login
Maximal index k of an odd Fibonacci number (A001519) such that A001519(k) = Fibonacci(2k-1) <= n (the 'lower' odd Fibonacci Inverse).
12

%I #17 Sep 08 2022 08:45:30

%S 1,2,2,2,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,

%T 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,

%U 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6

%N Maximal index k of an odd Fibonacci number (A001519) such that A001519(k) = Fibonacci(2k-1) <= n (the 'lower' odd Fibonacci Inverse).

%C Inverse of the odd Fibonacci sequence (A001519), nearly, since a(A001519(n))=n except for n=0 (see A130256 for another version). a(n)+1 is the number of odd Fibonacci numbers (A001519) <= n (for n >= 1).

%H G. C. Greubel, <a href="/A130255/b130255.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = floor((1 + arcsinh(sqrt(5)*n/2)/log(phi))/2).

%F a(n) = floor((1 + arccosh(sqrt(5)*n/2)/log(phi))/2).

%F a(n) = floor((1 + log_phi(sqrt(5)*n))/2) for n >= 1, where phi = (1 + sqrt(5))/2.

%F G.f.: g(x) = 1/(1-x)*Sum_{k>=1} x^Fibonacci(2k-1).

%F a(n) = floor((1/2)*(1 + log_phi(sqrt(5)*n + 1))) for n >= 1.

%e a(10)=3 because A001519(3) = 5 <= 10, but A001519(4) = 13 > 10.

%t Table[Floor[(1 +ArcSinh[Sqrt[5]*n/2]/Log[GoldenRatio])/2], {n, 1, 100}] (* _G. C. Greubel_, Sep 09 2018 *)

%o (PARI) phi=(1+sqrt(5))/2; vector(100, n, floor((1 +asinh(sqrt(5)*n/2)/log(phi))/2)) \\ _G. C. Greubel_, Sep 09 2018

%o (Magma) phi:=(1+Sqrt(5))/2; [Floor((1 +Argsinh(Sqrt(5)*n/2)/Log(phi))/2): n in [1..100]]; // _G. C. Greubel_, Sep 09 2018

%Y Cf. partial sums A130257. Other related sequences: A000045, A130233, A130237, A130239, A130256, A130259, A104160. Lucas inverse: A130241 - A130248.

%K nonn

%O 1,2

%A _Hieronymus Fischer_, May 24 2007, Jul 02 2007