All Questions
7 questions
1
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0
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41
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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 \...
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
$$
f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j}...
1
vote
0
answers
86
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Closed form for the family of polynomials
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $R(n,x)$ be the family of polynomials such that
$$
R(2n+1,x) = xR(n,x), \\
R(2n,x) = x(R(n,x+1) - R(n, x)), \\
R(0, x) = x
$$
Let $\...
1
vote
0
answers
59
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Simple recursion for the A329369 using Stirling numbers of both kinds
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with ...
3
votes
0
answers
89
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Recursion for reversed rows of the A373183 using unsigned Stirling numbers of the first kind
Let $\left[{n \atop k}\right]$ be unsigned Stirling numbers of the first kind. Here
$$
\left[{n \atop k}\right] = (n-1)\left[{n-1 \atop k}\right] + \left[{n-1 \atop k-1}\right], \\
\left[{n \atop 0}\...
3
votes
1
answer
324
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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 ...
9
votes
1
answer
826
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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 ...