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.

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A204515 a(n) = (2*n)! * (2*n+1)! / ((n+1)^2 * n!^3).

Original entry on oeis.org

1, 3, 40, 1050, 42336, 2328480, 163088640, 13913499600, 1401656256000, 162984589447680, 21497802046156800, 3172717285311974400, 518147911684085760000, 92790773980160256000000, 18083066033253630689280000, 3810158522787893903827200000
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 19 2014

Keywords

Comments

Central terms of the triangle A247500.

Crossrefs

Programs

  • Haskell
    a204515 n = a247500 (2 * n) n
    
  • Mathematica
    Table[((2n)!(2n+1)!)/((n+1)^2 n!^3),{n,0,20}] (* Harvey P. Dale, May 17 2019 *)
  • PARI
    a(n) = (2*n)! * (2*n+1)! / ((n+1)^2 * n!^3); \\ Michel Marcus, Feb 03 2022

Formula

a(n) = A248045(n+1) / (n+1).

A247499 a(n) = hypergeom([1, -n, -n-1], [2], 1).

Original entry on oeis.org

1, 2, 6, 25, 135, 896, 7042, 63841, 654901, 7491574, 94470926, 1301130777, 19423173211, 312256205652, 5376809244458, 98700795776641, 1923638785344457, 39661911384761866, 862362968121278038, 19717031047061570777, 472849461034147171791, 11866892471399392308232
Offset: 0

Views

Author

Peter Luschny, Oct 17 2014

Keywords

Crossrefs

Row sums of A247500.

Programs

  • Haskell
    a247499 = sum . a247500_row  -- Reinhard Zumkeller, Oct 19 2014
  • Maple
    A247499 := n -> hypergeom([1, -n, -n-1], [2], 1):
    seq(simplify(A247499(n)), n = 0..21);
  • Mathematica
    Table[Sum[n!/k!*Binomial[n+2, k+1]/(n+2), {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Jul 05 2018 *)

Formula

a(n) = n!*hypergeom([-n-1], [2], -1) - 1/((n+1)*(n+2)). (original name)
a(n) = Sum_{k=0..n} (n!/k!)*binomial(n+2, k+1)/(n+2).
From Vaclav Kotesovec, Jul 05 2018: (Start)
Recurrence: (n-3)*(n+2)*a(n) = 2*(n^3 - n^2 - 5*n - 1)*a(n-1) - (n-1)*(n^3 - n^2 - 3*n - 2)*a(n-2) + (n-2)^2*(n-1)^2*a(n-3).
a(n) ~ exp(2*sqrt(n) - n - 1/2) * n^(n - 1/4) / sqrt(2). (End)

Extensions

Name updated by Peter Luschny, Jan 09 2022
Showing 1-2 of 2 results.