# The Peak of Noncentral Stirling Numbers of the First Kind

Authors: Roberto B. Corcino; Cristina B. Corcino; Peter John B. Aranas
Journal: International Journal of Mathematics and Mathematical Sciences
 (2015)

---

## Abstract

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.


---

## Body

## 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 s
a
(
n
,
k
) and S
a
(
n
,
k
) which are defined by means of the following inverse relations:(1)
t
n
=
∑
k
=
0
n
‍
1
k
!
d
k
d
t
k
t
v
t
=
a
t
-
a
k
,
(2)
t
-
a
n
=
∑
k
=
0
n
‍
1
k
!
Δ
k
t
-
a
n
t
=
0
t
k
,where a, t are any real numbers, n is a nonnegative integer, and(3)
s
a
n
,
k
=
1
k
!
d
k
d
t
k
t
v
t
=
a
,
S
a
n
,
k
=
1
k
!
Δ
k
t
-
a
n
t
=
0
.The numbers satisfy the following recurrence relations:(4)
s
a
n
+
1
,
k
=
s
a
n
,
k
-
1
+
a
-
n
s
a
n
,
k
,
(5)
S
a
n
+
1
,
k
=
S
a
n
,
k
-
1
+
k
-
a
S
a
n
,
kand have initial conditions(6)
s
a
(
0,0
)
=
1
,
s
a
(
n
,
0
)
=
a
n
,
s
a
(
0
,
k
)
=
0
,
n
,
k
≠
0
,
S
a
(
0,0
)
=
1
,
S
a
(
n
,
0
)
=
-
a
n
,
S
a
(
0
,
k
)
=
0
,
n
,
k
≠
0
.It is worth mentioning that for a given negative binomial distribution Y and the sum X
=
X
1
+
X
2
+
⋯
+
X
k of k independent random variables following the logarithmic distribution, the numbers s
a
(
n
,
k
) appeared in the distribution of the sum W
=
X
+
Y, while the numbers S
a
(
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)
P
W
=
n
=
k
!
1
-
θ
-
s
-
log
⁡
1
-
θ
k
θ
n
n
!
-
1
n
-
k
s
-
s
n
,
k
,
P
[
Z
=
n
]
=
k
!
e
m
e
l
-
1
k
l
n
n
!
-
1
n
-
k
S
-
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)
K
n
,
r
(
1
)
=
r
+
log
⁡
n
-
1
r
-
1
-
1
r
+
o
1
.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.Letf
i
(
x
), i
=
1,2
,
…
,
n be differentiable functions and let F
n
(
x
)
=
∏
i
=
1
n
‍
f
i
(
x
). It can easily be verified that, for all n
≥
3,(9)
F
n
′
(
x
)
=
∑
1
≤
j
1
<
j
2
<
⋯
<
j
n
-
1
≤
n
i
∈
N
n
∖
{
j
1
,
j
2
,
…
,
j
n
-
1
}
f
i
′
(
x
)
∏
k
=
1
n
-
1
‍
f
j
k
(
x
)
.Now, consider the following derivative of (
ξ
j
+
a
)
n when n
=
1,2: (10)
d
d
ξ
ξ
j
+
a
1
=
j
,
d
d
ξ
ξ
j
+
a
2
=
(
ξ
j
+
a
)
j
+
(
ξ
j
+
a
-
1
)
j
.Then, for n
≥
3 and using (9), we get(11)
1
j
n
d
d
ξ
ξ
j
+
a
n
=
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
1
≤
n
-
1
∏
k
=
1
n
-
1
‍
ξ
+
a
-
j
k
j
.
‍Then, we have the following lemma.Lemma 1.
For any nonnegative integersn and k, one has(12)
1
j
n
d
k
d
ξ
k
ξ
j
+
a
n
=
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
k
!
∏
q
=
1
n
-
k
‍
ξ
+
a
-
j
q
j
.Proof.
We prove by induction onk. For k
=
0, (12) clearly holds. For k
=
1, (12) can easily be verified using (11). Suppose for m
≥
1, (13)
1
j
n
d
m
d
ξ
m
ξ
j
+
a
n
=
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
m
≤
n
-
1
‍
m
!
