Question

Why does this not hold? (partitioning sets into nonempty subsets)

score 66 · Answer 1 · 2018-04-04 19:19:12

66

Views

Why does this not hold? (partitioning sets into nonempty subsets)

Published on 04 Apr 2018 - 19:19

score 135 · Answer 2 · 2018-04-05 04:14:24

135

Views

Some strange multinomial averaging

Published on 05 Apr 2018 - 4:14

score 126 · Answer 3 · 2018-04-05 11:26:47

126

Views

Induction regarding Stirling numbers of the second kind.

Published on 05 Apr 2018 - 11:26

score 52 · Answer 4 · 2018-04-10 03:55:28

52

Views

Stirling Number Proof Help Needed

Published on 10 Apr 2018 - 3:55

score 67 · Answer 5 · 2018-04-10 22:08:46

67

Views

prove $S(n,1) = -1^{n-1}(n-1)!$ (Stirling numbers of the first kind)

Published on 10 Apr 2018 - 22:08

score 145 · Answer 6 · 2026-03-25 22:06:08

145

Views

Riemann $\zeta$-function and a series involving Stirling numbers of first kind.

Published on 25 Mar 2026 - 22:06

score 365 · Answer 7 · 2018-04-16 02:53:49

365

Views

How to use bijective proof to prove $S(n+1, m+1) = \sum\limits_{k=m}^n S(k, m) × \binom{n}{k}$

Published on 16 Apr 2018 - 2:53

score 185 · Answer 8 · 2018-04-25 09:28:23

185

Views

Simplifying a sum involving Stirling numbers of the second kind

Published on 25 Apr 2018 - 9:28

score 852 · Answer 9 · 2026-03-25 17:44:38

852

Views

Combinatorial and Algebraic Proof of an Identity involving Stirling Numbers of the second kind ${n+1\brace k+1}=\sum_i \binom{n}{i}{i\brace k}$

Published on 25 Mar 2026 - 17:44

score 213 · Answer 10 · 2018-05-06 11:23:48

213

Views

Stirling Numbers of the Second Kind $ \sum_{j= m}^n {j \brace m} (m+1)^{n-j} = {n+1\brace m+1} $

Published on 06 May 2018 - 11:23

score 114 · Answer 11 · 2018-05-06 13:53:23

Stirling Numbers $ \sum_{j= m}^{n-r} \binom{n}j {j \brace m} {n-j\brace r} = \binom{m} {m+r} {n\brace m+r} $ Combinatorical Proof or Algebraic

score 82 · Answer 12 · 2018-05-12 13:13:28

82

Views

Counting using Stirling numbers

Published on 12 May 2018 - 13:13

score 192 · Answer 13 · 2018-05-12 13:30:30

192

Views

Stirling numbers of the second kind for small numbers of partition

Published on 12 May 2018 - 13:30

score 133 · Answer 14 · 2026-03-22 19:38:42

133

Views

What is the relationship between sums of powers of $n$ and the stirling numbers?

Published on 22 Mar 2026 - 19:38

score 136 · Answer 15 · 2026-03-25 08:07:34

136

Views

Closed form for $\sum\limits_{k=0}^{b}k^na^k$

Published on 25 Mar 2026 - 8:07

List Question

Why does this not hold? (partitioning sets into nonempty subsets)

Some strange multinomial averaging

Induction regarding Stirling numbers of the second kind.

Stirling Number Proof Help Needed

prove $S(n,1) = -1^{n-1}(n-1)!$ (Stirling numbers of the first kind)

Riemann $\zeta$-function and a series involving Stirling numbers of first kind.

How to use bijective proof to prove $S(n+1, m+1) = \sum\limits_{k=m}^n S(k, m) × \binom{n}{k}$

Simplifying a sum involving Stirling numbers of the second kind

Combinatorial and Algebraic Proof of an Identity involving Stirling Numbers of the second kind ${n+1\brace k+1}=\sum_i \binom{n}{i}{i\brace k}$

Stirling Numbers of the Second Kind $ \sum_{j= m}^n {j \brace m} (m+1)^{n-j} = {n+1\brace m+1} $

Stirling Numbers $ \sum_{j= m}^{n-r} \binom{n}j {j \brace m} {n-j\brace r} = \binom{m} {m+r} {n\brace m+r} $ Combinatorical Proof or Algebraic

Counting using Stirling numbers

Stirling numbers of the second kind for small numbers of partition

What is the relationship between sums of powers of $n$ and the stirling numbers?

Closed form for $\sum\limits_{k=0}^{b}k^na^k$

Trending Questions

Popular # Hahtags

Popular Questions