%I #7 Aug 16 2024 21:21:06
%S 0,1,1,2,1,1,1,2,2,1,1,2,1,1,1,3,1,2,1,2,1,1,1,2,2,1,2,2,1,1,1,3,1,1,
%T 1,2,1,1,1,2,1,1,1,2,2,1,1,3,2,2,1,2,1,2,1,2,1,1,1,2,1,1,2,3,1,1,1,2,
%U 1,1,1,2,1,1,2,2,1,1,1,3,3,1,1,2,1,1,1,2,1,2,1,2,1,1,1,3,1,2,2,2,1,1,1,2,1
%N The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the dual Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.
%C First differs from A299090 at n = 128. Differs from A046951 and A159631 at n = 1, 36, 64, 72, ... .
%C When the exponents in the prime factorization of n are expanded as sums of distinct Fibonacci numbers using the dual Zeckendorf representation (A104326), we get a unique factorization of n in terms of distinct terms of A115975, i.e., n is represented as a product of prime powers (A246655) whose exponents are Fibonacci numbers. a(n) is the maximum exponent of these prime powers. Thus all the terms are Fibonacci numbers.
%H Amiram Eldar, <a href="/A375430/b375430.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A130312(1 + A051903(n)).
%F a(n) = A000045(A375431(n)).
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{k>=4} Fibonacci(k) * (1 - 1/zeta(Fibonacci(k)-1)) = 1.48543763231328442311... .
%e For n = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.
%t A130312[n_] := Module[{k = 0}, While[Fibonacci[k] <= n, k++]; Fibonacci[k-2]]; a[n_] := A130312[1 + Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]
%o (PARI) A130312(n) = {my(k = 0); while(fibonacci(k) <= n, k++); fibonacci(k-2);}
%o a(n) = if(n == 1, 0, A130312(1 + vecmax(factor(n)[,2])));
%Y Cf. A000045, A095791, A104326, A115975, A246655, A331109, A368781, A375428, A375431.
%Y Cf. A046951, A159631, A299090.
%K nonn,easy,base
%O 1,4
%A _Amiram Eldar_, Aug 15 2024