A386260 Maximum exponent in the prime factorization of the exponent of the highest power of 2 dividing 2*n.

1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1

Offset: 1

Author

Amiram Eldar, Jul 17 2025

Keywords

• nonn
• easy

Comments

The first occurrence of k = 1, 2, ... is at n = 2^(2^k-2) = A051191(k).

The asymptotic density of the occurrences of 1 in this sequence is 4 * Sum_{k squarefree > 1} (1/2^k - 1/2^(k+1)) = 0.862752712766... .

Links

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

Crossrefs

Cf. A001511, A051191, A051903.

Programs

Mathematica

a[n_] := Module[{v = IntegerExponent[n, 2] + 2}, If[v == 1, 0, Max[FactorInteger[v][[;;, 2]]]]]; Array[a, 100]

PARI

a(n) = my(v = valuation(4*n, 2)); if(v == 1, 0, vecmax(factor(v)[,2]));

Formula

a(n) = A051903(A001511(2*n)).

A051903(A001511(2*n-1)) = 0 for all n >= 1, and therefore the odd-indexed terms of A051903(A001511(n)) are omitted.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=2} A051903(k)/2^(k-1) = 1.14512095789925078232... . If the odd-indexed zero terms had not been omitted, the asymptotic mean would be half this value, 0.57256047894962539116... .