A100007 Number of unitary divisors of 2n-1 (d such that d divides 2n-1, GCD(d,(2n-1)/d)=1). Bisection of A034444.

1, 2, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 2, 2, 2, 2, 4, 4, 2, 4, 2, 2, 4, 2, 2, 4, 2, 4, 4, 2, 2, 4, 4, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 4, 2, 4, 4, 4, 2, 4, 2, 2, 8, 2, 2, 4, 2, 4, 4, 4, 2, 4, 2, 2, 4, 2, 4, 4, 2, 2, 4, 4, 4, 4, 2, 2, 4, 4, 2, 4, 4, 2, 8, 2, 2, 4, 2, 4, 4, 2, 2, 4, 4, 4, 4, 2, 2, 8, 2, 2, 4, 4, 4, 4, 4

Offset: 1

Author

N. J. A. Sloane, Nov 20 2004

Keywords

• nonn
• easy

Examples

			a(13)=2 because among the three divisors of 25 only 1 and 25 are unitary.
		

Crossrefs

Cf. A001620, A034444, A068068, A100008.

Programs

Maple

with(numtheory): for n from 1 to 120 do printf(`%d,`,2^nops(ifactors(2*n-1)[2])) od: # Emeric Deutsch, Dec 24 2004

Mathematica

a[n_] := 2^PrimeNu[2*n-1]; Array[a, 100] (* Amiram Eldar, Jan 28 2023 *)

PARI

a(n) = 2^omega(2*n-1); \\ Amiram Eldar, Jan 28 2023

Formula

From Ilya Gutkovskiy, Apr 28 2017: (Start)

a(n) = [x^(2*n-1)] Sum_{k>=1} mu(k)^2*x^k/(1 - x^k).

a(n) = 2^omega(2*n-1). (End)

From Amiram Eldar, Jan 28 2023: (Start)

a(n) = A034444(2*n-1) = A068068(2*n-1).

Sum_{k=1..n} a(k) ~ 4*n*((log(n) + 2*gamma - 1 + 7*log(2)/3 - 2*zeta'(2)/zeta(2)) / Pi^2, where gamma is Euler's constant (A001620). (End)

Extensions

More terms from Emeric Deutsch, Dec 24 2004