∏
q
=
1
n
-
m
‍
ξ
+
a
-
j
q
j
.Then,(14)
1
j
n
d
m
+
1
d
ξ
m
+
1
ξ
j
+
a
n
WW
=
m
!
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
m
≤
n
-
1
‍
d
d
ξ
∏
q
=
1
n
-
m
‍
ξ
+
a
-
j
q
j
,where the sum has n
n
-
m
=
n
(
n
-
1
)
(
n
-
2
)
⋯
(
n
-
m
+
1
)
/
m
! terms and its summand(15)
d
d
ξ
∏
q
=
1
n
-
m
‍
ξ
+
a
-
j
q
j
=
∑
i
1
<
i
2
<
⋯
<
i
n
-
m
-
1
‍
∏
q
=
1
n
-
m
-
1
‍
ξ
+
a
-
i
q
j
,
i
q
∈
{
j
1
,
j
2
,
…
,
j
n
-
m
} has n
-
m
n
-
m
-
1
=
(
n
-
m
)
(
n
-
m
-
1
)
!
/
(
n
-
m
-
1
)
!
=
n
-
m terms. Therefore, the expansion of (
d
m
+
1
/
d
ξ
m
+
1
)
(
ξ
j
+
a
)
n has a total of n
(
n
-
1
)
⋯
(
n
-
m
+
1
)
(
n
-
m
)
/
m
! terms of the form ∏
q
=
1
n
-
m
-
1
‍
(
ξ
+
(
a
-
j
q
)
/
j
). However, if the sum is evaluated over all possible combinations j
1
j
2
⋯
j
n
-
m
-
1 such that 0
≤
j
1
<
j
2
<
⋯
<
j
n
-
m
-
1
≤
n
-
1, then the sum has n
n
-
m
-
1
=
n
(
n
-
1
)
⋯
(
n
-
m
)
(
n
-
m
-
1
!
)
/
(
m
+
1
)
!
(
n
-
m
-
1
)
!
=
(
1
/
m
+
1
)
(
n
n
-
1
⋯
n
-
m
/
m
!
) distinct terms. It follows that every term ∏
q
=
1
n
-
m
-
1
‍
(
ξ
+
(
a
-
j
q
)
/
j
) appears m
+
1 times in the expansion of (
1
/
j
n
)
(
d
m
+
1
/
d
ξ
m
+
1
)
(
ξ
j
+
a
)
n. Thus we have(16)
1
j
n
d
m
+
1
d
ξ
m
+
1
ξ
j
+
a
n
WW
=
m
!
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
m
-
1
≤
n
-
1
‍
m
+
1
∏
q
=
1
n
-
m
‍
ξ
+
a
-
j
q
j
WW
=
(
m
+
1
)
!
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
m
-
1
≤
n
-
1
‍
∏
q
=
1
n
-
m
‍
ξ
+
a
-
j
q
j
.Lemma 2.
Lets
^
(
n
,
k
;
a
)
=
1
/
k
!
lim
ξ
→
0
⁡
(
(
∑
j
=
0
k
‍
(
-
1
)
k
-
j
k
j
(
d
k
/
d
ξ
k
)
(
ξ
j
+
a
)
n
)
/
k
!
). Then(17)
s
^
n
,
k
;
a
=
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
a
-
j
q
.Proof.
Using Lemma1,(18)
s
^
n
,
k
;
a
WW
=
1
k
!
lim
ξ
→
0
∑
j
=
0
k
‍
-
1
k
-
j
k
j
j
n
∑
0
≤
j
1
<
⋯
<
j
n
-
k
≤
n
-
1
‍
k
!
∏
n
-
k
‍
W
W
W
W
W
W
W
W
×
∏
q
=
1
n
-
k
‍
ξ
+
a
-
j
q
j
k
!
