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.

A307905 Coefficient of x^n in (1 + n*x + x^3)^n.

Original entry on oeis.org

1, 1, 4, 30, 304, 3875, 59631, 1076383, 22309120, 522262245, 13631508400, 392535959156, 12362973152751, 422774554883590, 15600699362473876, 617888566413340503, 26145122799198386944, 1177107512023013681429, 56185125998674634494980, 2834081165961033246374350
Offset: 0

Views

Author

Seiichi Manyama, May 05 2019

Keywords

Crossrefs

Programs

  • Maple
    f:= n -> coeff((1+n*x+x^3)^n,x,n):
    map(f, [$0..30]); # Robert Israel, Mar 27 2023
  • Mathematica
    Flatten[{1, Table[n^n * HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, -n/3}, {1/2, 1}, -27/(4*n^3)], {n, 1, 20}]}] (* Vaclav Kotesovec, May 05 2019 *)
  • PARI
    {a(n) = polcoef((1+n*x+x^3)^n, n)}
    
  • PARI
    {a(n) = sum(k=0, n\3, n^(n-3*k)*binomial(n,3*k)*binomial(3*k,k))}

Formula

a(n) = Sum_{k=0..floor(n/3)} n^(n-3*k) * binomial(n,3*k) * binomial(3*k,k).
a(n) ~ c * n^n, where c = Sum_{k>=0} 1/(k!*(2*k)!) = HypergeometricPFQ[{}, {1/2, 1}, 1/4] = 1.52106585051363080966025715155941607334728986626976774617... - Vaclav Kotesovec, May 05 2019

A307904 Coefficient of x^n in (1 + x + n*x^3)^n.

Original entry on oeis.org

1, 1, 1, 10, 49, 151, 901, 5881, 28225, 165565, 1158601, 6993196, 44201521, 320103070, 2200745821, 15118248601, 113390231809, 845797019077, 6250243032145, 48718551529210, 384815404148401, 3021055319338813, 24492293678972725, 202303201125303565, 1669594463059152961
Offset: 0

Views

Author

Seiichi Manyama, May 05 2019

Keywords

Crossrefs

Programs

  • Mathematica
    Table[HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, -n/3}, {1/2, 1}, -27*n/4], {n, 0, 25}] (* Vaclav Kotesovec, May 05 2019 *)
  • PARI
    {a(n) = polcoef((1+x+n*x^3)^n, n)}
    
  • PARI
    {a(n) = sum(k=0, n\3, n^k*binomial(n, 3*k)*binomial(3*k, k))}

Formula

a(n) = Sum_{k=0..floor(n/3)} n^k * binomial(n,3*k) * binomial(3*k,k).
log(a(n)) ~ (n/3 - 1/2)*log(n) + (log(3) - 2*log(2)/3)*n + (2*n)^(2/3)/3 - (2*n)^(1/3)/9. - Vaclav Kotesovec, May 05 2019
Showing 1-2 of 2 results.