A130239 Maximal index k of the square of a Fibonacci number such that Fib(k)^2 <= n (the 'lower' squared Fibonacci Inverse).

0, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 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, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset

0,2

Links

G. C. Greubel, Table of n, a(n) for n = 0..5000

Formula

a(n) = max(k | Fib(k)^2 <= n) = A130233(floor(sqrt(n))).

a(n) = floor(arcsinh(sqrt(5n)/2)/log(phi)), where phi=(1+sqrt(5))/2.

G.f.: (1/(1-x))*Sum_{k>=1} x^(Fib(k)^2).

Example

a(10) = 4 since Fib(4)^2 = 9 <= 10 but Fib(5)^2 = 25 > 10.

Mathematica

A130233[n_]:= Floor[Log[GoldenRatio, 3/2 +n*Sqrt[5]]];
Table[A130233[Floor[Sqrt[n]]], {n, 0, 120}] (* G. C. Greubel, Mar 18 2023 *)

Prog

(Magma)
A130233:= func< n | Floor(Log(3/2 + n*Sqrt(5))/Log((1+Sqrt(5))/2)) >;
[A130233(Floor(Sqrt(n))): n in [0..120]]; // G. C. Greubel, Mar 18 2023
(SageMath)
def A130233(n): return int(log(3/2 +n*sqrt(5), golden_ratio))
def A130239(n): return A130233(floor(sqrt(n)))
[A130239(n) for n in range(121)] # G. C. Greubel, Mar 18 2023

Crossrefs

Partial sums: A130240. Other related sequences: A000045, A130233, A130234, A130235, A130236, A130237, A130238, A130240, A130243, A130246, A130248, A130239, A130251, A130253, A130257, A130261.

Sequence in context: A357186 A316844 A331245 * A091092 A218461 A186189

Adjacent sequences: A130236 A130237 A130238 * A130240 A130241 A130242

Keyword

nonn

Author

Hieronymus Fischer, May 17 2007, May 28 2007

Status

approved