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The indices of the terms of A375430 in the Fibonacci sequence.
3

%I #9 Aug 16 2024 21:21:17

%S 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 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 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 The indices of the terms of A375430 in the Fibonacci sequence.

%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="/A375431/b375431.txt">Table of n, a(n) for n = 1..10000</a>

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

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

%F 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 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 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 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 (PARI) A072649(n) = {my(k = 0); while(fibonacci(k) <= n, k++); k-2;}

%o a(n) = if(n == 1, 0, A072649(1 + vecmax(factor(n)[,2])));

%Y Cf. A051903, A072649, A375429, A375430.

%K nonn,easy,base

%O 1,2

%A _Amiram Eldar_, Aug 15 2024