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 Computer Modeling in Engineering & Sciences

DOI: 10.32604/cmes.2022.019941

ARTICLE

Partial Bell Polynomials, Falling and Rising Factorials, Stirling Numbers, and Combinatorial Identities

1College of Mathematics and Physics, Inner Mongolia Minzu University, Tongliao, 028043, China
2School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, 454010, China
3School of Mathematical Sciences, Tiangong University, Tianjin, 300387, China
*Corresponding Authors: Bai-Ni Guo. Email: bai.ni.guo@gmail.com; Feng Qi. Email: qifeng618@gmail.com
Received: 25 October 2021; Accepted: 24 January 2022

Abstract: In the paper, the authors collect, discuss, and find out several connections, equivalences, closed-form formulas, and combinatorial identities concerning partial Bell polynomials, falling factorials, rising factorials, extended binomial coefficients, and the Stirling numbers of the first and second kinds. These results are new, interesting, important, useful, and applicable in combinatorial number theory.

Keywords: Connection; equivalence; closed-form formula; combinatorial identity; partial Bell polynomial; falling factorial; rising factorial; binomial coefficient; Stirling number of the first kind; Stirling number of the second kind; problem

1  Preliminaries

In this paper, we use the notation

={1,2,},   ={1,2,},0={0,1,2,},={0,±1,±2,},  =(,),   ={x+iy:x,y,i=1}.

The partial Bell polynomials, also known as the Bell polynomials of the second kind, in combinatorics can be denoted and defined by

𝐁n,k(x1,x2,,xnk+1)=1ink+1,i{0}N,i=1nk+1ii=n,i=1nk+1i=kn!i=1nk+1i!i=1nk+1(xii!)i(1)

for nkN0. See TheoremA on page 134 in [1]. The partial Bell polynomials satisfy the identity

𝐁n,k(abx1,ab2x2,,abnk+1xnk+1)=akbn𝐁n,k(x1,x2,,xnk+1)(2)

for nkN0. See page 135 in [1].

The double factorial of negative odd integers −(2k+1) is defined by

(2k1)!!=(1)k(2k1)!!=(1)k2kk!(2k)!,kN0.

The falling factorial zn and the rising factorial (z)n for nN0 and zC can be defined by

zn=k=0n1(zk)={z(z1)(zn+1),nN1,n=0(3)

and

(z)n==0n1(z+)={z(z+1)(z+n1),nN1,n=0

respectively. It is easy to verify that

(z)n=(1)nzn(4)

and

zn=(1)n(z)n.(5)

See page 167 in [2] and related texts in the paper [3].

The Stirling numbers of the first kind s(n, k) for nkN0 can be analytically generated (see page 51 in [1]) by

[ln(1+x)]kk!=n=ks(n,k)xnn!,|x|<1

and can be explicitly computed (see Corollary 2.3 in [4]) by

|s(n+1,k+1)|=n!1=kn112=k11112k1=2k211k1k=1k111k

for nkN. The Stirling numbers of the second kind S(n, k) for nkN0 can be analytically generated (see page 51 in [1]) by

(ex1)kk!=n=kS(n,k)xnn!

and can be explicitly computed (see TheoremA on page 204 in [1]) by

S(n,k)={(1)kk!=0k(1)(k)n,n>kN0;1,n=kN0.(6)

For more information on the Stirling numbers of the first and second kinds s(n, k) and S(n, k), please refer to the papers [5,6] and the monographs [7,8].

The extended binomial coefficient (zw) for z,wC is defined in [9] by

(zw)={Γ(z+1)Γ(w+1)Γ(zw+1),zN,w,zwN0,zN,wN or zwNzww!,zN,wN0zzw(zw)!,z,wN,zwN00,z,wN,zwN,zN,wZ

in terms of the falling factorial zw, which is defined by (3), and the classical Euler’s gamma function Γ(z), which can be defined (see Chapter3 in [10]) by

Γ(z)=limnn!nzk=0n(z+k),zC{0,1,2,}.

