A368473 Product of exponents of prime factorization of the exponentially 2^n-numbers (A138302).

1, 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, 4, 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

Offset: 1

Author

Amiram Eldar, Dec 26 2023

Keywords

• nonn
• easy

Comments

The terms of A005361 that are powers of 2 (A000079).

The first position of 2^k, for k = 0, 1, ..., is 1, 4, 15, 126, 1134, ..., which is the position of A085629(2^k) in A138302.

Links

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

Crossrefs

Cf. A000079, A005361, A085629, A138302, A271727.

Cf. A366538, A367169.

Similar sequences: A322327, A368472, A368474.

Programs

Mathematica

f[n_] := Module[{p = Times @@ FactorInteger[n][[;; , 2]]}, If[p == 2^IntegerExponent[p, 2], p, Nothing]]; Array[f, 150]

PARI

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

Formula

a(n) = A005361(A138302(n)).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (1/d) * Product_{p prime} (1 + Sum_{k>=0} 2^k/p^(2^k)) = 1.504710204899266020302..., where d = A271727 is the asymptotic density of the exponentially 2^n-numbers.