-
1
.Note that ∏
q
=
1
n
-
k
‍
(
ξ
+
(
a
-
j
q
)
/
j
)
=
(
1
/
j
n
-
k
)
∏
q
=
1
n
-
k
‍
(
ξ
j
+
a
-
j
q
). Hence, the expression at the right-hand side of (18) becomes(19)
1
k
!
lim
ξ
→
0
∑
j
=
0
k
‍
-
1
k
-
j
k
j
j
k
W
W
W
W
×
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
(
ξ
j
+
a
-
j
q
)
WW
=
1
k
!
∑
j
=
0
k
‍
-
1
k
-
j
k
j
j
k
W
W
W
W
W
×
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
a
-
j
q
,which boils down to(20)
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
a
-
j
q
,since(21)
1
k
!
∑
j
=
0
k
‍
-
1
k
-
j
k
j
j
k
=
S
k
,
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)
s
a
n
,
k
=
s
^
n
,
k
;
a
=
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
a
-
j
q
.Proof.
We know that(23)
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
+
1
≤
n
‍
∏
q
=
1
n
-
k
+
1
‍
a
-
j
qis equal to the sum of the products (
a
-
j
1
)
(
a
-
j
2
)
⋯
(
a
-
j
n
+
1
-
k
) where the sum is evaluated overall possible combinations j
1
j
2
⋯
j
n
+
1
-
k, j
i
∈
{
0,1
,
2
,
…
,
n
}. These possible combinations can be divided into two: the combinations with j
i
=
n for some i
∈
{
1,2
,
…
,
n
-
k
+
1
} and the combinations with j
i
≠
n for all i
∈
{
0,1
,
2
,
…
,
n
-
k
+
1
}. Thus (24)
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
+
1
≤
n
‍
∏
q
=
1
n
-
k
+
1
‍
a
-
j
qis equal to(25)
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
+
1
≤
n
-
1
‍
∏
q
=
1
n
-
k
+
1
‍
a
-
j
q
+
a
-
n
W
W
×
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
a
-
j
q
.This implies that(26)
s
^
n
+
1
,
k
;
a
=
s
^
n
,
k
-
1
;
a
+
a
-
n
s
^
n
,
k
;
a
.This is exactly the triangular recurrence relation in (4) for s
a
(
n
,
k
). This proves the theorem.The explicit formula in Theorem3 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 s
a
(
n
,
k
).A0-1
tableau, as defined in [4] by de Médicis and Leroux, is a pair φ
=
(
λ
,
f
), where(27)
λ
=
λ
1
≥
λ
2
≥
⋯
≥
λ
kis a partition of an integer m, and f
=
(
f
i
j
)
1
≤
j
≤
λ
i 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 f
14
=
f
15
=
f
23
=
f
31
=
f
42
=
1, f
i
j
=
0 elsewhere (
1
≤
j
≤
λ
i
): 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
=
(
a
i
)
i
≥
0, a
i
∈
Z
+
∪
{
0
}. If T
d
A
(
h
,
r
) is the set of A-tableaux with exactly r distinct columns whose lengths are in the set A
=
{
a
0
,
a
1
,
…
,
a
h
}, then |
T
d
A
(
h
,
r
)
|
=
h
+
1
r. Now, transforming each column c of an A-tableau in T
d
A
(
n
-
1
,
n
-
k
) into a column of length ω
(
|
c
|
), we obtain a new tableau which is called A
ω-tableau. If ω
(
|
c
|
)
=
|
c
|, then the A
ω-tableau is simply the A-tableau. Now, we define an A
ω
(
0,1
)-tableau to be a 0-1 tableau which is constructed by filling up the cells of an A
ω-tableau with 0’s and 1’s such that there is only one 1 in each column. We use T
d
A
ω
(
0,1
)
(
n
-
1
,
n
-
k
) to denote the set of such A
ω
(
0,1
)-tableaux.It can easily be seen that every(
n
-
k
) combination j
1
j
2
⋯
j
n
-
k of the set {
0,1
,
2
,
…
,
n
-
1
} can be represented geometrically by an element ϕ in T
d
A
(
n
-
1
,
n
-
k
) with j
i 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)
s
a
(
n
,
k
)
=
∑
ϕ
∈
T
d
A
n
-
1
,
n
-
k
‍
∏
c
∈
ϕ
‍
ω
c
.Thus, using (29), we can easily prove the following theorem.Theorem 4.
