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.

Showing 1-2 of 2 results.

A324590 a(n) = n!^(4*n) * Product_{k=1..n} binomial(n + 1/k^2, n).

Original entry on oeis.org

1, 2, 1080, 16133644800, 139256878046022696960000, 6288402750181849898716908922601472000000000, 8322157105451357856813375261666887975745751468393837363200000000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 09 2019

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> n!^(4*n)*mul(binomial(n+1/k^2, n), k=1..n):
    seq(a(n), n=0..7);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[n!^(4*n) * Product[Binomial[1/k^2 + n, n], {k, 1, n}], {n, 1, 8}]

Formula

a(n) ~ n!^(4*n) * n^(Pi^2/6) / A303670.
a(n) ~ n^(4*n^2 + 2*n + Pi^2/6) * (2*Pi)^(2*n) / exp(4*n^2 - 1/3 - gamma*Pi^2/6 + c), where gamma is the Euler-Mascheroni constant A001620 and c = A306774 = Sum_{k>=2} (-1)^k * Zeta(k) * Zeta(2*k) / k.
a(n) = n!^n * A324596(n).

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 24 2023

A324597 a(n) = n!^(4*n) * Product_{k=1..n} binomial(n + 1/k^3, n).

Original entry on oeis.org

1, 2, 918, 11592504000, 86712397842439769400000, 3472997049383321958747830928094241894400000, 4152034082374349458781848863476555783741415883758270213129361920000000
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 09 2019

Keywords

Comments

In general, for m > 1, Product_{k=1..n} binomial(n + 1/k^m, n) ~ n^Zeta(m) / c(m), where c(m) = Product_{j>=1} Gamma(1 + 1/j^m).
Equivalently, c(m) = -gamma * Zeta(m) + Sum_{k>=2} (-1)^k*Zeta(k)*Zeta(m*k)/k, where gamma is the Euler-Mascheroni constant A001620.

Crossrefs

Programs

  • Maple
    a:= n-> n!^(4*n)*mul(binomial(n+1/k^3, n), k=1..n):
    seq(a(n), n=0..7);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[n!^(4*n) * Product[Binomial[n + 1/j^3, n], {j, 1, n}], {n, 1, 8}]

Formula

a(n) ~ n!^(4*n) * n^Zeta(3) / (Product_{j>=1} Gamma(1 + 1/j^3)).
a(n) ~ n^(4*n^2 + 2*n + Zeta(3)) * (2*Pi)^(2*n) / exp(4*n^2 - 1/3 - gamma*Zeta(3) + c), where c = A306778 = Sum_{k>=2} (-1)^k*Zeta(k)*Zeta(3*k)/k.

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 24 2023
Showing 1-2 of 2 results.