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 1-4 of 4 results.

A348085 a(n) = [x^n] Product_{k=1..2*n} 1/(1 - (2*k-1) * x).

Original entry on oeis.org

1, 4, 170, 13776, 1652442, 262842580, 52116296024, 12380577235040, 3427841258566890, 1083931844930932140, 385417972804020879450, 152219732613102667656000, 66113646914860527721527960, 31319437721634527178263452656
Offset: 0

Views

Author

Seiichi Manyama, Sep 28 2021

Keywords

Crossrefs

Programs

  • PARI
    a(n) = polcoef(1/prod(k=1, 2*n, 1-(2*k-1)*x+x*O(x^n)), n);
    
  • PARI
    a(n) = if(n==0, 1, -sum(k=0, 2*n-1, (-1)^k*(2*k+1)^(3*n-1)*binomial(2*n-1, k))/(2^(2*n-1)*(2*n-1)!));

Formula

a(n) = A039755(3*n-1,2*n-1) for n > 0.
a(n) = (-1/(2^(2*n-1) * (2*n-1)!)) * Sum_{k=0..2*n-1} (-1)^k * (2*k+1)^(3*n-1) * binomial(2*n-1,k) for n > 0.
a(n) ~ 3^(3*n - 1/2) * n^(n - 1/2) / (sqrt(2*Pi*(1-c)) * (3 - 2*c)^n * c^(2*n - 1/2) * exp(n)), where c = -LambertW(-3*exp(-3/2)/2) = 0.62578253420128292... - Vaclav Kotesovec, Oct 02 2021
From Seiichi Manyama, May 16 2025: (Start)
a(n) = Sum_{k=0..n} 2^k * binomial(3*n-1,k+2*n-1) * Stirling2(k+2*n-1,2*n-1) for n > 0.
a(n) = Sum_{k=0..n} (-2)^k * (4*n-1)^(n-k) * binomial(3*n-1,k+2*n-1) * Stirling2(k+2*n-1,2*n-1) for n > 0. (End)

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))).

A383882 a(n) = [x^n] Product_{k=1..4*n} 1/(1 - k*x).

Original entry on oeis.org

1, 10, 750, 106470, 22350954, 6220194750, 2157580085700, 896587036640680, 434225240080346858, 240175986308550372366, 149377949042637543000150, 103192471874508023383125750, 78394850841083734162487127720, 64957213308036504429927388238088, 58298851680969051596827194829579744
Offset: 0

Views

Author

Vaclav Kotesovec, May 13 2025

Keywords

Comments

In general, for m>=1, Stirling2((m+1)*n, m*n) ~ (-1)^(m*n) * (m+1)^((m+1)*n) * n^(n - 1/2) / (sqrt(2*Pi*(1 + w(m))) * exp(n) * m^(m*n + 1/2) * w(m)^(m*n) * (1 + 1/m + w(m))^n), where w(m) = LambertW(-(1 + 1/m)/exp(1 + 1/m)).

Crossrefs

Cf. A007820 (m=1), A348084 (m=2), A383881 (m=3).
Cf. A217913.

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[1/(1-k*x), {k, 1, 4*n}], {x, 0, n}], {n, 0, 15}]
    Table[StirlingS2[5*n, 4*n], {n, 0, 15}]
    Table[SeriesCoefficient[1/(Pochhammer[1 - 1/x, 4*n]*x^(4*n)), {x, 0, n}], {n, 0, 15}]

Formula

a(n) = Stirling2(5*n,4*n).
a(n) ~ 5^(5*n) * n^(n - 1/2) / (sqrt(2*Pi*(1 + w)) * exp(n) * 4^(4*n + 1/2) * w^(4*n) * (5/4 + w)^n), where w = LambertW(-5/(4*exp(5/4))).

A384129 Number of permutations of 3*n objects with exactly 2*n cycles.

Original entry on oeis.org

1, 3, 85, 4536, 357423, 37312275, 4853222764, 756111184500, 137272511800831, 28460103232088385, 6634460278534540725, 1717750737160208150400, 489078062391738506912340, 151874660255802127280374140, 51082995429153110239690350120, 18500755859447038660174079965500
Offset: 0

Views

Author

Seiichi Manyama, May 20 2025

Keywords

Crossrefs

Programs

  • PARI
    a(n) = abs(stirling(3*n, 2*n, 1));

Formula

a(n) = A132393(3*n,2*n) = |Stirling1(3*n,2*n)|.
a(n) = (3*n)! * [x^(3*n)] log(1 - x)^(2*n) / (2*n)!.
a(n) ~ 3^(4*n - 1/2) * w^(3*n) * n^(n - 1/2) / (sqrt(Pi*(w-1)) * 2^(2*n + 1/2) * exp(n) * (3*w-2)^n), where w = -LambertW(-1, -2*exp(-2/3)/3) = 1.4293552275170056487... - Vaclav Kotesovec, May 23 2025
Showing 1-4 of 4 results.