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.

A340971 a(n) = Sum_{k=0..n} n^k * binomial(n,k) * binomial(2*k,k).

Original entry on oeis.org

1, 3, 33, 721, 23649, 1032801, 56317969, 3682424775, 280767441537, 24456613613401, 2395993939827201, 260764460901476643, 31213273328323059169, 4075382667781540713807, 576394007453263029232497
Offset: 0

Views

Author

Seiichi Manyama, Feb 01 2021

Keywords

Crossrefs

Main diagonal of A340970.

Programs

  • Mathematica
    a[0] = 1; a[n_] := Sum[n^k * Binomial[n, k] * Binomial[2*k, k], {k, 0, n}]; Array[a, 15, 0] (* Amiram Eldar, Feb 01 2021 *)
  • PARI
    a(n) = sum(k=0, n, n^k*binomial(n, k)*binomial(2*k, k));
    
  • PARI
    a(n) = polcoef(1/sqrt((1-x)*(1-(4*n+1)*x)+x*O(x^n)), n);
    
  • PARI
    a(n) = polcoef((1+(2*n+1)*x+(n*x)^2)^n, n);

Formula

a(n) = [x^n] 1/sqrt((1-x)*(1-(4*n+1)*x)).
a(n) = [x^n] (1+(2*n+1)*x+(n*x)^2)^n.
a(n) = n! * [x^n] exp((2*n+1)*x) * BesselI(0,2*n*x). - Ilya Gutkovskiy, Feb 01 2021
a(n) ~ exp(1/4) * 4^n * n^(n - 1/2) / sqrt(Pi). - Vaclav Kotesovec, Nov 13 2021
From Seiichi Manyama, Aug 19 2025: (Start)
a(n) = (1/4)^n * Sum_{k=0..n} (4*n+1)^k * binomial(2*k,k) * binomial(2*(n-k),n-k).
a(n) = Sum_{k=0..n} (-n)^k * (4*n+1)^(n-k) * binomial(n,k) * binomial(2*k,k). (End)

A383133 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(n*k,k) * n^k.

Original entry on oeis.org

1, 0, 17, 1889, 412225, 151448249, 84430503361, 66535567456546, 70456680210155009, 96530372235620300465, 166169585125820280654001, 351113456811120647774884511, 893491183170443755035588745153, 2695374684029443253628238600963667, 9511442599320236554084097413617603681
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 17 2025

Keywords

Crossrefs

Programs

  • Mathematica
    Unprotect[Power]; 0^0 = 1; Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[n k, k] n^k, {k, 0, n}], {n, 0, 14}]

Formula

a(n) = [x^n] ((1 + n*x)^n - x)^n.
a(n) ~ exp(n - 1/2) * n^(2*n - 1/2) / sqrt(2*Pi). - Vaclav Kotesovec, Apr 19 2025
Showing 1-2 of 2 results.