of distinct parts, and each part is congruent to 0 modulo m, 1 modulo m or m − 1 modulo m.

### Theorem 2.2.17

Let n ∈ ℕ and m ≥ 3.

$pe,m′(n)-po,m′(n)={(-1)k,n=Pm+2,k or Qm+2,k for some k∈ℕ;0,otherwise.$
Proof

In the infinite product expansion of

$(q;qm)∞(qm-1;qm)∞(qm;qm)∞,$

the coefficients of qN has +1 contribution from a partition of N that consists of an even number of parts, where each part is congruent to 0 modulo m, 1 modulo m or m − 1 modulo m. It has −1 contribution from a partition of N that consists of an odd number of parts, where each part is congruent to 0 modulo m, 1 modulo m or m − 1 modulo m. By Theorem 2.1.1, our result follows immediately.

### Remark 2.2.18

In the case m = 3, Theorem 2.2.17 is reduced to the interesting observation made by Legendre on the Pentagonal Number Theorem [1, p.2].

2.3. Recurrence Relations and Convolution Sums for Restricted Divisor Sums

### Definition 2.3.1

Let r ∈ ℕ0. Let n, m ∈ ℕ. The function σr,m(n) is defined by

$σr,m(n):=∑{d∈ℕ,d∣n,d≡r mod m}d.$

### Definition 2.3.2

Let n, m ∈ ℕ. The restricted divisor sum $σm′(n)$ is defined by

$σm′(n):=σm-1,m(n)+σ0,m(n)+σ1,m(n).$

The restricted Lambert series for σr,m(n) is

$∑n=1∞σr,m(n)qn=∑{n∣n≡r mod m}nqn1-qn.$

The restricted Lambert series for $σm′(n)$ is

$∑n=1∞σm′(n)qn=∑{n∣n≡0 mod m, n≡1 mod m, n≡m-1 mod m}nqn1-qn.$

The following theorems connect $pm′(n), σm′(n)$ and the generalized nth-gonal numbers.

### Theorem 2.3.3

Let m ≥ 3 and n ∈ ℕ.

$σm′(n)=∑k=0∞(-1)k+1(Pm+2,k·pm′(n-Pm+2,k)+Qm+2,k·pm′(n-Qm+2k)).$
Proof

Let m ∈ ℕ such that m ≥ 3. Define the function Gm(x) as follows:

$Gm(x)=∏k=1∞(1-xkm)(1-xkm-(m-1))(1-xkm-1).$

Taking the logarithm of Gm(x), we get

$ln(Gm(x))=∑k=1∞(ln(1-xkm)+ln(1-xkm-(m-1))+ln(1-xkm-1)).$

Differentiating and then multiplying by x, we get

$-xGm′(x)Gm(x)=∑k=1∞(kmxkm1-xkm+(km-(m-1))xkm-(m-1)1-xkm-(m-1)+(km-1)xkm-11-xkm-1).$

By (2.11),

$1Gm(x)=∑n=0∞pm′(n)xn.$

By (2.12),

$xGm′(x)=∑n=0∞nem+2,nxn.$

Putting (2.13), (2.16), (2.17) into (2.15), we get

$(∑n=0∞pm′(n)xn) (∑n=0∞nem+2,nxn)=-∑n=1∞σm′(n)xn.$

We get the desired result by comparing coefficients of xn on both sides of the equation.

### Theorem 2.3.4

Let m ≥ 3 and n ∈ ℕ.

$σm′(n)=-nem+2,n+∑k=0n-1(-1)k+1 (σm′(n-Pm+2,k)+σm′(n-Qm+2,k)).$
Proof

By (2.13), (2.15), (2.17), we get

$-∑n=0∞nem+2,nxn=(∑n=0∞em+2,nxn) (∑n=0∞σm′(n)xn).$

We get the desired result by comparing coefficients of xn on both sides of the equation.

### Theorem 2.3.5

Let m ≥ 3 and n ∈ ℕ.

$npm′(n)=∑k=0∞σm′(k)pm′(n-k).$
Proof

Based on the definition of Gm(x) in (2.14), let the function Fm(x) be

$Fm(x)=1Gm(x).$

Then

$Fm′(x)=-Gm′(x)(Gm(x))2=-Gm′(x)Gm(x)·1Gm(x)=-Gm′(x)Gm(x)·Fm(x).$

By (2.13), (2.15),

$-Gm′(x)Gm(x)=∑n=1∞σm′(n)xn-1.$

By (2.16),

$Fm′(x)=∑n=0∞npm′(n)xn-1.$

Putting (2.16), (2.20), (2.21) into (2.19) and multiplying by x, we get

$∑n=0∞npm′(n)xn=(∑n=0∞σm′(n)xn) (∑n=0∞pm′(n)xn).$

We get the desired result by comparing coefficients of xn on both sides of the equation (2.22).

### Remark 2.3.6

In the case m = 3, Theorem 2.3.5 is identical to (1.1).

