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Revision History for A181546

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A181546

a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)^4.
(history; published version)

#17 by Harvey P. Dale at Sat May 22 19:27:34 EDT 2021

STATUS

editing

approved

#16 by Harvey P. Dale at Sat May 22 19:27:30 EDT 2021

MATHEMATICA

Table[Sum[Binomial[n-k, k]^4, {k, 0, Floor[n/2]}], {n, 0, 30}] (* Harvey P. Dale, May 22 2021 *)

STATUS

approved

editing

#15 by Joerg Arndt at Sun Jan 27 03:04:07 EST 2019

STATUS

proposed

approved

#14 by Michel Marcus at Sun Jan 27 02:10:00 EST 2019

STATUS

editing

proposed

#13 by Michel Marcus at Sun Jan 27 02:09:54 EST 2019

CROSSREFS

Cf. A000032, A000045.

STATUS

proposed

editing

Discussion

Sun Jan 27

02:10

Michel Marcus: ok

#12 by Seiichi Manyama at Sun Jan 27 02:00:48 EST 2019

STATUS

editing

proposed

#11 by Seiichi Manyama at Sun Jan 27 02:00:37 EST 2019

COMMENTS

Conjecture: Given F(n,L) = Sum_{k=0..[n/2]} C(n-k,k)^L, then Limit_{n->oo} F(n+1,L)/F(n,L) = (Fibonacci(L)*sqrt(5) + Lucas(L))/2 for L>=0 where Fibonacci(n) = A000045(n) and Lucas(n) = A000032(n).

for L>=0 where Fibonacci(n) = A000045(n) and Lucas(n) = A000032(n).

STATUS

proposed

editing

#10 by Seiichi Manyama at Sun Jan 27 01:59:38 EST 2019

STATUS

editing

proposed

#9 by Seiichi Manyama at Sun Jan 27 01:58:29 EST 2019

COMMENTS

Conjecture: Given F(n,L) = Sum_{k=0..[n/2]} C(n-k,k)^L, then Limit_{n->oo} F(n+1,L)/F(n,L) = (Fibonacci(L)*sqrt(5) + Lucas(L))/2

* Limit_{n->oo} F(n+1,L)/F(n,L) = (Fibonacci(L)*sqrt(5) + Lucas(L))/2

STATUS

proposed

editing

Discussion

Sun Jan 27

01:59

Seiichi Manyama: @Michel Marcus:  See the comment of A181545.

#8 by Michel Marcus at Sun Jan 27 01:53:21 EST 2019

STATUS

editing

proposed