All Questions
5 questions with no upvoted or accepted answers
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
$$
\left[{n \atop k}\right] = (n-1)\left[{n-1 \atop k}\right] + \left[{n-1 \atop k-1}\right], \\
\left[{n \atop 0}\...
2
votes
0
answers
126
views
Generalized identity with Stirling numbers of the second kind and falling factorials
It is known that Striling numbers of the second kind satisfy the relation
$$
\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)\...
1
vote
0
answers
41
views
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 \...
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 $\...
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 ...