A386259 Exponent of the highest power of 2 dividing the maximum exponent in the prime factorization of n; a(1) = 0.

0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 2, 2, 0, 0, 1, 0, 0, 0

Offset: 1

Author

Amiram Eldar, Jul 17 2025

Keywords

• nonn
• easy

Comments

First differs from A386258 at n = 36.

The first occurrence of k = 1, 2, ... is at n = 2^(2^k) = A001146(k).

The asymptotic density of the occurrences of 1 in this sequence is Sum_{k>=0} (1/zeta(4*k+3) - 1/zeta(4*k+2)) = 0.23327987006010127485... .

Links

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

Crossrefs

Cf. A001146, A001511, A007814, A051903, A386258.

Programs

Mathematica

a[n_] := IntegerExponent[Max[FactorInteger[n][[;; , 2]]], 2]; a[1] = 0; Array[a, 100]

PARI

a(n) = if(n == 1, 0, valuation(vecmax(factor(n)[,2]), 2));

Formula

a(n) = A007814(A051903(n)) for n >= 2.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} A001511(k) * (1/zeta(2*k+1)-1/zeta(2*k)) = 0.32062532266909439596... .