Question

Fibonacci numbers as weighted averages of powers of 5 -- a bijective/combinatorial proof?

score 112 · Answer 1 · 2019-05-10 12:43:15

112

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Fibonacci numbers as weighted averages of powers of 5 -- a bijective/combinatorial proof?

Published on 10 May 2019 - 12:43

score 79 · Answer 2 · 2019-05-13 10:43:09

79

Views

assigning p tasks to k people of n

Published on 13 May 2019 - 10:43

score 66 · Answer 3 · 2026-03-25 01:20:32

66

Views

Number of equal pairs from a set of size n in range 1-nlog

Published on 25 Mar 2026 - 1:20

score 138 · Answer 4 · 2026-03-25 18:52:28

138

Views

Games and Dyadic Rationals

Published on 25 Mar 2026 - 18:52

score 117 · Answer 5 · 2019-05-17 10:37:30

117

Views

Combinatorial proof for an identity

Published on 17 May 2019 - 10:37

score 189 · Answer 6 · 2019-05-22 13:17:03

189

Views

Combinatorial proof for $\sum_{k=0}^p {p+q\choose k} {p+q-k\choose p-k}=2^p {p+q \choose p}$

Published on 22 May 2019 - 13:17

score 59 · Answer 7 · 2019-05-23 15:46:03

59

Views

Proof of $\sum_{k=n-1}^{n+p-1} {k \choose p}={n+p \choose p}$ using the equation $x_1 + x_2 + \dots + x_n = p$

Published on 23 May 2019 - 15:46

score 239 · Answer 8 · 2026-03-25 21:49:03

239

Views

Ways of distributing passengers in ships

Published on 25 Mar 2026 - 21:49

score 178 · Answer 9 · 2019-05-26 06:57:14

178

Views

Prove identity $\sum^n_{k = 0} \binom {r+k} {r} = \binom{r+n+1} {r+1}$ using lattice paths

Published on 26 May 2019 - 6:57

score 148 · Answer 10 · 2019-05-29 11:14:54

148

Views

Proof of $\sum_{k=1}^{i} \binom{n-k}{n-i} = \frac{n \cdot \binom{n-1}{n-i}}{n+1-i}$

Published on 29 May 2019 - 11:14

score 112 · Answer 11 · 2019-06-06 12:09:50

112

Views

Proof needed for the combinatorial identity $\sum_{a=0}^n\binom{n+a}{a}/{2^{n+a}}=1$

Published on 06 Jun 2019 - 12:09

score 48 · Answer 12 · 2019-06-08 04:11:01

48

Views

Can any counting situation that works for one side of an identity work for the other (combinatorial proof)

Published on 08 Jun 2019 - 4:11

score 348 · Answer 13 · 2019-06-13 03:26:14

348

Views

Combinatorial Proof of $3^n - 2^n = \sum_{i=0}^{n-1}2^i3^{n-1-i}$

Published on 13 Jun 2019 - 3:26

score 179 · Answer 14 · 2019-06-17 08:26:17

179

Views

Combinatorial proof of $\frac{2^{2n}}{n+1}$ $\leq$ ${2n}\choose{n}$

Published on 17 Jun 2019 - 8:26

score 1k · Answer 15 · 2019-06-17 20:18:04

1k

Views

How to prove formula for power sum

Published on 17 Jun 2019 - 20:18

List Question

Fibonacci numbers as weighted averages of powers of 5 -- a bijective/combinatorial proof?

assigning p tasks to k people of n

Number of equal pairs from a set of size n in range 1-nlog

Games and Dyadic Rationals

Combinatorial proof for an identity

Combinatorial proof for $\sum_{k=0}^p {p+q\choose k} {p+q-k\choose p-k}=2^p {p+q \choose p}$

Proof of $\sum_{k=n-1}^{n+p-1} {k \choose p}={n+p \choose p}$ using the equation $x_1 + x_2 + \dots + x_n = p$

Ways of distributing passengers in ships

Prove identity $\sum^n_{k = 0} \binom {r+k} {r} = \binom{r+n+1} {r+1}$ using lattice paths

Proof of $\sum_{k=1}^{i} \binom{n-k}{n-i} = \frac{n \cdot \binom{n-1}{n-i}}{n+1-i}$

Proof needed for the combinatorial identity $\sum_{a=0}^n\binom{n+a}{a}/{2^{n+a}}=1$

Can any counting situation that works for one side of an identity work for the other (combinatorial proof)

Combinatorial Proof of $3^n - 2^n = \sum_{i=0}^{n-1}2^i3^{n-1-i}$

Combinatorial proof of $\frac{2^{2n}}{n+1}$ $\leq$ ${2n}\choose{n}$

How to prove formula for power sum

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