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

1,2

Comments

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

Links

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

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... .

Example

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.

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]

Prog

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

Crossrefs

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

Sequence in context: A328162 A375767 A375769 * A067132 A336500 A328401

Adjacent sequences: A375426 A375427 A375428 * A375430 A375431 A375432

Keyword

nonn,easy,base

Author

Amiram Eldar, Aug 15 2024

Status

approved