We locate the peak of the distribution of noncentral Stirling numbers of the first kind
by determining the value of the index corresponding to the maximum value of the distribution.
1. Introduction
In 1982, Koutras [1] introduced the noncentral Stirling numbers of the first and second kind as a natural extension of the definition of the classical Stirling numbers, namely, the expression of the factorial (x)n in terms of powers of x and vice versa. These numbers are, respectively, denoted by sa(n,k) and Sa(n,k) which are defined by means of the following inverse relations:(1)tn=∑k=0n1k!dkdtktvt=at-ak,(2)t-an=∑k=0n1k!Δkt-ant=0tk,where a, t are any real numbers, n is a nonnegative integer, and(3)san,k=1k!dkdtktvt=a,San,k=1k!Δkt-ant=0.The numbers satisfy the following recurrence relations:(4)san+1,k=san,k-1+a-nsan,k,(5)San+1,k=San,k-1+k-aSan,kand have initial conditions(6)sa(0,0)=1,sa(n,0)=an,sa(0,k)=0,n,k≠0,Sa(0,0)=1,Sa(n,0)=-an,Sa(0,k)=0,n,k≠0.It is worth mentioning that for a given negative binomial distribution Y and the sum X=X1+X2+⋯+Xk of k independent random variables following the logarithmic distribution, the numbers sa(n,k) appeared in the distribution of the sum W=X+Y, while the numbers Sa(n,k) appeared in the distribution of the sum Z=X^+Y^ where X^ is the sum of k independent random variables following the truncated Poisson distribution away from zero and Y^ is a Poisson random variable. More precisely, the probability distributions of W and Z are given, respectively, by(7)PW=n=k!1-θ-s-log1-θkθnn!-1n-ks-sn,k,P[Z=n]=k!emel-1klnn!-1n-kS-m/l(n,k).For a more detailed discussion of noncentral Stirling numbers, one may see [1].
Determining the location of the maximum of Stirling numbers is an interesting problem to consider. In [2], Mezö obtained results for the so-called r-Stirling numbers which are natural generalizations of Stirling numbers. He showed that the sequences of r-Stirling numbers of the first and second kinds are strictly log-concave. Using the theorem of Erdös and Stone [3] he was able to establish that the largest index for which the sequence of r-Stirling numbers of the first kind assumes its maximum is given by the approximation(8)Kn,r(1)=r+logn-1r-1-1r+o1.
Following the methods of Mezö, we establish strict log-concavity and hence unimodality of the sequence of noncentral Stirling numbers of the first kind and, eventually, obtain an estimating index at which the maximum element of the sequence of noncentral Stirling numbers of the first kind occurs.
2. Explicit Formula
In this section, we establish an explicit formula in symmetric function form which is necessary in locating the maximum of noncentral Stirling numbers of the first kind.
Let fi(x), i=1,2,…,n be differentiable functions and let Fn(x)=∏i=1nfi(x). It can easily be verified that, for all n≥3,(9)Fn′(x)=∑1≤j1<j2<⋯<jn-1≤ni∈Nn∖{j1,j2,…,jn-1}fi′(x)∏k=1n-1fjk(x).Now, consider the following derivative of (ξj+a)n when n=1,2: (10)ddξξj+a1=j,ddξξj+a2=(ξj+a)j+(ξj+a-1)j.Then, for n≥3 and using (9), we get(11)1jnddξξj+an=∑0≤j1<j2<⋯<jn-1≤n-1∏k=1n-1ξ+a-jkj.Then, we have the following lemma.
Lemma 1.
For any nonnegative integers n and k, one has(12)1jndkdξkξj+an=∑0≤j1<j2<⋯<jn-k≤n-1k!∏q=1n-kξ+a-jqj.
Proof.
