A113415 Expansion of Sum_{k>0} x^k/(1-x^(2k))^2.

1, 1, 3, 1, 4, 3, 5, 1, 8, 4, 7, 3, 8, 5, 14, 1, 10, 8, 11, 4, 18, 7, 13, 3, 17, 8, 22, 5, 16, 14, 17, 1, 26, 10, 26, 8, 20, 11, 30, 4, 22, 18, 23, 7, 42, 13, 25, 3, 30, 17, 38, 8, 28, 22, 38, 5, 42, 16, 31, 14, 32, 17, 55, 1, 44, 26, 35, 10, 50, 26, 37, 8, 38, 20, 65, 11, 50, 30, 41

Offset: 1

Author

Michael Somos, Oct 29 2005

Keywords

• nonn

Comments

Arithmetic mean between the number of odd divisors (A001227) and their sum (A000593). This fact was essentially found by the algorithmic search of Jon Maiga's Sequence Machine, and is easily seen to be correct when compared to the PARI-program given by the original author. - Antti Karttunen, Dec 07 2021

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000
• Jon Maiga, Computer-generated formulas for A113415, Sequence Machine.

Crossrefs

Cf. A000265, A000593, A001227, A001511, A003602, A048673, A064989, A069734, A336840, A349371, A349915 (Dirichlet inverse).

Quadrisection of A349916.

Programs

Mathematica

Array[DivisorSum[#, If[OddQ[#], (# + 1)/2, 0] &] &, 79] (* Michael De Vlieger, Dec 08 2021 *)

PARI

a(n)=if(n<1, 0, sumdiv(n, d, if(d%2, (d+1)/2)))

Formula

G.f.: Sum_{k>0} x^k/(1-x^(2k))^2 = Sum_{k>0} k x^(2k-1)/(1-x^(2k-1)).

a(n) = (1/2) * Sum_{d|n} (d+1)*(d mod 2). - Wesley Ivan Hurt, Nov 25 2021 [From PARI prog]

From Antti Karttunen, Dec 07 2021: (Start)

All these formulas, except the last, were found by the Sequence Machine in some form or another:

a(n) = (1/2) * (A000593(n)+A001227(n)).

a(n) = A069734(A000265(n)). [See either Rutherford's or Luschny's formula in A069734]

a(n) = A349371(n) / A001511(n).

a(n) = A349371(A000265(n)) = A336840(A064989(n)).

a(n) = a(2*n) = a(A000265(n)) = A349916(4*n).

(End)