The number ofA
ω
(
0,1
)-tableaux in T
d
A
ω
(
0,1
)
(
n
-
1
,
n
-
k
) where A
=
{
0,1
,
2
,
…
,
n
-
1
} such that ω
(
|
c
|
)
=
a
-
|
c
| is equal to s
a
(
n
,
k
).Letϕ be an A-tableau in T
d
A
(
n
-
1
,
n
-
k
) with A
=
{
0,1
,
2
,
…
,
n
-
1
}, and(30)
ω
A
ϕ
=
∏
c
∈
ϕ
‍
ω
c
=
∏
i
=
1
n
-
k
‍
a
-
j
i
,
j
i
∈
{
0,1
,
2
,
…
,
n
-
1
}
.If a
=
a
1
+
a
2 for some a
1 and a
2, then, with ω
*
(
j
)
=
a
2
-
j,(31)
ω
A
ϕ
=
∏
i
=
1
n
-
k
‍
a
1
+
ω
*
j
i
=
∑
r
=
0
n
-
k
‍
a
1
n
-
k
-
r
∑
j
i
≤
q
1
<
q
2
<
⋯
<
q
r
≤
j
n
-
k
‍
∏
i
=
1
r
‍
ω
*
q
i
.Suppose B
ϕ is the set of all A-tableaux corresponding to ϕ such that for each ψ
∈
B
ϕ eitherψ has no column whose weight is a
1, or
ψ has one column whose weight is a
1, or
⋮
ψ has n
-
k columns whose weights are a
1. Then, we may write (32)
ω
A
(
ϕ
)
=
∑
ψ
∈
B
ϕ
‍
ω
A
(
ψ
)
.Now, if r columns in ψ have weights other than a
1, then(33)
ω
A
(
ψ
)
=
a
1
n
-
k
-
r
∏
i
=
1
r
‍
ω
*
q
i
,where q
1
,
q
2
,
…
,
q
r
∈
{
j
1
,
j
2
,
…
,
j
n
-
k
}. Hence, (29) may be written as(34)
s
a
(
n
,
k
)
=
∑
ϕ
∈
T
d
A
(
n
-
1
,
n
-
k
)
‍
∑
ψ
∈
B
ϕ
‍
ω
A
ψ
.Note that for each r, there correspond n
-
k
r tableaux with r distinct columns having weights w
*
(
q
i
), q
i
∈
{
j
1
,
j
2
,
…
,
j
n
-
k
}. Since T
d
A
(
n
-
1
,
n
-
k
) has n
k elements, for each ϕ
∈
T
d
A
(
n
-
1
,
n
-
k
), the total number of A-tableaux ψ corresponding to ϕ is (35)
n
k
n
-
k
relements. However, only n
r tableaux in B
ϕ with r distinct columns of weights other than a
1 are distinct. Hence, every distinct tableau ψ appears (36)
n
k
n
-
k
r
n
r
=
n
-
r
ktimes in the collection. Consequently, we obtain(37)
s
a
n
,
k
=
∑
r
=
0
n
-
k
‍
n
-
r
k
a
1
n
-
k
-
r
∑
ψ
∈
B
r
‍
∏
c
∈
ψ
‍
ω
*
c
,where B
r 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)
s
a
(
n
,
k
)
=
∑
j
=
k
n
‍
j
k
a
1
j
-
k
∑
ψ
*
∈
B
n
-
j
‍
∏
c
∈
ψ
*
‍
ω
*
c
.Clearly, B
n
-
j
=
T
d
A
(
n
-
1
,
n
-
j
). Thus, using (22), we obtain the following theorem.Theorem 5.
The numberss
a
(
n
,
k
) satisfy the following identity:(39)
s
a
(
n
,
k
)
=
∑
j
=
k
n
‍
j
k
a
1
j
-
k
s
a
2
n
,
j
,where a
=
a
1
+
a
2  for some numbers a
1 and a
2.The next theorem contains certain convolution-type formula fors
a
(
n
,
k
) which will be proved using the combinatorics of A-tableau.Theorem 6.