We prove by induction on k. For k=0, (12) clearly holds. For k=1, (12) can easily be verified using (11). Suppose for m≥1, (13)1jndmdξmξj+an=∑0≤j1<j2<⋯<jn-m≤n-1m!∏q=1n-mξ+a-jqj.Then,(14)1jndm+1dξm+1ξj+anWW=m!∑0≤j1<j2<⋯<jn-m≤n-1ddξ∏q=1n-mξ+a-jqj,where the sum has nn-m=n(n-1)(n-2)⋯(n-m+1)/m! terms and its summand(15)ddξ∏q=1n-mξ+a-jqj=∑i1<i2<⋯<in-m-1∏q=1n-m-1ξ+a-iqj,iq∈{j1,j2,…,jn-m} has n-mn-m-1=(n-m)(n-m-1)!/(n-m-1)!=n-m terms. Therefore, the expansion of (dm+1/dξm+1)(ξj+a)n has a total of n(n-1)⋯(n-m+1)(n-m)/m! terms of the form ∏q=1n-m-1(ξ+(a-jq)/j). However, if the sum is evaluated over all possible combinations j1j2⋯jn-m-1 such that 0≤j1<j2<⋯<jn-m-1≤n-1, then the sum has nn-m-1=n(n-1)⋯(n-m)(n-m-1!)/(m+1)!(n-m-1)!=(1/m+1)(nn-1⋯n-m/m!) distinct terms. It follows that every term ∏q=1n-m-1(ξ+(a-jq)/j) appears m+1 times in the expansion of (1/jn)(dm+1/dξm+1)(ξj+a)n. Thus we have(16)1jndm+1dξm+1ξj+anWW=m!∑0≤j1<j2<⋯<jn-m-1≤n-1m+1∏q=1n-mξ+a-jqjWW=(m+1)!∑0≤j1<j2<⋯<jn-m-1≤n-1∏q=1n-mξ+a-jqj.
Lemma 2.
Let s^(n,k;a)=1/k!limξ→0((∑j=0k(-1)k-jkj(dk/dξk)(ξj+a)n)/k!). Then(17)s^n,k;a=∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-ka-jq.
Proof.
Using Lemma 1,(18)s^n,k;aWW=1k!limξ→0∑j=0k-1k-jkjjn∑0≤j1<⋯<jn-k≤n-1k!∏n-kWWWWWWWW×∏q=1n-kξ+a-jqjk!-1.Note that ∏q=1n-k(ξ+(a-jq)/j)=(1/jn-k)∏q=1n-k(ξj+a-jq). Hence, the expression at the right-hand side of (18) becomes(19)1k!limξ→0∑j=0k-1k-jkjjkWWWW×∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-k(ξj+a-jq)WW=1k!∑j=0k-1k-jkjjkWWWWW×∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-ka-jq,which boils down to(20)∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-ka-jq,since(21)1k!∑j=0k-1k-jkjjk=Sk,k=1,where S(n,k) denote the Stirling numbers of the second kind.
Theorem 3.
The noncentral Stirling numbers of the first kind equal(22)san,k=s^n,k;a=∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-ka-jq.
Proof.
We know that(23)∑0≤j1<j2<⋯<jn-k+1≤n∏q=1n-k+1a-jqis equal to the sum of the products (a-j1)(a-j2)⋯(a-jn+1-k) where the sum is evaluated overall possible combinations j1j2⋯jn+1-k, ji∈{0,1,2,…,n}. These possible combinations can be divided into two: the combinations with ji=n for some i∈{1,2,…,n-k+1} and the combinations with ji≠n for all i∈{0,1,2,…,n-k+1}. Thus (24)∑0≤j1<j2<⋯<jn-k+1≤n∏q=1n-k+1a-jqis equal to(25)∑0≤j1<j2<⋯<jn-k+1≤n-1∏q=1n-k+1a-jq+a-nWW×∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-ka-jq.This implies that(26)s^n+1,k;a=s^n,k-1;a+a-ns^n,k;a.This is exactly the triangular recurrence relation in (4) for sa(n,k). This proves the theorem.
