Question

A formula for $1^m+2^m+3^m+\ldots+n^m$ using binomial coefficients

score 135 · Answer 1 · 2021-09-04 19:48:19

135

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A formula for $1^m+2^m+3^m+\ldots+n^m$ using binomial coefficients

Published on 04 Sep 2021 - 19:48

score 224 · Answer 2 · 2021-09-07 00:03:21

224

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Number of surjections and generating function

Published on 07 Sep 2021 - 0:03

score 40 · Answer 3 · 2026-03-28 07:36:28

40

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Distributing balls into bins for large numbers: approximating probabilities

Published on 28 Mar 2026 - 7:36

score 70 · Answer 4 · 2021-09-12 14:37:17

70

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Stirling numbers of the first kind explicit product in reverse for $|s(n,k)|$?

Published on 12 Sep 2021 - 14:37

score 280 · Answer 5 · 2021-09-14 03:07:25

280

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How to pronounce Stirling Numbers of Second Kind ${n\brace k}$?

Published on 14 Sep 2021 - 3:07

score 102 · Answer 6 · 2021-09-18 05:31:12

102

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Prove that $\left\{{n+1}\atop{k+1}\right\}=\sum_{i=k}^{n}{\left(k+1\right)^{n-i}\left\{{i}\atop{k}\right\}}$

Published on 18 Sep 2021 - 5:31

score 131 · Answer 7 · 2021-09-18 06:26:06

131

Views

Prove that $\left\{{n+k+1}\atop{k}\right\}=\sum_{i=1}^{k}{i\left\{{n+i}\atop{i}\right\}}$

Published on 18 Sep 2021 - 6:26

score 142 · Answer 8 · 2021-09-19 13:01:03

142

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How to show this identity $\prod_{q=1}^k\frac{1}{1-qz}=\sum_{j=1}^{k}jz\prod_{q=1}^j\frac{1}{1-qz}+1$ avoiding a proof by induction

Published on 19 Sep 2021 - 13:01

score 96 · Answer 9 · 2026-03-29 15:54:56

96

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Does the (Stirling number of the second kind) equality ${2n\brace 2} = 2^{2n-1}-1$ hold?

Published on 29 Mar 2026 - 15:54

score 61 · Answer 10 · 2026-03-29 16:00:47

61

Views

How to relate the number of all applications $X\to Y$ with the Stirling numbers of the second kind?

Published on 29 Mar 2026 - 16:00

score 106 · Answer 11 · 2026-03-29 15:59:34

106

Views

Inequality with Stirling numbers of second kind

Published on 29 Mar 2026 - 15:59

score 107 · Answer 12 · 2021-10-16 13:15:56

107

Views

Alternating sum of numbers of surjective functions

Published on 16 Oct 2021 - 13:15

score 87 · Answer 13 · 2021-10-21 06:09:42

87

Views

A calculation about the sum of the product of Binomial and Stirling numbers of the first kind

Published on 21 Oct 2021 - 6:09

score 82 · Answer 14 · 2021-10-21 11:29:49

82

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definition of stirling numbers

Published on 21 Oct 2021 - 11:29

score 424 · Answer 15 · 2026-03-28 00:27:45

424

Views

Find closed formula for the exponential generating function of sequence counting permutations with 2 cycles.

Published on 28 Mar 2026 - 0:27

List Question

A formula for $1^m+2^m+3^m+\ldots+n^m$ using binomial coefficients

Number of surjections and generating function

Distributing balls into bins for large numbers: approximating probabilities

Stirling numbers of the first kind explicit product in reverse for $|s(n,k)|$?

How to pronounce Stirling Numbers of Second Kind ${n\brace k}$?

Prove that $\left\{{n+1}\atop{k+1}\right\}=\sum_{i=k}^{n}{\left(k+1\right)^{n-i}\left\{{i}\atop{k}\right\}}$

Prove that $\left\{{n+k+1}\atop{k}\right\}=\sum_{i=1}^{k}{i\left\{{n+i}\atop{i}\right\}}$

How to show this identity $\prod_{q=1}^k\frac{1}{1-qz}=\sum_{j=1}^{k}jz\prod_{q=1}^j\frac{1}{1-qz}+1$ avoiding a proof by induction

Does the (Stirling number of the second kind) equality ${2n\brace 2} = 2^{2n-1}-1$ hold?

How to relate the number of all applications $X\to Y$ with the Stirling numbers of the second kind?

Inequality with Stirling numbers of second kind

Alternating sum of numbers of surjective functions

A calculation about the sum of the product of Binomial and Stirling numbers of the first kind

definition of stirling numbers

Find closed formula for the exponential generating function of sequence counting permutations with 2 cycles.

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