cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A328373 Expansion of Sum_{k>=1} x^(k^2) * (1 + x^(2*k^2)) / (1 - x^(2*k^2))^2.

Original entry on oeis.org

1, 0, 3, 1, 5, 0, 7, 0, 10, 0, 11, 3, 13, 0, 15, 1, 17, 0, 19, 5, 21, 0, 23, 0, 26, 0, 30, 7, 29, 0, 31, 0, 33, 0, 35, 10, 37, 0, 39, 0, 41, 0, 43, 11, 50, 0, 47, 3, 50, 0, 51, 13, 53, 0, 55, 0, 57, 0, 59, 15, 61, 0, 70, 1, 65, 0, 67, 17, 69, 0, 71, 0, 73, 0, 78, 19, 77, 0, 79, 5, 91
Offset: 1

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Author

Ilya Gutkovskiy, Oct 14 2019

Keywords

Comments

Sum of odd divisors d of n such that n/d is square.

Crossrefs

Cf. A000593, A010052, A035316, A036554 (positions of 0's), A056911 (fixed points), A076752, A193356, A328372.

Programs

  • Magma
    a:=[];for n in [1..81] do  v:=[d:d in Divisors(n)| IsOdd(d) and IsSquare(n div d)]; if #v ne 0  then Append(~a,&+v); else Append(~a,0); end if; end for; a; // Marius A. Burtea, Oct 14 2019
  • Mathematica
    nmax = 81; CoefficientList[Series[Sum[x^(k^2) (1 + x^(2 k^2))/(1 - x^(2 k^2))^2, {k, 1, Floor[Sqrt[nmax]] + 1}], {x, 0, nmax}], x] // Rest
    Table[DivisorSum[n, # &, OddQ[#] && IntegerQ[(n/#)^(1/2)] &], {n, 1, 81}]
    f[p_, e_] := If[p == 2, Boole @ EvenQ[e], If[EvenQ[e], (p^(e + 2) - 1)/(p^2 - 1), (p^(e + 2) - p)/(p^2 - 1)]]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Oct 16 2020 *)
  • PARI
    a(n) = sumdiv(n, d, if ((d%2) && issquare(n/d), d)); \\ Michel Marcus, Oct 14 2019
    

Formula

G.f.: Sum_{k>=1} (2*k - 1) * (theta_3(x^(2*k - 1)) - 1) / 2.
G.f.: Sum_{i>=1} Sum_{j>=1} phi(i) * x^(i*j^2) / (1 + x^(i*j^2)).
Dirichlet g.f.: (1 - 2^(1 - s)) * zeta(s-1) * zeta(2*s).
a(n) = Sum_{d|n} A193356(d) * A010052(n/d).
Sum_{k=1..n} a(k) ~ Pi^4 * n^2 / 360. - Vaclav Kotesovec, Oct 14 2019
Multiplicative with a(2^e) = 0 if e is odd, and 1 if e is even, and for p > 2, a(p^e) = (p^(e + 2) - p)/(p^2 - 1) if e is odd, and (p^(e + 2) - 1)/(p^2 - 1) if e is even. - Amiram Eldar, Oct 16 2020