A375769 The indices of the terms of A375768 in the Fibonacci sequence.

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

Offset

1,2

Comments

First differs from A375767 at n = 2448.

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(A375768(n)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (2/zeta(2) + Sum_{k>=3} (k * (1/zeta(Fibonacci(k)+1) - 1/zeta(Fibonacci(k)))) / d = 2.4999593748274972257073..., where d = 1/zeta(4) + Sum_{k>=5} (1/zeta(Fibonacci(k)+1) - 1/zeta(Fibonacci(k))) = 0.94462177878047854647... is the density of A369939.

If the chosen index for 1 is 1 instead of 2, then the asymptotic mean is (1/zeta(2) + Sum_{k>=3} (k * (1/zeta(Fibonacci(k)+1) - 1/zeta(Fibonacci(k)))) / d = 1.85639269500896710302009... .

Mathematica

fibQ[n_] := Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}]; A130233[n_] := Module[{k = 2}, While[Fibonacci[k] <= n, k++]; k-1]; s[n_] := Module[{e = Max[FactorInteger[n][[;; , 2]]]}, If[fibQ[e], A130233[e], Nothing]]; s[1] = 0; Array[s, 100]

Prog

(PARI) isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
A130233(n) = {my(k = 2); while(fibonacci(k) <= n, k++); k-1; }
lista(kmax) = {my(e); print1(0, ", "); for(k = 2, kmax, e = vecmax(factor(k)[, 2]); if(isfib(e), print1(A130233(e), ", "))); }

Crossrefs

Cf. A000045, A130233, A369939, A375767, A375768.

Sequence in context: A161189 A328162 A375767 * A375429 A067132 A336500

Adjacent sequences: A375766 A375767 A375768 * A375770 A375771 A375772

Keyword

nonn,easy

Author

Amiram Eldar, Aug 27 2024

Status

approved