The explicit formula in Theorem 3 is necessary in locating the peak of the distribution of noncentral Stirling numbers of the first kind. Besides, this explicit formula can also be used to give certain combinatorial interpretation of sa(n,k).
A 0-1 tableau, as defined in [4] by de Médicis and Leroux, is a pair φ=(λ,f), where(27)λ=λ1≥λ2≥⋯≥λkis a partition of an integer m, and f=(fij)1≤j≤λi is a “filling” of the cells of corresponding Ferrers diagram of shape λ with 0’s and 1’s, such that there is exactly one 1 in each column. Using the partition λ=(5,3,3,2,1) we can construct 60 distinct 0-1 tableaux. One of these 0-1 tableaux is given in the following figure with f14=f15=f23=f31=f42=1, fij=0 elsewhere (1≤j≤λi): Also, as defined in [4], an A-tableau is a list ϕ of column c of a Ferrers diagram of a partition λ (by decreasing order of length) such that the lengths |c| are part of the sequence A=(ai)i≥0, ai∈Z+∪{0}. If TdA(h,r) is the set of A-tableaux with exactly r distinct columns whose lengths are in the set A={a0,a1,…,ah}, then |TdA(h,r)|=h+1r. Now, transforming each column c of an A-tableau in TdA(n-1,n-k) into a column of length ω(|c|), we obtain a new tableau which is called Aω-tableau. If ω(|c|)=|c|, then the Aω-tableau is simply the A-tableau. Now, we define an Aω(0,1)-tableau to be a 0-1 tableau which is constructed by filling up the cells of an Aω-tableau with 0’s and 1’s such that there is only one 1 in each column. We use TdAω(0,1)(n-1,n-k) to denote the set of such Aω(0,1)-tableaux.
It can easily be seen that every (n-k) combination j1j2⋯jn-k of the set {0,1,2,…,n-1} can be represented geometrically by an element ϕ in TdA(n-1,n-k) with ji as the length of (n-k-i+1)th column of ϕ where A={0,1,2,…,n-1}. Hence, with ω(|c|)=a-|c|, (22) may be written as(29)sa(n,k)=∑ϕ∈TdAn-1,n-k∏c∈ϕωc.Thus, using (29), we can easily prove the following theorem.
Theorem 4.
The number of Aω(0,1)-tableaux in TdAω(0,1)(n-1,n-k) where A={0,1,2,…,n-1} such that ω(|c|)=a-|c| is equal to sa(n,k).
Let ϕ be an A-tableau in TdA(n-1,n-k) with A={0,1,2,…,n-1}, and(30)ωAϕ=∏c∈ϕωc=∏i=1n-ka-ji,ji∈{0,1,2,…,n-1}.If a=a1+a2 for some a1 and a2, then, with ω*(j)=a2-j,(31)ωAϕ=∏i=1n-ka1+ω*ji=∑r=0n-ka1n-k-r∑ji≤q1<q2<⋯<qr≤jn-k∏i=1rω*qi.Suppose Bϕ is the set of all A-tableaux corresponding to ϕ such that for each ψ∈Bϕ either
ψ has no column whose weight is a1, or
ψ has one column whose weight is a1, or
⋮
ψ has n-k columns whose weights are a1. Then, we may write (32)ωA(ϕ)=∑ψ∈BϕωA(ψ).
