A375429 - OEIS

%I #9 Aug 16 2024 21:20:53

%S 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,

%T 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,

%U 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

%N The indices of the terms of A375428 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="/A375429/b375429.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A130233(A375428(n)).

%F a(n) = A130233(A051903(n)).

%F 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... .

%F 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... .

%e 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.

%t 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]

%o (PARI) A130233(n) = {my(k = 2); while(fibonacci(k) <= n, k++); k-1;}

%o a(n) = if(n == 1, 0, A130233(vecmax(factor(n)[,2])));

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

%K nonn,easy,base

%O 1,2

%A _Amiram Eldar_, Aug 15 2024