On page 206 in [1] and on page 165 in [7], there are two relations

zn==0ns(n,)z,zC,nN0(7)

and

n!(zn)==0ns(n,)z,nN0,zC.(8)

The falling factorial zn and the rising factorial (z)n can be represented by

zn=n!(zn)=Γ(z+1)Γ(zn+1)and(z)n=(1)nn!(zn)=Γ(z+n)Γ(z)

for zC and nN0.

In this paper, we will collect, discuss, and find out several connections, equivalences, closed-form formulas, and combinatorial identities concerning partial Bell polynomials 𝐁n,k, falling factorials zn, rising factorials (z)n, extended binomial coefficients (zw), and the Stirling numbers of the first and second kinds s(n, k) and S(n, k).

2  Equivalences

Among the Stirling numbers of the first and second kinds s(n, k) and S(n, k), the falling factorial αn, and extended binomial coefficient (αn), there are the following beautiful equivalences.

Theorem 2.1. For nkN0 and αC, we have

=kns(n,)αS(,k)=(1)kk!=0k(1)(k)αn(9)

and

=kns(n,)αS(,k)=(1)kn!k!=0k(1)(k)(αn).(10)

Proof. In Remark 3.1 of [11], the formula

𝐁n,k(1,1λ,(1λ)(12λ),,=0nk(1λ))=(1)kk!=0k(1)(k)q=0n1(qλ)(11)

for nkN0 and λC was concluded. An equivalence of the formula (11) is

𝐁n,k(α1,α2,,αnk+1)=(1)kk!=0k(1)(k)αn(12)

for nkN0 and αC, which was proved in Theorems 2.1 and 4.1 of [2].

The formulas (11) and (12) can be rewritten respectively as

𝐁n,k(1,1λ,(1λ)(12λ),,=0nk(1λ))={(1)kλnn!k!=0k(1)(k)(/λn),λ0S(n,k),λ=0

and

𝐁n,k(α1,α2,,αnk+1)=(1)kn!k!=0k(1)(k)(αn)(13)

for nkN0 and α,λC, as done in Remark 7 of [12].

Considering the formulas (6), (7), and (8) in the formulas (12) or (13), we can derive

𝐁n,k(α1,α2,,αnk+1)=j=kns(n,j)αjS(j,k)(14)

for nkN0 and αC.

Combining (12), (13), and (14) results in

𝐁n,k(α1,α2,,αnk+1)=(1)kk!=0k(1)(k)αn=(1)kn!k!=0k(1)(k)(αn)==kns(n,)αS(,k)(15)

for nkN0 and αC. The equivalences in (9) and (10) are thus proved. The proof of Theorem 2.1 is complete.

3  Simpler Closed-Form Formulas

When taking α=±1,±2,12 in (9) and (10) respectively, we can derive several simpler closed-form combinatorial identities.

Theorem 3.1. For nkN0, we have

j=kns(n,j)1jS(j,k)=(0nk), (16)

j=kns(n,j)2jS(j,k)=n!k!(knk)22kn, (17)

j=kns(n,j)(12)jS(j,k)=(1)n+k[2(nk)1]!!2n(2nk12(nk)), (18)

j=kns(n,j)(1)jS(j,k)=(1)nn!k!(n1k1), (19)

and

j=kns(n,j)(2)jS(j,k)=(1)n+kn!k!=0k(1)(k)(n+21n).(20)

Proof. By the definition (1), we can easily deduce that, for nkN0,

𝐁n,k(1,0,,0)=𝐁n,k(11,12,,1nk+1)=j=kns(n,j)1jS(j,k)=(0nk),(21)

where we used the relation (15). See also pages 167–168 in [2]. The identity (16), which recovers the first one in (28) below, is thus proved.

In Theorem 5.1 of [13] and in Section 3 of [14], the formula

𝐁n,k(x,1,0,,0)=12nkn!k!(knk)x2kn(22)

was established for nkN0, where we assumed (00)=1 and (pq)=0 for q>pN0. Making use of the identity (2), we can derive from the formula (22) that

𝐁n,k(2,2,0,,0)=𝐁n,k(21,22,,2nk+1)=j=kns(n,j)2jS(j,k)=n!k!(knk)22kn(23)

for nkN0. The identity (17) is thus verified.

