A174199 Bisection of A137921.

1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 3, 5, 3, 5, 5, 5, 3, 6, 3, 6, 5, 5, 3, 7, 5, 5, 6, 6, 3, 7, 3, 6, 6, 5, 7, 8, 3, 5, 6, 8, 3, 8, 3, 7, 8, 5, 3, 9, 5, 7, 6, 7, 3, 9, 6, 8, 6, 5, 3, 11, 3, 5, 9, 7, 7, 8, 3, 7, 6, 10, 3, 11, 3, 5, 9, 7, 7, 8, 3, 10, 8, 5, 3, 11, 7, 5, 6, 9, 3, 12, 6, 7, 6, 5, 7, 11, 3, 8, 10, 10, 3, 9, 3, 9, 11, 5, 3, 12, 3, 9, 6, 10, 3, 9, 7, 7, 10, 5, 7, 14

Offset: 1

Author

N. J. A. Sloane, Nov 26 2010

Keywords

• nonn

Comments

Since the other bisection (A099774) is known - it is the bisection of A000005 - the obvious problem is to find a formula for this sequence.

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000005, A001620, A099774, A137921.

Programs

PARI

a(n) = sumdiv(2*n, d, ((2*n)%(d+1) != 0)); \\ Amiram Eldar, Jan 18 2024

Formula

From Amiram Eldar, Jan 18 2024: (Start)

a(n) = A137921(2*n).

Sum_{k=1..n} a(k) ~ (n/2) * (3*log(n) + 6*gamma - 7 + log(2)), where gamma is Euler's constant (A001620). (End)