A375766 The maximum exponent in the prime factorization of the numbers whose exponents in their prime factorization are all Fibonacci numbers.

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

Offset: 1

Author

Amiram Eldar, Aug 27 2024

Keywords

• nonn
• easy

Comments

First differs from A375768 at n = 2448.

All the terms are Fibonacci numbers by definition.

Links

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

Crossrefs

Cf. A000045, A051903, A115063, A375274, A375767, A375768.

Programs

Mathematica

fibQ[n_] := Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}]; s[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[AllTrue[e, fibQ], Max[e], Nothing]]; s[1] = 0; Array[s, 100]

PARI

isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
lista(kmax) = {my(e, ans); print1(0, ", "); for(k = 2, kmax, e = factor(k)[,2]; ans = 1; for(i = 1, #e, if(!isfib(e[i]), ans = 0; break)); if(ans, print1(vecmax(e), ", ")));}

Formula

a(n) = A051903(A115063(n)).

a(n) = A000045(A375767(n)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (1/zeta(2) + Sum_{k>=3} (Fibonacci(k) * (d(k) - d(k-1)))) / A375274 = 1.52546070254904121983..., where d(k) = Product_{p prime} ((1-1/p)*(1 + Sum_{i=2..k} 1/p^Fibonacci(i))) for k >= 3, and d(2) = 1/zeta(2).