In the proof of Theorem 3.2 in [2], it was obtained that

𝐁n,k(121,122,,12nk+1)=j=kns(n,j)(12)jS(j,k)=(1)n+k[2(nk)1]!!2n(2nk12(nk))(24)

for nkN0. The identity (18) is thus proved.

Replacing α by α in (14) and utilizing (5) give

𝐁n,k((α)1,(α)2,,(α)nk+1)=(1)nj=kns(n,j)(α)jS(j,k)(25)

for nkN0 and αC. Taking α=12 in the formula (25) and making use of the identity

𝐁n,k((1)!!,1!!,3!!,,[2(nk)1]!!)=[2(nk)1]!!(2nk12(nk))

in Section 1.5 of [15] and in Theorem 1.2 of [16], we derive

𝐁n,k((12)1,(12)2,,(12)nk+1)=(1)k2n𝐁n,k((1)!!,1!!,3!!,,[2(nk)1]!!)=(1)nj=kns(n,j)(12)jS(j,k)=(1)k2n[2(nk)1]!!(2nk12(nk))

for nkN0. The identity (18) is proved again.

Employing the relation (25) and using the identities

𝐁n,k(1!,2!,3!,,(nk+1)!)=n!k!(n1k1)

and

𝐁n,k(2!,3!,,(nk+2)!)=n!k!=0k(1)k(k)(n+21n)

in Sections 1.3 and 1.9 of [15] and in Lemma 6 of [17], we acquire

𝐁n,k((1)1,(1)2,,(1)nk+1)=𝐁n,k(1!,2!,,(nk+1)!)=(1)nj=kns(n,j)(1)jS(j,k)=n!k!(n1k1)(26)

and

𝐁n,k((2)1,(2)2,,(2)nk+1)=𝐁n,k(2!,3!,,(nk)!)=(1)nj=kns(n,j)(2)jS(j,k)=n!k!=0k(1)k(k)(n+21n).(27)

The identities (19) and (20) are thus derived. The proof of Theorem 3.1 is complete.

Remark 3.1. We can regard those identities from (17) to (20) in Theorem 3.1 as generalizations of the orthogonality relations

j=0ns(n,j)S(j,k)=j=0nS(n,j)s(j,k)=(0nk),nkN0(28)

listed on page 171 in [7].

Theorem 3.2. For nN0 and αC, we have

k=0n(1)kk!𝐁n,k(α1,α2,,αnk+1)=k=0n=0k(1)(k)αn=n!k=0n=0k(1)(k)(αn)=k=0n(1)kk!j=kns(n,j)αjS(j,k)=αn(29)

and

k=0n(1)kk!𝐁n,k((α)1,(α)2,,(α)nk+1)=k=0n=0k(1)(k)(α)n=(1)nn!k=0n=0k(1)(k)(αn)=(1)nk=0n(1)kk!j=kns(n,j)(α)jS(j,k)=(α)n.(30)

Proof. Combining (12) and (14) yields

(1)kk!j=kns(n,j)αjS(j,k)==0k(1)(k)αn,nkN0.

Accordingly, similar to arguments in Lemma 2.2 of [18], we acquire

k=0n(1)kk!j=kns(n,j)αjS(j,k)=k=0n=0k(1)(k)αn==1n(1)[k=n(k)]αn==1n(1)(n+1+1)m=0ns(n,m)(α)m=m=0ns(n,m)αm=1n(1)(n+1+1)m=m=0ns(n,m)(α)m=αn

for nN, where we used the relation

k=n(k)=(n+1+1),,nN,(31)

which is a special case x = 0 and r= of the identity

k=0n(k+xr)=(n+x+1r+1)(xr+1)

in the formula (1.48) on pages 27–28 of [8], we used the relation (7) twice, and we used the equality

=1n(1)(n+1+1)m==1m(1)(n+1+1)m=(1)m,m,nN,(32)

which is a special case r = 1 and p = m of the identity

k=0rpr+1(1)k(rp+1k)(rpkr)p=1

in the formula (X.5) on page 132 of [8]. Further applying relations in (15), we conclude those relations in (29).

Replacing α by α in (29), using the identities (4), (5), and (2) in sequence, and simplifying lead to (30). The proof of Theorem 3.2 is thus complete.

