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%I #8 Aug 28 2024 00:57:11
%S 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,
%T 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,
%U 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
%N The indices of the terms of A375766 in the Fibonacci sequence.
%C First differs from A375769 at n = 2448.
%C Since 1 appears twice in the Fibonacci sequence (1 = Fibonacci(1) = Fibonacci(2)), its index here is chosen to be 2.
%H Amiram Eldar, <a href="/A375767/b375767.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A130233(A375766(n)).
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (2/zeta(2) + Sum_{k>=3} (k * (d(k) - d(k-1)))) / A375274 = 2.49917281727849805875..., where d(k) = Product_{p prime} ((1-1/p)*(1 + Sum_{i=2..k} 1/p^Fibonacci(i))) for k >= 3, and d(2) = 1/zeta(2).
%F If the chosen index for 1 is 1 instead of 2, then the asymptotic mean is (1/zeta(2) + Sum_{k>=3} (k * (d(k) - d(k-1)))) / A375274 = 1.85541131398927903176... .
%t 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 = FactorInteger[n][[;; , 2]]}, If[AllTrue[e, fibQ], A130233[Max[e]], Nothing]]; s[1] = 0; Array[s, 100]
%o (PARI) isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
%o A130233(n) = {my(k = 2); while(fibonacci(k) <= n, k++); k-1;}
%o lista(kmax) = {my(e, ans); print1(0, ", "); for(k = 2, kmax, e = factor(k)[,2]; ans = 1; for(i = 1, #e, if(!isfib(e[i]), ans = 0; break)); if(ans, print1(A130233(vecmax(e)), ", ")));}
%Y Cf. A000045, A130233, A375274, A375766, A375769.
%K nonn,easy
%O 1,2
%A _Amiram Eldar_, Aug 27 2024