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.

A305653 Expansion of Product_{k>=1} 1/(1 - x^k)^((k+1)*binomial(k+2,3)/2).

Original entry on oeis.org

1, 1, 7, 27, 98, 323, 1085, 3471, 10998, 33874, 102737, 305849, 897899, 2597822, 7423408, 20957775, 58524868, 161741013, 442705279, 1200718351, 3228796864, 8611973548, 22793714865, 59887897679, 156252738062, 404964879419, 1042884107691, 2669317020743, 6792321636929
Offset: 0

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Author

Ilya Gutkovskiy, Jun 07 2018

Keywords

Comments

Euler transform of A002415, shifted left one place.

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, add(a(n-j)*add(d^2*
          (d+2)*(d+1)^2/12, d=numtheory[divisors](j)), j=1..n)/n)
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Jun 07 2018
  • Mathematica
    nmax = 28; CoefficientList[Series[Product[1/(1 - x^k)^((k + 1) Binomial[k + 2, 3]/2), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 28; CoefficientList[Series[Exp[Sum[x^k (1 + x^k)/(k (1 - x^k)^5), {k, 1, nmax}]], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d^2 (d + 1)^2 (d + 2)/12, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 28}]

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^A002415(k+1).
G.f.: exp(Sum_{k>=1} x^k*(1 + x^k)/(k*(1 - x^k)^5)).
a(n) ~ exp(Zeta'(-1)/6 - Zeta(3) / (4*Pi^2) + 149*Zeta(5) / (32*Pi^4) + 15876 * Zeta(3) * Zeta(5)^2 / Pi^12 - 666792 * Zeta(5)^3 / Pi^14 + 108884466432 * Zeta(5)^5 / (5*Pi^24) + Zeta'(-3)/3 + (-7*(7/2)^(1/6) * Pi / (384*sqrt(3)) - 21 * 2^(5/6) * sqrt(3) * 7^(1/6) * Zeta(3) * Zeta(5) / Pi^7 + 3087 * sqrt(3) * (7/2)^(1/6) * Zeta(5)^2 / (2*Pi^9) - 30339036 * 2^(5/6) * sqrt(3) * 7^(1/6) * Zeta(5)^4 / Pi^19) * n^(1/6) + ((7/2)^(1/3) * Zeta(3) / (2*Pi^2) - 21 * (7/2)^(1/3) * Zeta(5) / (2*Pi^4) + 254016 * 2^(2/3) * 7^(1/3) * Zeta(5)^3 / Pi^14) * n^(1/3) + (sqrt(7/6) * Pi / 12 - 756 * sqrt(42) * Zeta(5)^2 / Pi^9) * sqrt(n) + (9 * 2^(1/3) * 7^(2/3) * Zeta(5) / Pi^4) * n^(2/3) + (2 * (2/7)^(1/6) * sqrt(3) * Pi) / 5 * n^(5/6)) * Pi^(1/90) / (2^(247/270) * 3^(34/45) * 7^(23/270) * n^(79/135)). - Vaclav Kotesovec, Jun 08 2018