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.

A383881 a(n) = [x^n] Product_{k=1..3*n} 1/(1 - k*x).

Original entry on oeis.org

1, 6, 266, 22275, 2757118, 452329200, 92484925445, 22653141490980, 6466506598695390, 2108114165258886708, 772778072287000494520, 314641228029527540596455, 140880584836935832288402135, 68799366730032076856334789900, 36392216443342587869022660451080, 20728132932716479897744043460870000
Offset: 0

Views

Author

Vaclav Kotesovec, May 13 2025

Keywords

Crossrefs

Programs

  • Magma
    [&+[Abs(StirlingSecond(4*n, 3*n))]: n in [0..15]]; // Vincenzo Librandi, May 21 2025
  • Mathematica
    Table[SeriesCoefficient[Product[1/(1-k*x), {k, 1, 3*n}], {x, 0, n}], {n, 0, 15}]
    Table[StirlingS2[4*n, 3*n], {n, 0, 15}]
    Table[SeriesCoefficient[(-1)^n/(Pochhammer[1 - 1/x, 3*n]*x^(3*n)), {x, 0, n}], {n, 0, 15}]

Formula

a(n) = Stirling2(4*n,3*n).
a(n) ~ (-1)^(3*n) * 4^(4*n) * n^(n - 1/2) / (sqrt(2*Pi*(1 + w)) * exp(n) * 3^(3*n + 1/2) * w^(3*n) * (4/3 + w)^n), where w = LambertW(-4/(3*exp(4/3))).

A384130 Number of permutations of 4*n objects with exactly 3*n cycles.

Original entry on oeis.org

1, 6, 322, 32670, 4899622, 973941900, 241276443496, 71603372991150, 24764667228756390, 9781650150525639540, 4344363139637533397580, 2143082171052546774398348, 1162585907585797437278546956, 687872810620417599693839111880, 440840269604491448260396623711300
Offset: 0

Views

Author

Seiichi Manyama, May 20 2025

Keywords

Comments

In general, for m>=1, abs(Stirling1((m+1)*n, m*n)) ~ (m+1)^((m+2)*n - 1/2) * w(m)^((m+1)*n) * n^(n - 1/2) / (sqrt(2*Pi*(w(m)-1)) * exp(n) * m^(m*n) * ((m+1)*w(m) - m)^n), where w(m) = -LambertW(-1, -m*exp(-m/(m+1))/(m+1)). - Vaclav Kotesovec, May 23 2025

Crossrefs

Programs

  • Magma
    [&+[Abs(StirlingFirst(4*n, 3*n))]: n in [0..15]]; // Vincenzo Librandi, May 21 2025
  • Mathematica
    a[n_]:=Abs[StirlingS1[4 n,3 n]] Table[a[n],{n,0,15}] (* Vincenzo Librandi, May 21 2025 *)
  • PARI
    a(n) = abs(stirling(4*n, 3*n, 1));
    

Formula

a(n) = A132393(4*n,3*n) = |Stirling1(4*n,3*n)|.
a(n) = (4*n)! * [x^(4*n)] (-log(1 - x))^(3*n) / (3*n)!.
a(n) ~ 2^(10*n - 3/2) * n^(n - 1/2) * w^(4*n) / (sqrt(Pi*(w-1)) * 3^(3*n) * exp(n) * (4*w-3)^n), where w = -LambertW(-1, -3*exp(-3/4)/4) = 1.3002007416590685881... - Vaclav Kotesovec, May 23 2025
Showing 1-2 of 2 results.