The numberss
a
(
n
,
k
) have convolution formula(40)
s
a
(
m
+
j
,
n
)
=
∑
k
=
0
n
‍
s
a
m
,
k
s
a
-
m
j
,
n
-
k
.Proof.
Suppose thatϕ
1 is a tableau with exactly m
-
k distinct columns whose lengths are in the set A
1
=
{
0,1
,
2
,
…
,
m
-
1
} and ϕ
2 is a tableau with exactly j
-
n
+
k distinct columns whose lengths are in the set A
2
=
{
m
,
m
+
1
,
m
+
2
,
…
,
m
+
j
-
1
}. Then ϕ
1
∈
T
d
A
1
(
m
-
1
,
m
-
k
) and ϕ
2
∈
T
d
A
2
(
j
-
1
,
j
-
n
+
k
). Notice that by joining the columns of ϕ
1 and ϕ
2, 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
∈
T
d
A
(
m
+
j
-
1
,
m
+
j
-
n
). Hence(41)
∑
ϕ
∈
T
d
A
(
m
+
j
-
1
,
m
+
j
-
n
)
‍
ω
A
(
ϕ
)
WW
=
∑
k
=
0
n
‍
∑
ϕ
1
∈
T
d
A
1
(
m
-
1
,
m
-
k
)
‍
ω
A
1
(
ϕ
1
)
W
W
W
×
∑
ϕ
2
∈
T
d
A
2
(
j
-
1
,
j
-
n
+
k
)
‍
ω
A
2
(
ϕ
2
)
.Note that(42)
∑
ϕ
2
∈
T
d
A
2
(
j
-
1
,
j
-
n
+
k
)
‍
ω
A
2
(
ϕ
2
)
WW
=
∑
m
≤
g
1
<
g
2
<
⋯
<
g
j
-
n
+
k
≤
m
+
j
-
1
‍
∏
q
=
1
j
-
n
+
k
‍
(
a
-
g
q
)
WW
=
∑
0
≤
g
1
<
g
2
<
⋯
<
g
n
-
k
-
j
≤
j
-
1
‍
∏
q
=
1
j
-
n
+
k
‍
(
a
-
(
m
+
g
q
)
)
WW
=
s
a
-
m
(
j
,
n
-
k
)
.
Also, using (29), we have(43)
∑
ϕ
1
∈
T
d
A
1
(
m
-
1
,
m
-
k
)
‍
ω
A
1
(
ϕ
1
)
=
s
a
(
m
,
k
)
∑
ϕ
∈
T
d
A
(
m
+
j
-
1
,
m
+
j
-
n
)
‍
ω
A
(
ϕ
)
=
s
a
(
m
+
j
,
n
)
.Thus,(44)
s
a
(
m
+
j
,
n
)
=
∑
k
=
0
n
‍
s
a
(
m
,
k
)
s
a
-
m
(
j
,
n
-
k
)
.The following theorem gives another form of convolution formula.Theorem 7.
The numberss
a
(
n
,
k
) satisfy the second form of convolution formula (45)
s
a
(
n
+
1
,
m
+
j
+
1
)
=
∑
k
=
0
n
‍
s
a
(
k
,
n
)
s
a
-
(
k
+
1
)
(
n
-
k
,
j
)
.Proof.
