A375428 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 2, 2, 1, 1, 1, 3, 1

Offset: 1

Author

Amiram Eldar, Aug 15 2024

Keywords

• nonn
• easy
• base

Comments

Differs from A095691 and A365552 at n = 1, 32, 36, 64, 72, 96, 100, ... . Differs from A368105 at n = 1, 36, 72, 100, 108, ... .

When the exponents in the prime factorization of n are expanded as sums of distinct Fibonacci numbers using the Zeckendorf representation (A014417), 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 = 16 = 2^4, the Zeckendorf representation of 4 is 101, i.e., 4 = Fibonacci(2) + Fibonacci(4) = 1 + 3. Therefore 16 = 2^(1+3) = 2^1 * 2^3, and a(16) = 3.
		

Links

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

Crossrefs

Cf. A000045, A014417, A051903, A087172, A115975, A246655, A318465, A368781, A375429, A375430.

Programs

Mathematica

A087172[n_] := Module[{k = 2}, While[Fibonacci[k] <= n, k++]; Fibonacci[k-1]]; a[n_] := A087172[Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]

PARI

A087172(n) = {my(k = 2); while(fibonacci(k) <= n, k++); fibonacci(k-1);}
a(n) = if(n == 1, 0, A087172(vecmax(factor(n)[,2])));

Formula

a(n) = A087172(A051903(n)) for n >= 2.

a(n) = A000045(A375429(n)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2 - 1/zeta(2) + Sum_{k>=4} Fibonacci(k) * (1 - 1/zeta(Fibonacci(k))) = 1.64419054900327345836... .