Remark 3.2. The last equality in (29) can be rewritten as

k=0n(1)kk!j=kns(n,j)αjS(j,k)=j=0ns(n,j)αjk=0jS(j,k)[(1)kk!]=αn.(33)

Theorem 12.1 on page 171 of [7] reads that, if bα and ak are a collection of constants independent of n, then

an=α=0nS(n,α)bαif and only ifbn=k=0ns(n,k)ak.

Applying Theorem 12.1 on page 171 in [7] to the second equality in (33), we find

k=0nS(n,k)αk=αnj=0nS(n,j)[(1)jj!].

Considering the explicit formula (6) and utilizing (31) and (32), we arrive at

k=0nS(n,k)[(1)kk!]=k=1n=1k(1)(k)n==1n(1)nk=n(k)==1n(1)n(n+1+1)=(1)n

for nN. Therefore, we obtain

k=0nS(n,k)αk=(α)n,nN0,αC,

which is a recovery of the well-known relation

k=0nS(n,k)αk=αn,nN0,αC

in the equation (1.27) on page 19 of [10].

4  Several Combinatorial Identities

In items (3.163) and (3.164) on pages 91–92 of [8], we find two identities

k=0n(nk)(k/2)=n(n11)2n2

and

k=0n(1)k(nk)(k/2)=(1)2n2[(2n11)(2n1)].

Lemma 2.2 in [18] reads that

=0k(1)(k)(/2n)={0,k>nN0;(1)nk![2(nk)1]!!(2n)!!(2nk12(nk)),nkN0.(34)

We can also find some discussions and alternative proofs for these three identities at the sites https://math.stackexchange.com/q/1098257 and https://math.stackexchange.com/q/4235171.

Theorem 4.1. For n,kN0, the identities

=0k(1)(k)(n)={0,k>nN0;(1)kk!n!(0nk),nkN0, (35)

=0k(1)(k)(2n)={0,k>nN0;(1)k(knk)22kn,nkN0, (36)

=0k(1)(k)(n)={0,k>nN0;(1)n+k(n1k1),nkN0, (37)

=0k(1)(k)(2n)={0,k>nN0;(1)n=0k(1)(k)(n+21n),nkN0, (38)

and the identity (34) are valid.

Proof. For the case nkN0, these identities follow from Theorem 3.1, the equivalence (10), and simplifying.

For the case k>nN0, making use of the relation 8 and utilizing the explicit formula (6), we acquire

=0k(1)(k)(αn)=1n!=0k(1)(k)[n!(αn)]=1n!=0k(1)(k)q=0ns(n,q)(α)q=1n!q=0ns(n,q)αq=0k(1)(k)q=(1)kk!n!q=0ns(n,q)αqS(q,k)

for all k,nN0 and αC, where 00 was regarded as 1. Therefore, it is clear that

=0k(1)(k)(αn)=0,k>nN0,αC.

The proof of Theorem 4.1 is complete.

Remark 4.1. The identity (35) can be simplified as

=0k(1)(k)(n)={0,kn(1)k,n=k

for n,kN0.

The identities (36), (37), and (39) in Theorem 4.1 are probably new.

Theorem 4.2. For nN0, we have

k=0n(1)kk!(0nk)=(1)nn!, (39)

k=0n(1)k(knk)22k=(1)n2n(n+1), (40)

k=0n(1)k(n1k1)={1,n=0;1,n=1;0,n2, (41)

k=0n(1)k(n+2k1n)(n+1k+1)={1,n=0;2,n=1;1,n=2;0,n3, (42)

=0n(1)(/2n)(n+1+1)=(1)n(2n1)!!(2n)!!, (43)

k=0nk![2(nk)1]!!(2nk12(nk))=(2n1)!!, (44)

and

k=1n(2nk1n1)k(k+1)2k=n22n.(45)

Proof. From (29), we conclude that

k=0n(1)kk!𝐁n,k(α1,α2,,αnk+1)=αn.(46)

for nN0 and αC.

Substituting (21) into (46) gives

k=0n(1)kk!(0nk)=1n=(1)nn!.

The identity (39) is thus proved.

Substituting (23) into (46) results in

k=0n(1)k(knk)22k=2n2nn!=(1)n2n(n+1).

The identity (40) is verified.