Letϕ
1 be a tableau with k
-
m columns whose lengths are in A
1
=
0,1
,
…
,
k
-
1,
ϕ
2 be a tableau with n
-
k
-
j columns whose lengths are in A
2
=
{
k
+
1
,
…
,
n
}.Then ϕ
1
∈
T
d
A
1
(
k
-
1
,
k
-
m
); ϕ
2
∈
T
d
A
2
(
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 ofs
a
(
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 polynomiala
1
x
+
a
2
x
2
+
⋯
+
a
n
x
n has only real zeros then(46)
a
k
2
≥
a
k
+
1
a
k
-
1
k
k
-
1
n
-
k
+
1
n
-
k
(
k
=
2
,
…
,
n
-
1
)
.Now, consider the following polynomial:(47)
∑
k
=
0
n
‍
s
a
n
,
k
t
+
a
k
.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
=
0
n
‍
s
a
(
n
,
k
)
t
k are a
,
a
-
1
,
…
,
a
-
n
+
1. Applying Newton’s Inequality completes the proof of the following theorem.Theorem 9.
The sequence{
s
a
(
n
,
k
)
}
k
=
0
n is strictly log-concave and, hence, unimodal.By replacingt with -
t, the relation in (1) may be written as(48)
〈
t
〉
n
=
∑
k
=
0
n
‍
-
1
n
-
k
s
a
n
,
k
t
+
a
k
,where 〈
t
〉
n
=
t
(
t
+
1
)
(
t
+
2
)
⋯
(
t
+
n
-
1
). Note that, from Theorem 3 with a
<
0, (49)
s
a
n
,
k
=
-
1
n
-
k
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
b
+
j
q
,where b
=
-
a
>
0. Now, we define the signless noncentral Stirling number of the first kind, denoted by |
s
a
(
n
,
k
)
|, as(50)
s
a
n
,
k
=
-
1
n
-
k
s
a
n
,
k
=
∑
0
≤
j
1
<
j
2
<
⋯
<
j
n
-
k
≤
n
-
1
‍
∏
q
=
1
n
-
k
‍
b
+
j
q
.To introduce the main result of this paper, we need to state first the following theorem of Erdös and Stone [3].Theorem 10 (see [3]).
Letu
1
<
u
2
<
⋯ be an infinite sequence of positive real numbers such that(51)
∑
i
=
1
∞
‍
1
u
i
=
∞
,
∑
i
=
1
∞
‍
1
u
i
2
<
∞
.Denote by ∑
n
,
k
‍ the sum of the product of the first n of them taken k at a time and denote by K
n the largest value of k for which ∑
n
,
k
‍ assumes its maximum value. Then (52)
K
n
=
n
-
∑
i
=
1
n
‍
1
u
i
-
∑
i
=
1
n
‍
1
u
i
2
1
+
1
u
i
-
1
+
o
(
1
)
.We also need to recall the asymptotic expansion of harmonic numbers which is given by(53)
1
1
+
1
2
+
⋯
+
1
n
=
log
⁡
⁡
n
+
γ
+
o
1
,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{
|
s
a
(
n
,
k
)
|
}
k
=
0
n.Theorem 11.
The largest index for which the sequence{
|
s
a
(
n
,
k
)
|
}
k
=
0
n assumes its maximum is given by the approximation (54)
k
a
,
n
=
log
⁡
⁡
b
+
n
b
+
o
1
,where [
x
] is the integer part of x and b
=
-
a, a
<
0.Proof.
Using Theorem10 and by (50), we see that |
s
a
(
n
,
k
)
|
=
∑
n
-
1
,
n
-
k
‍. Denoting by k
a
,
n for which ∑
n
,
n
-
k
‍ is maximum and with u
1
=
b
+
0
,
u
2
=
b
+
1
,
…
,
u
n
-
1
=
b
+
n
-
1 we have(55)
k
a
,
n
=
∑
i
=
0
n
-
1
‍
1
b
+
i
-
∑
i
=
0
n
-
1
‍
1
b
+
i
2
1
+
1
b
+
i
-
1
+
o
1
=
∑
i
=
0
n
-
1
‍
1
b
+
i
-
∑
i
=
0
n
-
1
‍
1
b
+
i
b
+
i
+
1
+
o
1
=
∑
i
=
0
n
-
1
‍
1
b
+
i
+
1
+
o
1
.But using (53), we see that(56)
∑
i
=
0
n
-
1
‍
1
b
+
i
+
1
=
log
⁡
⁡
(
b
+
n
)
-
log
⁡
⁡
b
.From this we get(57)
k
a
,
n
=
log
⁡
b
+
n
b
+
o
1
.For the case in whicha
>
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 fora
≥
n is given by the approximation(58)
k
a
,
n
=
log
⁡
a
+
1
a
-
n
+
1
+
o
1
.Proof.
