A375429 The indices of the terms of A375428 in the Fibonacci sequence.

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

Offset: 1

Author

Amiram Eldar, Aug 15 2024

Keywords

• nonn
• easy
• base

Comments

Since 1 appears twice in the Fibonacci sequence (1 = Fibonacci(1) = Fibonacci(2)), its index here is chosen to be 2.

Examples

			For n = 16 = 2^4, the Zeckendorf representation of 4 is 101, i.e., 4 = Fibonacci(2) + Fibonacci(4). Therefore 16 = 2^(Fibonacci(2) + Fibonacci(4)) = 2^Fibonacci(2) * 2^Fibonacci(4), and a(16) = 4.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A051903, A130233, A375428, A375431.

Programs

Mathematica

A130233[n_] := Module[{k = 2}, While[Fibonacci[k] <= n, k++]; k-1]; a[n_] := A130233[Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]

PARI

A130233(n) = {my(k = 2); while(fibonacci(k) <= n, k++); k-1;}
a(n) = if(n == 1, 0, A130233(vecmax(factor(n)[,2])));

Formula

a(n) = A130233(A375428(n)).

a(n) = A130233(A051903(n)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3 - 1/zeta(2) + Sum_{k>=4} (1 - 1/zeta(Fibonacci(k))) = 2.59996215929231584366... .

If the chosen index for 1 is 1 instead of 2, then the asymptotic mean is 3 - 2/zeta(2) + Sum_{k>=4} (1 - 1/zeta(Fibonacci(k))) = 1.99203505743828921499... .