A375431 The indices of the terms of A375430 in the Fibonacci sequence.

0, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 3, 3, 2, 3, 3, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 4, 3, 3, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 4, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4, 4, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 2, 4, 2, 3, 3, 3, 2, 2, 2, 3, 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 = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3). Therefore 8 = 2^(Fibonacci(2) + Fibonacci(3)) = 2^Fibonacci(2) * 2^Fibonacci(3), and a(8) = 3.
		

Links

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

Crossrefs

Cf. A051903, A072649, A375429, A375430.

Programs

Mathematica

A072649[n_] := Module[{k = 0}, While[Fibonacci[k] <= n, k++]; k-2]; a[n_] := A072649[1 + Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]

PARI

A072649(n) = {my(k = 0); while(fibonacci(k) <= n, k++); k-2;}
a(n) = if(n == 1, 0, A072649(1 + vecmax(factor(n)[,2])));

Formula

a(n) = A072649(1 + A051903(n)) for n >= 2.

a(n) = A072649(A375430(n)) + 1 for n >= 2.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3 - 1/zeta(2) + Sum_{k>=5} (1 - 1/zeta(Fibonacci(k)-1)) = 2.47666161947309359914... .

If the chosen index for 1 is 1 instead of 2, then the asymptotic mean is 3 - 2/zeta(2) + Sum_{k>=5} (1 - 1/zeta(Fibonacci(k)-1)) = 1.86873451761906697048... .