From the definition, fora
≥
n,s
a
(
n
,
k
)
>
0 and by Theorem 3, s
a
(
n
,
k
) is the sum of the products (
a
-
j
1
)
(
a
-
j
2
)
⋯
(
a
-
j
n
-
k
) where j
i’s are taken from the set {
0,1
,
2
,
…
,
n
-
1
}. By Theorem 10, s
a
(
n
,
k
)
=
∑
n
,
n
-
k
‍. Thus with u
1
=
a
,
u
2
=
a
-
1
,
…
,
u
n
-
1
=
a
-
n
+
1 we have(59)
k
a
,
n
=
∑
i
=
0
n
-
1
‍
1
a
-
i
-
∑
i
=
0
n
-
1
‍
1
a
-
i
2
1
+
1
a
-
i
-
1
+
o
1
=
∑
i
=
0
n
-
1
‍
1
a
-
i
-
∑
i
=
0
n
-
1
‍
1
a
-
i
a
-
i
+
1
+
o
1
=
∑
i
=
0
n
-
1
‍
1
a
-
i
+
1
+
o
1
.Again, using (53), we get(60)
k
a
,
n
=
log
⁡
a
+
1
a
-
n
+
1
+
o
1
.Example 13.
The maximum element of the sequence{
s
-
1
(
9
,
k
)
}
k
=
0
9 occurs at (Table 1)(61)
k
-
1,9
=
log
⁡
⁡
1
+
8
1
+
o
1
=
log
⁡
⁡
9
+
o
1
≈
2
.Table 1
Values ofs
-
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
8409500Example 14.
The maximum element of the sequence{
s
10
(
10
,
k
)
}
k
=
0
10 occurs at (Table 2)(62)
k
10,10
=
log
⁡
⁡
10
+
1
10
-
10
+
1
+
o
1
=
log
⁡
⁡
11
+
o
1
≈
2
.Table 2
Values ofs
10
(
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
8409500We know that the classical Stirling numbers of the first kind are special cases ofs
a
(
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)
s
n
,
k
=
∑
1
≤
i
1
<
i
2
<
⋯
<
i
n
-
k
≤
n
-
1
‍
i
1
i
2
⋯
i
n
-
k
.Using Theorem 10, |
s
(
n
,
k
)
|
=
∑
n
-
1
,
n
-
k
‍. We use k
n to denote the largest value of k for which ∑
n
-
1
,
n
-
k
‍ is maximum. With u
1
=
1
,
u
2
=
2
,
…
,
u
n
-
1
=
n
-
1 we have(64)
k
n
=
1
+
∑
i
=
1
n
-
1
‍
1
i
-
∑
i
=
1
n
-
1
‍
1
i
2
1
+
1
i
-
1
+
o
1
=
1
+
∑
i
=
1
n
-
1
‍
1
i
-
∑
i
=
1
n
-
1
‍
1
i
-
1
i
+
1
+
o
1
=
1
+
∑
i
=
1
n
-
1
‍
1
i
+
1
+
o
1
.Using (53), we see that(65)
∑
i
=
1
n
-
1
‍
1
1
+
i
=
log
⁡
⁡
n
-
log
⁡
⁡
1
+
γ
+
o
1
.Therefore, we have(66)
k
n
=
log
⁡
⁡
n
+
γ
+
o
1
.Example 15.
It is shown in Table3 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
=
0
7 occurs at(67)
k
7
=
log
⁡
⁡
7
+
γ
+
o
1
=
1.95
+
0.5772
+
o
1
=
2.53
+
o
1
≈
2
.Moreover, when n
=
10, the maximum value occurs at(68)
k
10
=
log
⁡
⁡
10
+
γ
+
o
1
=
2.30
+
0.5772
+
o
1
=
2.8772
+
o
1
≈
3
.Table 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
269325Recently, 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.



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*Source: 982812-2015-10-12.xml*