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Revision History for A130235

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Showing entries 1-10 | older changes

A130235

Partial sums of the 'lower' Fibonacci Inverse A130233.
(history; published version)

#19 by Peter Luschny at Fri Mar 17 17:07:52 EDT 2023

STATUS

reviewed

approved

#18 by Michel Marcus at Fri Mar 17 15:50:24 EDT 2023

STATUS

proposed

reviewed

#17 by G. C. Greubel at Fri Mar 17 15:34:07 EDT 2023

STATUS

editing

proposed

#16 by G. C. Greubel at Fri Mar 17 15:34:01 EDT 2023

LINKS

G. C. Greubel, <a href="/A130235/b130235.txt">Table of n, a(n) for n = 0..5000</a>

FORMULA

a(n) = Sum_{0<=k<=0..n} A130233(k) = (n+1)*A130233(n) - Fib(A130233(n)+2) + 1.

PROG

(Magma)

m:=120;

f:= func< x | (&+[x^Fibonacci(j): j in [1..Floor(3*Log(3*m+1))]])/(1-x)^2 >;

R<x>:=PowerSeriesRing(Rationals(), m+1);

[0] cat Coefficients(R!( f(x) )); // G. C. Greubel, Mar 17 2023

(SageMath)

m=120

def f(x): return sum( x^fibonacci(j) for j in range(1, int(3*log(3*m+1))))/(1-x)^2

def A130235_list(prec):

P.<x> = PowerSeriesRing(ZZ, prec)

return P( f(x) ).list()

A130235_list(m) # G. C. Greubel, Mar 17 2023

STATUS

approved

editing

#15 by Charles R Greathouse IV at Tue Apr 14 10:24:31 EDT 2020

STATUS

proposed

approved

#14 by Vaclav Kotesovec at Tue Apr 14 07:53:46 EDT 2020

STATUS

editing

proposed

#13 by Vaclav Kotesovec at Tue Apr 14 07:52:54 EDT 2020

MATHEMATICA

nmax = 90; CoefficientList[Series[Sum[x^Fibonacci[k], {k, 1, 1 + Log[3/2 + Sqrt[5]*nmax]/Log[GoldenRatio]}]/(1-x)^2, {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 14 2020 *)

STATUS

proposed

editing

Discussion

Tue Apr 14

07:53

Vaclav Kotesovec: More efficient program.

#12 by Vaclav Kotesovec at Tue Apr 14 07:39:29 EDT 2020

STATUS

editing

proposed

#11 by Vaclav Kotesovec at Tue Apr 14 07:39:20 EDT 2020

MATHEMATICA

nmax = 6090; CoefficientList[Series[Sum[x^Fibonacci[k], {k, 1, nmax}]/(1-x)^2, {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 14 2020 *)

#10 by Vaclav Kotesovec at Tue Apr 14 07:29:26 EDT 2020

MATHEMATICA

nmax = 60; CoefficientList[Series[Sum[x^Fibonacci[k], {k, 1, 20nmax}]/(1-x)^2, {x, 0, 60nmax}], x] (* Vaclav Kotesovec, Apr 14 2020 *)

Discussion

Tue Apr 14

07:38

Vaclav Kotesovec: Colin, first terms are identical, but for n=63, a(n)=497, and your g.f. gives 496. Conjectures are wrong.