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.

A317021 Expansion of Product_{k>=1} 1/(1 - x^k)^((3*k-1)*binomial(k+2,3)/2).

Original entry on oeis.org

1, 1, 11, 51, 216, 861, 3477, 13367, 50377, 184667, 664484, 2345230, 8142476, 27825576, 93750686, 311682789, 1023547782, 3322634928, 10669887669, 33916213669, 106776876109, 333111724130, 1030264525744, 3160359629535, 9618807643826, 29057370625281, 87153154537437
Offset: 0

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Author

Ilya Gutkovskiy, Jul 19 2018

Keywords

Comments

Euler transform of A002419.

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, add(add(
          (3*d-1)*binomial(d+2, 3)/2*d, d=numtheory
          [divisors](j))*a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, Jul 19 2018
  • Mathematica
    nmax = 26; CoefficientList[Series[Product[1/(1 - x^k)^((3 k - 1) Binomial[k + 2, 3]/2), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 26; CoefficientList[Series[Exp[Sum[x^k (1 + 5 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) (d + 2) (3 d - 1)/12, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 26}]

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^A002419(k).
G.f.: exp(Sum_{k>=1} x^k*(1 + 5*x^k)/(k*(1 - x^k)^5)).
a(n) ~ 1/(2^(1987/2160) * 3^(713/1080) * 7^(173/2160) * n^(1253/2160) * Pi^(7/360)) * exp(-1/72 + (1/12-Zeta'(-1))/6 - Zeta(3)/(30 * Pi^2) + (111 * Zeta(5))/(200 * Pi^4) - (7056 * Zeta(3) * Zeta(5)^2)/Pi^12 - (592704 * Zeta(5)^3)/(5 * Pi^14) + (43016085504 * Zeta(5)^5)/(5 * Pi^24) + (2 * Zeta'(-3))/3 + ((-7 * (7/2)^(1/6) * Pi)/(3200 * 3^(2/3)) + (14 * 2^(5/6) * 3^(1/3) * 7^(1/6) * Zeta(3) * Zeta(5))/Pi^7 + (1029 * 2^(5/6) * 3^(1/3) * 7^(1/6) * Zeta(5)^2)/(5 * Pi^9) - (17978688 * 2^(5/6) * 3^(1/3) * 7^(1/6) * Zeta(5)^4)/Pi^19) * n^(1/6) + (-((7/6)^(1/3) * Zeta(3))/(2 * Pi^2) - (7 * 3^(2/3) * (7/2)^(1/3) * Zeta(5))/(5 * Pi^4) + (75264 * 6^(2/3) * 7^(1/3) * Zeta(5)^3)/Pi^14) * n^(1/3) + ((sqrt(7/2) * Pi)/60 - (1008 * sqrt(14) * Zeta(5)^2)/Pi^9) * sqrt(n) + ((6 * 6^(1/3) * 7^(2/3) * Zeta(5))/Pi^4) * n^(2/3) + ((2 * (2/7)^(1/6) * 3^(2/3) * Pi)/5) * n^(5/6)). - Vaclav Kotesovec, Jul 28 2018