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A375766
The maximum exponent in the prime factorization of the numbers whose exponents in their prime factorization are all Fibonacci numbers.
3
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
1,4
COMMENTS
First differs from A375768 at n = 2448.
All the terms are Fibonacci numbers by definition.
LINKS
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).
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]
PROG
(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), ", "))); }
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Aug 27 2024
STATUS
approved