Utilizing the relations (2) and (5), we can reformulate the identity (26) as

𝐁n,k(11,12,,1nk+1)=(1)nn!k!(n1k1).

Substituting this equality into (46) arrives at

k=0n(1)k(n1k1)=(1)n1nn!={1,n=0;1,n=1;0,n2.

The formula (41) follows.

Utilizing the relations (2) and (5), we can reformulate the identity (27) as

𝐁n,k(21,22,,2nk+1)=(1)nn!k!=0k(1)k(k)(n+21n).

Substituting this equality into (46) and employing (31) reveal

(1)n2nn!=k=0n=0k(1)(k)(n+21n)==0n(1)(n+21n)k=n(k)==0n(1)(n+21n)(n+1+1)

and

(1)n2nn!={1,n=0;2,n=1;1,n=2;0,n3.

The fourth equality (42) in Theorem 4.2 is thus proved.

Employing (31), we can rearrange the identity (43) as

(1)n(2n1)!!(2n)!!==0n(1)(/2n)k=n(k)==0n(1)(/2n)(n+1+1).

The equality (43) is deduced.

In Theorem 3.2 of [2], on page 5 in [15], and in Theorem 4.2 of [19], there is the equality

=0k(1)(k)2n=(1)nk![2(nk)1]!!2n(2nk12(nk))

for nk0. Hence, we obtain

(1)n2nk=0nk![2(nk)1]!!(2nk12(nk))=k=0n=0k(1)(k)2n==0n(1)[k=n(k)]2n==0n(1)(n+1+1)m=0ns(n,m)(2)m=m=0ns(n,m)2m=0n(1)(n+1+1)m=m=0ns(n,m)(12)m=12n=(1)n(2n1)!!2n

for nN, where we assumed 00 = 1 and used (7), (31), and (32). Hence, we acquire (44) and

k=0n=0k(1)(k)2n=(1)n(2n1)!!2n.

Combining the last one with the relation

k=0n=0k(1)(k)2n=n!k=0n=0k(1)(k)(/2n),

which is obtained by applying α=12 to the second equality in (29), yields the identity (43) again.

Substituting (24) into (46) leads to

k=0n(1)kk!(1)n+k[2(nk)1]!!2n(2nk12(nk))=12n=(2n1)!!2n

which is a recovery of the formula (44).

For kN, let sk and Sk be two sequences independent of n such that nkN. Theorem 4.4 on page 528 in [20] reads that

sn=k=1n(knk)Skif and only if(1)nnSn=k=1n(2nk1n1)(1)kksk.(47)

Letting Sn = ( −1)n22n and sn = ( −1)n2n(n+1), considering (40), applying the inversion theorem expressed by (47), and simplifying figure out the identity (45).

Remark 4.2. The formula (44) is also alternatively established in the proof of Theorem 3.2 in [18] and in Remark 5.3 of [21].

Remark 4.3. The identity (34) established in Lemma 2.2 of [18] and recovered in Theorem 4.1, the identity (36) in Theorem 4.1, and the formula (43) in Theorem 4.2 were announced at https://math.stackexchange.com/a/4268339 and https://math.stackexchange.com/a/4268341 online.

Remark 4.4. In Remark 3.4 of [18], applying the inversion theorem expressed by (47), we obtained

k=1n(1)k(knk)(2k1k)=(1)n2n1,nN

and

k=1n(1)k(2nk1n1)2(k+1)/2ksin3(k+1)π4=2nn,nN.

5  Several Problems and Numerical Demonstrations

Can one find out simpler closed-form formulas like those in Theorem 3.1 for the quantities

j=knS(n,j)(1)js(j,k),j=knS(n,j)(±2)js(j,k),j=knS(n,j)(±12)js(j,k)

for nkN0?

By the methods used in this paper, can one find out more combinatorial identities like those in Theorems (4.1) and 4.2?

In general, can one find explicit and closed-form formulas of the quantities

Bn,k(α1,α2,...,αnk+1),l=0k(1)l(kl) αl n,l=0k(1)l(kl)(αln),l=kns(n,l)αlS(l,k),l=knS(n,l)αls(l,k)

for some special values αC{0,±1,±2,±12}?

