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

%I A375431 #9 Aug 16 2024 21:21:17
%S A375431 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,
%T A375431 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,
%U A375431 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
%N A375431 The indices of the terms of A375430 in the Fibonacci sequence.
%C A375431 Since 1 appears twice in the Fibonacci sequence (1 = Fibonacci(1) = Fibonacci(2)), its index here is chosen to be 2.
%H A375431 Amiram Eldar, <a href="/A375431/b375431.txt">Table of n, a(n) for n = 1..10000</a>
%F A375431 a(n) = A072649(1 + A051903(n)) for n >= 2.
%F A375431 a(n) = A072649(A375430(n)) + 1 for n >= 2.
%F A375431 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... .
%F A375431 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... .
%e A375431 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.
%t A375431 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]
%o A375431 (PARI) A072649(n) = {my(k = 0); while(fibonacci(k) <= n, k++); k-2;}
%o A375431 a(n) = if(n == 1, 0, A072649(1 + vecmax(factor(n)[,2])));
%Y A375431 Cf. A051903, A072649, A375429, A375430.
%K A375431 nonn,easy,base
%O A375431 1,2
%A A375431 _Amiram Eldar_, Aug 15 2024