Now, if r columns in ψ have weights other than a1, then(33)ωA(ψ)=a1n-k-r∏i=1rω*qi,where q1,q2,…,qr∈{j1,j2,…,jn-k}. Hence, (29) may be written as(34)sa(n,k)=∑ϕ∈TdA(n-1,n-k)∑ψ∈BϕωAψ.Note that for each r, there correspond n-kr tableaux with r distinct columns having weights w*(qi), qi∈{j1,j2,…,jn-k}. Since TdA(n-1,n-k) has nk elements, for each ϕ∈TdA(n-1,n-k), the total number of A-tableaux ψ corresponding to ϕ is (35)nkn-krelements. However, only nr tableaux in Bϕ with r distinct columns of weights other than a1 are distinct. Hence, every distinct tableau ψ appears (36)nkn-krnr=n-rktimes in the collection. Consequently, we obtain(37)san,k=∑r=0n-kn-rka1n-k-r∑ψ∈Br∏c∈ψω*c,where Br denotes the set of all tableaux ψ having r distinct columns whose lengths are in the set {0,1,2,…,n-1}. Reindexing the double sum, we get (38)sa(n,k)=∑j=knjka1j-k∑ψ*∈Bn-j∏c∈ψ*ω*c.Clearly, Bn-j=TdA(n-1,n-j). Thus, using (22), we obtain the following theorem.
Theorem 5.
The numbers sa(n,k) satisfy the following identity:(39)sa(n,k)=∑j=knjka1j-ksa2n,j,where a=a1+a2 for some numbers a1 and a2.
The next theorem contains certain convolution-type formula for sa(n,k) which will be proved using the combinatorics of A-tableau.
Theorem 6.
The numbers sa(n,k) have convolution formula(40)sa(m+j,n)=∑k=0nsam,ksa-mj,n-k.
Proof.
Suppose that ϕ1 is a tableau with exactly m-k distinct columns whose lengths are in the set A1={0,1,2,…,m-1} and ϕ2 is a tableau with exactly j-n+k distinct columns whose lengths are in the set A2={m,m+1,m+2,…,m+j-1}. Then ϕ1∈TdA1(m-1,m-k) and ϕ2∈TdA2(j-1,j-n+k). Notice that by joining the columns of ϕ1 and ϕ2, we obtain an A-tableau ϕ with m+j-n distinct columns whose lengths are in the set A={0,1,2,…,m+j-1}; that is, ϕ∈TdA(m+j-1,m+j-n). Hence(41)∑ϕ∈TdA(m+j-1,m+j-n)ωA(ϕ)WW=∑k=0n∑ϕ1∈TdA1(m-1,m-k)ωA1(ϕ1)WWW×∑ϕ2∈TdA2(j-1,j-n+k)ωA2(ϕ2).Note that(42)∑ϕ2∈TdA2(j-1,j-n+k)ωA2(ϕ2)WW=∑m≤g1<g2<⋯<gj-n+k≤m+j-1∏q=1j-n+k(a-gq)WW=∑0≤g1<g2<⋯<gn-k-j≤j-1∏q=1j-n+k(a-(m+gq))WW=sa-m(j,n-k).
Also, using (29), we have(43)∑ϕ1∈TdA1(m-1,m-k)ωA1(ϕ1)=sa(m,k)∑ϕ∈TdA(m+j-1,m+j-n)ωA(ϕ)=sa(m+j,n).Thus,(44)sa(m+j,n)=∑k=0nsa(m,k)sa-m(j,n-k).
The following theorem gives another form of convolution formula.
Theorem 7.
The numbers sa(n,k) satisfy the second form of convolution formula (45)sa(n+1,m+j+1)=∑k=0nsa(k,n)sa-(k+1)(n-k,j).
Proof.
Let
ϕ1 be a tableau with k-m columns whose lengths are in A1=0,1,…,k-1,
ϕ2 be a tableau with n-k-j columns whose lengths are in A2={k+1,…,n}.
Then ϕ1∈TdA1(k-1,k-m); ϕ2∈TdA2(n-k-1,n-k-j). Using the same argument above, we can easily obtain the convolution formula.3. The Maximum of Noncentral Stirling Numbers of the First Kind
We are now ready to locate the maximum of sa(n,k). First, let us consider the following theorem on Newton’s inequality [5] which is a good tool in proving log-concavity or unimodality of certain combinatorial sequences.
Theorem 8.
