A375767 -id:A375767 - cp's OEIS frontend

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

Amiram Eldar, Aug 27 2024

First differs from A375768 at n = 2448.

All the terms are Fibonacci numbers by definition.

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

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

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]

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), ", ")));}

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).

A375769 The indices of the terms of A375768 in the Fibonacci sequence.

Original entry on oeis.org

0, 2, 2, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 4, 3, 2, 4, 3, 2, 2, 2, 5, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 2, 2, 3, 3, 2, 3, 2, 4, 2, 4, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 5, 2, 3, 3, 3, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2

Offset: 1

Amiram Eldar, Aug 27 2024

First differs from A375767 at n = 2448.

Since 1 appears twice in the Fibonacci sequence (1 = Fibonacci(1) = Fibonacci(2)), its index here is chosen to be 2.

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

Cf. A000045, A130233, A369939, A375767, A375768.

fibQ[n_] := Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}]; A130233[n_] := Module[{k = 2}, While[Fibonacci[k] <= n, k++]; k-1]; s[n_] := Module[{e = Max[FactorInteger[n][[;; , 2]]]}, If[fibQ[e], A130233[e], Nothing]]; s[1] = 0; Array[s, 100]

isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
A130233(n) = {my(k = 2); while(fibonacci(k) <= n, k++); k-1;}
lista(kmax) = {my(e); print1(0, ", "); for(k = 2, kmax, e = vecmax(factor(k)[,2]); if(isfib(e), print1(A130233(e), ", ")));}

a(n) = A130233(A375768(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (2/zeta(2) + Sum_{k>=3} k * (1/zeta(Fibonacci(k)+1) - 1/zeta(Fibonacci(k)))) / d = 2.4999593748274972257073..., where d = 1/zeta(4) + Sum_{k>=5} (1/zeta(Fibonacci(k)+1) - 1/zeta(Fibonacci(k))) = 0.94462177878047854647... is the density of A369939.
If the chosen index for 1 is 1 instead of 2, then the asymptotic mean is (1/zeta(2) + Sum_{k>=3} k * (1/zeta(Fibonacci(k)+1) - 1/zeta(Fibonacci(k)))) / d = 1.85639269500896710302009... .

cp's OEIS Frontend

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

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