A374327 The maximal exponent in the prime factorization of the numbers whose maximal exponent in their prime factorization is a power of 2.

1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2

Offset: 1

Author

Amiram Eldar, Jul 04 2024

Keywords

• nonn
• easy

Comments

First differs from {A369933(n+1), n>=1} at n = 378.

The first occurrence of 2^k, for k = 0, 1, ..., is at 1, 3, 14, 224, 57307, ..., which is the position of 2^(2^k) at A369938.

Links

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

Crossrefs

Cf. A051903, A369933, A369938.

Similar sequences: A374324, A374325, A374326, A374328.

Programs

Mathematica

f[n_] := Module[{e = If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]}, If[e == 2^IntegerExponent[e, 2], e, Nothing]]; Array[f, 150]

PARI

lista(kmax) = {my(e); for(k = 2, kmax, e = vecmax(factor(k)[, 2]); if(e >> valuation(e, 2) == 1, print1(e, ", ")));}

Formula

a(n) = 2^A374328(n).

a(n) = A051903(A369938(n)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=0} 2^k * d(k) / Sum_{k>=0} d(k) = 1.41151462942556759486..., where d(k) = 1/zeta(2^k+1) - 1/zeta(2^k) for k>=1, and d(0) = 1/zeta(2).