If the polynomial a1x+a2x2+⋯+anxn has only real zeros then(46)ak2≥ak+1ak-1kk-1n-k+1n-k(k=2,…,n-1).
Now, consider the following polynomial:(47)∑k=0nsan,kt+ak.This polynomial is just the expansion of the factorial 〈t〉n=t(t+1)(t+2)⋯(t+n-1) which has real roots 0,-1,-2,…,-n+1. If we replace t by t-a, we see at once that the roots of the polynomial ∑k=0nsa(n,k)tk are a,a-1,…,a-n+1. Applying Newton’s Inequality completes the proof of the following theorem.
Theorem 9.
The sequence {sa(n,k)}k=0n is strictly log-concave and, hence, unimodal.
By replacing t with -t, the relation in (1) may be written as(48)〈t〉n=∑k=0n-1n-ksan,kt+ak,where 〈t〉n=t(t+1)(t+2)⋯(t+n-1). Note that, from Theorem 3 with a<0, (49)san,k=-1n-k∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-kb+jq,where b=-a>0. Now, we define the signless noncentral Stirling number of the first kind, denoted by |sa(n,k)|, as(50)san,k=-1n-ksan,k=∑0≤j1<j2<⋯<jn-k≤n-1∏q=1n-kb+jq.
To introduce the main result of this paper, we need to state first the following theorem of Erdös and Stone [3].
Theorem 10 (see [<xref ref-type="bibr" rid="B5">3</xref>]).
Let u1<u2<⋯ be an infinite sequence of positive real numbers such that(51)∑i=1∞1ui=∞,∑i=1∞1ui2<∞.Denote by ∑n,k the sum of the product of the first n of them taken k at a time and denote by Kn the largest value of k for which ∑n,k assumes its maximum value. Then (52)Kn=n-∑i=1n1ui-∑i=1n1ui21+1ui-1+o(1).
We also need to recall the asymptotic expansion of harmonic numbers which is given by(53)11+12+⋯+1n=logn+γ+o1,where γ is the Euler-Mascheroni constant.
The following theorem contains a formula that determines the value of the index corresponding to the maximum of the sequence {|sa(n,k)|}k=0n.
Theorem 11.
The largest index for which the sequence {|sa(n,k)|}k=0n assumes its maximum is given by the approximation (54)ka,n=logb+nb+o1,where [x] is the integer part of x and b=-a, a<0.
Proof.
Using Theorem 10 and by (50), we see that |sa(n,k)|=∑n-1,n-k. Denoting by ka,n for which ∑n,n-k is maximum and with u1=b+0,u2=b+1,…,un-1=b+n-1 we have(55)ka,n=∑i=0n-11b+i-∑i=0n-11b+i21+1b+i-1+o1=∑i=0n-11b+i-∑i=0n-11b+ib+i+1+o1=∑i=0n-11b+i+1+o1.But using (53), we see that(56)∑i=0n-11b+i+1=log(b+n)-logb.From this we get(57)ka,n=logb+nb+o1.
For the case in which a>0 we will only consider the sequence of noncentral Stirling numbers of the first kind for which a≥n.
Theorem 12.
The maximizing index for which the maximum noncentral Stirling number occurs for a≥n is given by the approximation(58)ka,n=loga+1a-n+1+o1.
Proof.
From the definition, for a≥n,sa(n,k)>0 and by Theorem 3, sa(n,k) is the sum of the products (a-j1)(a-j2)⋯(a-jn-k) where ji’s are taken from the set {0,1,2,…,n-1}. By Theorem 10, sa(n,k)=∑n,n-k. Thus with u1=a,u2=a-1,…,un-1=a-n+1 we have(59)ka,n=∑i=0n-11a-i-∑i=0n-11a-i21+1a-i-1+o1=∑i=0n-11a-i-∑i=0n-11a-ia-i+1+o1=∑i=0n-11a-i+1+o1.Again, using (53), we get(60)ka,n=loga+1a-n+1+o1.
Example 13.
