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

Author

Hieronymus Fischer, May 17 2007, May 28 2007

Keywords

• nonn

Examples

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

Links

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

Crossrefs

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

Programs

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

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 *)

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

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).