For better understanding the above problems, by the Wolfram Mathematica 12, we numerically compute the quantity

Q(k,n;α)==knS(n,)αs(,k)

for 0kn9 and list their values as

Q(0,0;α)=1,Q(0,n;α)=0,1n9;Q(1,1;α)=α,Q(1,2;α)=(α1)α,Q(1,3;α)=(α1)α(2α1),Q(1,4;α)=(α1)α(6α26α+1),Q(1,5;α)=(α1)α(2α1)(12α212α+1),Q(1,6;α)=(α1)α(120α4240α3+150α230α+1),Q(1,7;α)=(α1)α(2α1)(360α4720α3+420α260α+1),Q(1,8;α)=(α1)α(5040α615120α5+16800α48400α3+1806α2126α+1),Q(1,9;α)=(α1)α(2α1)(20160α660480α5+65520α430240α3+5292α2252α+1);Q(2,2;α)=α2,Q(2,3;α)=3(α1)α2,Q(2,4;α)=(α1)α2(11α7),Q(2,5;α)=5(α1)α2(10α212α+3),Q(2,6;α)=(α1)α2(274α3476α2+239α31),Q(2,7;α)=7(α1)α2(252α4570α3+430α2120α+9),Q(2,8;α)=(α1)α2(13068α536324α4+36560α315940α2+2771α127),Q(2,9;α)=3(α1)α2(36528α6120288α5+151368α490300α3+25550α22940α+85);Q(3,3;α)=α3,Q(3,4;α)=6(α1)α3,Q(3,5;α)=5(α1)α3(7α5),Q(3,6;α)=15(α1)α3(15α220α+6),Q(3,7;α)=7(α1)α3(232α3443α2+257α43),Q(3,8;α)=14(α1)α3(938α42310α3+1965α2660α+69),Q(3,9;α)=(α1)α3(118124α5354628α4+395660α3199690α2+43595α3025);Q(4,4;α)=α4,Q(4,5;α)=10(α1)α4,Q(4,6;α)=5(α1)α4(17α13),Q(4,7;α)=35(α1)α4(21α230α+10),Q(4,8;α)=7(α1)α4(967α31973α2+1257α243),Q(4,9;α)=42(α1)α4(1602α44200α3+3885α21470α+185);Q(5,5;α)=α5,Q(5,6;α)=15(α1)α5,Q(5,7;α)=35(α1)α5(5α4),Q(5,8;α)=70(α1)α5(28α242α+15),Q(5,9;α)=21(α1)α5(1069α32291α2+1559α331);Q(6,6;α)=α6,Q(6,7;α)=21(α1)α6,Q(6,8;α)=14(α1)α6(23α19),Q(6,9;α)=126(α1)α6(36α256α+21);Q(7,7;α)=α7,Q(7,8;α)=28(α1)α7,Q(7,9;α)=42(α1)α7(13α11);Q(8,8;α)=α8,Q(8,9;α)=36(α1)α8,Q(9,9;α)=α9.

If fixing k = 4, 5 and n = 7, 8 and regarding α as a real variable on the interval [ −9, 9], then the graphs plotted by the Wolfram Mathematica 12 are showed in Figs. 1 and 2.

Figure 1: The graphs of Q(4,7;α) and Q(4,8;α) for α[9,9]

Figure 2: The graphs of Q(5,7;α) and Q(5,8;α) for α[9,9]

6  Conclusions

In this paper, we collected, discussed, and found out significant connections, equivalences, closed-form formulas, and combinatorial identities concerning partial Bell polynomials 𝐁n,k, falling factorials zn, rising factorials (z)n, extended binomial coefficients (zw), and the Stirling numbers of the first and second kinds s(n, k) and S(n, k). These results are new, interesting, important, useful, and applicable in combinatorial number theory and other areas, as done in the papers [2227] and closely related references therein.

Acknowledgement: The authors thank anonymous referees for their careful corrections, helpful suggestions, and valuable comments on the original version of this paper.

Funding Statement: This work was supported in part by the National Natural Science Foundation of China (Grant No.12061033), by the Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region (Grants No. NJZY20119), and by the Natural Science Foundation of Inner Mongolia (Grant No. 2019MS01007), China.

Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study.

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