The maximum element of the sequence {s-1(9,k)}k=09 occurs at (Table 1)(61)k-1,9=log1+81+o1=log9+o1≈2.
Values of s-1(n,k).
n/k
0
1
2
3
0
1
1
1
2
2
1
3
6
6
1
4
24
35
10
1
5
120
225
85
15
6
720
1624
735
175
7
5040
13132
6769
1960
8
40320
109584
118124
67284
9
362880
1026576
1172700
723680
10
3628800
10628640
12753576
8409500
Example 14.
The maximum element of the sequence {s10(10,k)}k=010 occurs at (Table 2)(62)k10,10=log10+110-10+1+o1=log11+o1≈2.
Values of s10(n,k).
n/k
0
1
2
3
0
1
1
10
1
2
90
19
1
3
720
242
27
1
4
5040
2414
431
34
5
30240
19524
5000
635
6
151200
127860
44524
8175
7
604800
662640
305956
77224
8
1814400
2592720
1580508
537628
9
3628800
6999840
5753736
2655764
10
3628800
10628640
12753576
8409500
We know that the classical Stirling numbers of the first kind are special cases of sa(n,k) by taking a=0. However, formulas in Theorems 11 and 12 do not hold when a=0. Hence, these formulas are not applicable to determine the maximum of the classical Stirling numbers. Here, we derive a formula that determines the value of the index corresponding to the maximum of the signless Stirling numbers of the first kind.
The signless Stirling numbers of the first kind [6] are the sum of all products of n-k different integers taken from {1,2,3,…,n-1}. That is, (63)sn,k=∑1≤i1<i2<⋯<in-k≤n-1i1i2⋯in-k.Using Theorem 10, |s(n,k)|=∑n-1,n-k. We use kn to denote the largest value of k for which ∑n-1,n-k is maximum. With u1=1,u2=2,…,un-1=n-1 we have(64)kn=1+∑i=1n-11i-∑i=1n-11i21+1i-1+o1=1+∑i=1n-11i-∑i=1n-11i-1i+1+o1=1+∑i=1n-11i+1+o1.Using (53), we see that(65)∑i=1n-111+i=logn-log1+γ+o1.Therefore, we have(66)kn=logn+γ+o1.
Example 15.
It is shown in Table 3 that the maximum value of |s(n,k)| when n=7 occurs at k=2. Using (66), it can be verified that the maximum element of the sequence {|s(7,k)|}k=07 occurs at(67)k7=log7+γ+o1=1.95+0.5772+o1=2.53+o1≈2.Moreover, when n=10, the maximum value occurs at(68)k10=log10+γ+o1=2.30+0.5772+o1=2.8772+o1≈3.
Values of |s(n,k)| for 0≤n≤10.
s(n,k)
0
1
2
3
4
5
0
1
1
0
1
2
0
1
1
3
0
2
3
1
4
0
6
11
6
1
5
0
24
50
35
10
1
6
0
120
274
225
85
15
7
0
720
1764
1624
735
175
8
0
540
13068
13132
6769
1960
9
0
40320
109584
118124
67284
22449
10
0
362880
1026576
1172700
723680
269325
Recently, a paper by Cakić et al. [7] established explicit formulas for multiparameter noncentral Stirling numbers which are expressible in symmetric function forms. One may then try to investigate the location of the maximum value of these numbers using the Erdös-Stone theorem.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The authors wish to thank the referees for reading the paper thoroughly.
KoutrasM.Noncentral Stirling numbers and some applicationsMezöI.On the maximum of r-Stirling numbersErdösP.On a conjecture of Hammersleyde MédicisA.LerouxP.Generalized Stirling numbers, convolution formulae and p,q-analoguesLiebE. H.Concavity properties and a generating function for Stirling numbersComtetL.CakićN. P.El-DesoukyB. S.MilovanovićG. V.Explicit formulas and combinatorial identities for generalized Stirling numbers