A369933 The maximal exponent in the prime factorization of the exponentially 2^n numbers (A138302).

0, 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

Offset: 1

Author

Amiram Eldar, Feb 06 2024

Keywords

• nonn
• easy

Comments

Differs from A368473 at n = 1, 32, 89, 126, 159, ... .

Links

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

Crossrefs

Cf. A013661, A051903, A271727, A138302, A368473, A369934.

Programs

Mathematica

pow2Q[n_] := n == 2^IntegerExponent[n, 2]; f[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[AllTrue[e, pow2Q], Max @@ e, Nothing]]; f[1] = 0; Array[f, 150]

PARI

ispow2(n) = n >> valuation(n, 2) == 1;
lista(kmax) = {my(e); print1(0, ", "); for(k = 2, kmax, e = factor(k)[, 2]; if(ispow2(vecprod(e)), print1(vecmax(e), ", "))); }

Formula

a(n) = A051903(A138302(n)).

a(n) = 2^A369934(n), for n >= 2.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (1/zeta(2) + Sum_{k>=1} (2^k * (d(k) - d(k-1)))) / A271727 = 1.40540547368932408503..., where d(k) = Product_{p prime} (1 - 1/p^3 + Sum_{i=2..k} (1/p^(2^i)-1/p^(2^i+1))) for k >= 1, and d(0) = 1/zeta(2).