A375430 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.

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, 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, 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

Offset: 1

Author

Amiram Eldar, Aug 15 2024

Keywords

• nonn
• easy
• base

Comments

First differs from A299090 at n = 128. Differs from A046951 and A159631 at n = 1, 36, 64, 72, ... .

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.

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A095791, A104326, A115975, A246655, A331109, A368781, A375428, A375431.

Cf. A046951, A159631, A299090.

Programs

Mathematica

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]

PARI

A130312(n) = {my(k = 0); while(fibonacci(k) <= n, k++); fibonacci(k-2);}
a(n) = if(n == 1, 0, A130312(1 + vecmax(factor(n)[,2])));

Formula

a(n) = A130312(1 + A051903(n)).

a(n) = A000045(A375431(n)).

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... .