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A375431
The indices of the terms of A375430 in the Fibonacci sequence.
3
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, 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, 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
OFFSET
1,2
COMMENTS
Since 1 appears twice in the Fibonacci sequence (1 = Fibonacci(1) = Fibonacci(2)), its index here is chosen to be 2.
LINKS
FORMULA
a(n) = A072649(1 + A051903(n)) for n >= 2.
a(n) = A072649(A375430(n)) + 1 for n >= 2.
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... .
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... .
EXAMPLE
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.
MATHEMATICA
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]
PROG
(PARI) A072649(n) = {my(k = 0); while(fibonacci(k) <= n, k++); k-2; }
a(n) = if(n == 1, 0, A072649(1 + vecmax(factor(n)[, 2])));
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
Amiram Eldar, Aug 15 2024
STATUS
approved