3. Identities for Bell Polynomials

### 3.1. Preliminaries

Let (x1, x2, …) be a sequence of real numbers. The partial exponential Bell polynomials are polynomials given by

$Bn,k(x1,x2,…,xn-k+1)=∑π(n,k)n!j1!j2!…jn-k+1!(x11!)j1 (x22!)j2⋯(xn-k+1(n-k+1)!)jn-k+1$

where π(n, k) is the positive integer sequence (j1, j2, jnk+1) satisfies the following equations:

$j1+j2+⋯+jn-k+1=k,j1+2j2+⋯+(n-k+1)jn-k+1=n.$

For n ≥ 1, the nth-complete Bell polynomial Bn(x1, …, xn) is the following:

$Bn(x1…xn)=∑k=1nBn,k(x1,…,xn-k+1).$

The complete exponential Bell polynomials can also be defined by power series expansion as follows:

$exp (∑m=1∞xmtmm!)=∑n=0∞Bn(x1,…,xn)tnn!,$

where B0 ≡ 1. Alternatively, the complete Bell polyomials can be recursively defined by

$Bn+1(x1,…,xn+1)=∑i=0n(ni) Bn-i(x1,…,xn-i)xi+1.$

One interesting property of the Bell polynomials is that there exists an inversion formula in the following sense. If we define

$yn=Bn(x1,x2,…,xn),$

then

$xn=∑k=1n(-1)k-1(k-1)!·Bn,k(y1,…,yn-k+1).$

For detailed properties of such inverse formulas, see the paper written by Chaou et. al [5]. Bell polynomials were first introduced by Bell [2]. The books written by Comtet [7] and Riordan [10] serve as excellent references for the numerous applications of Bell polynomials in combinatorics. Recently, there has been extensive research in finding identities on (partial/complete) Bell polynomials. The paper written by W. Wang and T. Wang [12] provides many interesting identities for partial Bell polynomials. Bouroubi and Benyahia-Tani [3] and the author [8] proved some new identities for complete Bell polynomials based on Ramanujan’s congruences.

### 3.2. Identities for Complete Bell Polynomials and Some Corollaries

We recall the notations $σm′(n),pm′(n)$ and eg,n used in Section 2.2 and Section 2.3.

Theorem 3.2.1

Let n ∈ ℕ and m ≥ 3.

$Bn(d1,d2,…,dn)=n!·em+2,n$

where$dn=-(n-1)!·σm′(n)$.

Proof

By (2.14), we take the logarithm of Gm(x) and use the formal power series expansion of ln(1 − x) to get

$ln(Gm(x))=∑k=1∞(ln(1-xkm)+ln(1-xkm-(m-1))+ln(1-xkm-1))$$=-∑k=1∞∑j=1∞(xkm)jj-∑k=1∞∑j=1∞(xkm-(m-1))jj-∑k=1∞∑j=1∞(xkm-1)jj=-∑k=1∞σm′(n)nxn=∑k=1∞(-(n-1)!·σm′(n))xnn!.$

Taking exponential on both sides of (3.6) to get

$Gm(x)=exp (∑k=1∞dnxnn!)=∑n=0∞Bn(d1,…,dn)xnn!$

where $dn=-(n-1)!·σm′(n)$ and the last equality is due to (3.1). By (2.12),

$Gm(x)=∑n=0∞em+2,nxn.$

Now the result is clear by comparing (3.7) and (3.8).

### Theorem 3.2.2

Let n ∈ ℕ and m ≥ 3.

$Bn(c1,c2,…,cn)=n!·pm′(n)$

where$cn=(n-1)!·σm′(n)$.

Proof

It is essentially the same as the proof of Theorem 3.2.1. Let the function Fm(x) be

$Fm(x)=1Gm(x).$

By taking logarithm of Fm(x) and using the formal power series of ln(1−x), we get

$ln(Fm(x))=-∑k=1∞(ln(1-xkm)+ln(1-xkm-(m-1))+ln(1-xkm-1))=∑k=1∞((n-1)!·σm′(n))xnn!.$

We get the result as desired by taking exponential on both sides of (3.9), applying (3.1) and generating function of $pm′(n)$ in (2.11).

### Remark 3.2.3

It is worthwhile to notice that Theorem 2.3.4 (resp. Theorem 2.3.5) can be proved by applying Theorem 3.2.1 (resp. Theorem 3.2.2) and the convolution properties of complete Bell polynomials shown in (3.2). More precisely, by Theorem 3.2.2, the equation (3.2) becomes

$(n+1)!·pm′(n+1)=∑i=0n(ni) (n-i)!·pm′(n-i)·i!·σm′(i+1)(n+1)pm′(n+1)=∑i=0npm′(n-i)σm′(i+1).$

Now it is obvious that (3.10) is equivalent to Theorem 2.3.5. Likewise, Theorem 2.3.4 can be proved by using Theorem 3.2.1 and the equation (3.2).

By the inversion formulas of Bell polynomials as stated in (3.3) and (3.4), we immediately obtain the following corollaries due to Theorem 3.2.1 and Theorem 3.2.2 respectively.

### Corollary 3.2.4

Let n ∈ ℕ and m ≥ 3.

$σm′(n)=1(n-1)!∑k=1∞(-1)k(k-1)!·Bn,k(1!·em+2,1,2!·em+2.2,⋯,(n-k+1)!·em+2,n-k+1).$

### Corollary 3.2.5

Let n ∈ ℕ and m ≥ 3.

$σm′(n)=1(n-1)!∑k=1∞(-1)k-1(k-1)!·Bn,k(1!·pm′(1),2!·pm′(2),…,(n-k+1)!·pm′(n-k+1)).$

It might come as a surprise that RHS of the formula in Corollary 3.2.4 is equal to the RHS of the formula in Corollary 3.2.5 as the former one leads to a simple computation (many terms em+2,n are zeros) while the latter one gives a rather complicated computation due to the terms $pm′(n)$.

Acknowledgements

The author is grateful to the editor and the referees for carefully reading the paper and pointing out some mistakes in the first draft of it. Their comments were helpful to improve the quality of the article. The author is supported by UAEU Startup Grant 2016 (G00002235) from United Arab Emirates University.

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