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Tagged with stirling-numbers sequences-and-series
14 questions
1
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0
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38
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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 \...
- 4,948
4
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2
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219
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how to prove identity for nth derivative of $(\text{arctanh}(x))^j$?
this question asked on MSE
I worked on integral problem and I got that
$$ \int_0^1 \frac{x^n}{\ln \left(\frac{1-x}{1+x} \right) } dx=-\frac{2}{(n+1)!}\sum_{j=1}^{n+1}F(n,j) \eta'(-j)$$
where $\eta(x)$ ...
- 513
2
votes
1
answer
129
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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
$$
f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j}...
- 4,948
1
vote
0
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86
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Closed form for the family of polynomials
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $R(n,x)$ be the family of polynomials such that
$$
R(2n+1,x) = xR(n,x), \\
R(2n,x) = x(R(n,x+1) - R(n, x)), \\
R(0, x) = x
$$
Let $\...
- 4,948
1
vote
0
answers
59
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Simple recursion for the A329369 using Stirling numbers of both kinds
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with ...
- 4,948
3
votes
0
answers
89
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Recursion for reversed rows of the A373183 using unsigned Stirling numbers of the first kind
Let $\left[{n \atop k}\right]$ be unsigned Stirling numbers of the first kind. Here
$$
\left[{n \atop k}\right] = (n-1)\left[{n-1 \atop k}\right] + \left[{n-1 \atop k-1}\right], \\
\left[{n \atop 0}\...
- 4,948
3
votes
1
answer
223
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How to find the coefficient of $x^k$ in the expression $\prod_{p=2}^n (1+xp) $
I got this general formula for $ n\in N$ (I showed it here)
$$\int_0^1 \left(\frac{x}{1-x} \ln x \right)^n dx=n \sum_{p=0}^{n-1}a(n,p+1) (-1)^{n-p} \zeta(p+2)+n! $$
where $a(n,k)$ is the coefficient ...
- 513
1
vote
3
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183
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Evaluating a sinusoidal series
Define the sequence of functions
$$f_n(x)=\sum_{m=n}^\infty(-1)^m\frac{x^{2m}}{(2m+1)!} {m \choose n} $$
Is there a closed form expression for arbitrary $n$? It is clear that the result should assume ...
- 151
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 ...
- 4,948
0
votes
1
answer
378
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Simplify a double summation involving binomial coeficient
$$T(N,K)=\sum_{i=2}^{K}\sum_{j=2}^{i}(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}$$
Is it possible to evaluate the sum for $K=10^7$ efficiently. If we manage to remove one of the sums, it will be ...
- 221
16
votes
1
answer
584
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What is this sequence?
This is again a question that I asked at Stack Exchange, but got no answer so far, so I am trying here.
Let:
$$ a_n=\sum_{k\ge0}(k+1) {n+2\brack k+2}(n+2)^kB_k$$
$B_k$ is the Bernoulli number. ${n\...
- 505
15
votes
1
answer
733
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Positivity of a finite sum involving Stirling numbers
In my research in theoretical physics, I have arrived at some coefficients $a_{n,m}$ depending on two integers, $n\geq 1$ and $0\leq m\leq n$:
$$
a_{n,m}=\sum_{j=0}^{n-1} {2j \choose j+m} \left(\frac{...
- 183
2
votes
3
answers
529
views
How this expression leads to the given sequence
Here given is a sequence from OEIS.
The sequence is triangle of coefficients from fractional iteration of $e^x - 1$. Few terms are:
1, 1, 3, 1, 13, 18, 1, 50, 205, 180, 1, 201, 1865, 4245, 2700, 1, ...
- 21
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 ...
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