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Closed form for $a(2^m(2k+1))$

Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\dotsc,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
Notamathematician's user avatar
2 votes
0 answers
126 views

Generalized identity with Stirling numbers of the second kind and falling factorials

It is known that Striling numbers of the second kind satisfy the relation $$ \sum\limits_{k=0}^{n}{n \brace k}(x)_k = x^n. $$ where $(x)_n$ is the falling factorials such that $$ (x)_n = x(x-1)(x-2)\...
Notamathematician's user avatar
2 votes
1 answer
222 views

A question on signed Stirling numbers of the first kind

Let $(x)_0=1$ and $(x)_n=x(x-1)\cdots(x-n+1)$ for $n=1,2,3,\ldots$. The signed Stirling numbers of the first kind, $s(n,k)$ with $n\ge k\ge0$, are defined by $$(x)_n=\sum_{k=0}^ns(n,k)x^k.$$ Question. ...
Zhi-Wei Sun's user avatar
  • 15.6k
2 votes
1 answer
129 views

Recursion for the sum with Stirling numbers of both kinds

Let $s(n,k)$ be a (signed) Stirling number of the first kind. Let $n \brace k$ be a Stirling number of the second kind. Let $$ f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j}...
Notamathematician's user avatar
1 vote
0 answers
86 views

Closed form for the family of polynomials

Let $s(n,k)$ be a (signed) Stirling number of the first kind. Let $R(n,x)$ be the family of polynomials such that $$ R(2n+1,x) = xR(n,x), \\ R(2n,x) = x(R(n,x+1) - R(n, x)), \\ R(0, x) = x $$ Let $\...
Notamathematician's user avatar
1 vote
0 answers
59 views

Simple recursion for the A329369 using Stirling numbers of both kinds

Let $s(n,k)$ be a (signed) Stirling number of the first kind. Let $n \brace k$ be a Stirling number of the second kind. Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with ...
Notamathematician's user avatar
3 votes
0 answers
89 views

Recursion for reversed rows of the A373183 using unsigned Stirling numbers of the first kind

Let $\left[{n \atop k}\right]$ be unsigned Stirling numbers of the first kind. Here $$ \left[{n \atop k}\right] = (n-1)\left[{n-1 \atop k}\right] + \left[{n-1 \atop k-1}\right], \\ \left[{n \atop 0}\...
Notamathematician's user avatar
0 votes
1 answer
96 views

Prime divisibility of Stirling numbers of first kind

Prove that $p\mid\genfrac[]0{}{p^w}k$ where $p$ is an odd prime, $w \in \mathbb{N}$, $1<k<p^w$ and $k \neq p^v$ for some positive integer $v<w$. This has to be already done I just can't find ...
Benjamin L. Warren's user avatar
3 votes
1 answer
324 views

Sum with Stirling numbers of the second kind

Let $wt(n)$ be A000120, number of $1$'s in binary expansion of $n$ (or the binary weight of $n$) and $$n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_{wt(n)}+1}))\dots)$$ Then we have an integer sequence given ...
Notamathematician's user avatar
2 votes
1 answer
145 views

Estimation of a sum involving Stirling's number of second kind and binomial coefficient

Let $S(n, j)$ be Stirling's number of second kind. Let $p\in [0,1]$ and $m \in N$. Bound from above the following sum: $$ \sum_{j=0}^m S(n,j) {m \choose j}\, j! \, p^j $$
volond's user avatar
  • 97
1 vote
1 answer
258 views

Sum of divisors of Stirling numbers of the second kind

In this post we denote the Stirling number of the second kind as ${n\brace k}$, I add as reference the article Stirling numbers of the second kind from the encyclopedia Wikipedia. And we denote the ...
user142929's user avatar
1 vote
0 answers
63 views

On the equation involving Stirling numbers of the second kind ${n\brace a}{m\brace b}={k\brace c}$, and its solutions satisfying certain requirements

In this post we denote the Stirling numbers of the second kind as ${r\brace s}$ and we consider the proposal to ask if the equation of the title has infinitely many solutions $${n\brace a}{m\brace b}={...
user142929's user avatar
5 votes
0 answers
775 views

A conjecture about the degrees of special polynomials

Define the congruence "modulo m" on exponential Taylor series as $$ \sum_{n=0}^\infty \frac{a_n}{n!}x^n \equiv \sum_{n=0}^\infty \frac{b_n}{n!} x^n \mod m \iff \forall n: \frac{a_n-b_n}{m}\in \mathbb{...
Danil Krotkov's user avatar
2 votes
1 answer
338 views

Simple approximation to a sum involving Stirling numbers?

I have also posted this question at https://math.stackexchange.com/questions/486917/simple-approximation-to-a-sum-involving-stirling-numbers. I have an exact answer to a problem, which is the function:...
user1748601's user avatar
9 votes
1 answer
826 views

An infinite set of identities using Stirling numbers 1st kind - are they all zero?

I have the following set of series involving the Stirling numbers 1'st kind and binomials, which can be understood as a set of dot-products of row- and column-vectors of two infinite matrices (where R ...
Gottfried Helms's user avatar