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.

A383254 Expansion of 1/sqrt( (1-x) * (1-5*x)^3 ).

Original entry on oeis.org

1, 8, 51, 300, 1695, 9348, 50729, 272128, 1447155, 7643880, 40156281, 210019428, 1094338401, 5684293020, 29446107975, 152181330480, 784880109315, 4040712839880, 20768844586025, 106595697483700, 546389531720445, 2797395801163260, 14306735857573995
Offset: 0

Views

Author

Seiichi Manyama, May 05 2025

Keywords

Crossrefs

Programs

  • Magma
    [&+[(2*k+1)*Binomial(2*k, k)*Binomial(n+1, k+1): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, May 18 2025
  • Mathematica
    Table[Sum[(2*k+1)* Binomial[2*k, k]*Binomial[n+1,k+1],{k,0,n}],{n,0,28}] (* Vincenzo Librandi, May 18 2025 *)
  • PARI
    a(n) = sum(k=0, n, (2*k+1)*binomial(2*k, k)*binomial(n+1, k+1));
    

Formula

a(n) = Sum_{k=0..n} (2*k+1) * binomial(2*k,k) * binomial(n+1,k+1).
D-finite with recurrence n*a(n) +2*(-3*n-1)*a(n-1) +5*n*a(n-2)=0. - R. J. Mathar, May 05 2025
a(n) ~ 5^(n + 1/2) * sqrt(n/Pi). - Vaclav Kotesovec, May 05 2025
From Seiichi Manyama, Aug 19 2025: (Start)
a(n) = (1/4)^n * Sum_{k=0..n} 5^k * (2*k+1) * binomial(2*k,k) * binomial(2*(n-k),n-k).
a(n) = Sum_{k=0..n} (-1)^k * 5^(n-k) * binomial(2*k,k) * binomial(n+1,n-k). (End)

A383503 Expansion of 1/sqrt( (1-x) * (1-x-4*x^3)^3 ).

Original entry on oeis.org

1, 2, 3, 10, 23, 42, 97, 218, 435, 918, 1977, 4062, 8393, 17590, 36303, 74614, 154211, 317334, 650505, 1335054, 2736453, 5595950, 11439475, 23370270, 47681965, 97217882, 198110199, 403383026, 820820215, 1669405626, 3393344257, 6893850650, 13999109715, 28414742790
Offset: 0

Views

Author

Seiichi Manyama, May 05 2025

Keywords

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 35); Coefficients(R!( 1/Sqrt( (1-x) * (1-x-4*x^3)^3 ))); // Vincenzo Librandi, May 06 2025
  • Mathematica
    Table[Sum[(2*k+1)*Binomial[2*k,k]*Binomial[n-2*k+1,k+1],{k,0,Floor[n/3]}],{n,0,35}] (* Vincenzo Librandi, May 06 2025 *)
  • PARI
    a(n) = sum(k=0, n\3, (2*k+1)*binomial(2*k, k)*binomial(n-2*k+1, k+1));
    

Formula

a(n) = Sum_{k=0..floor(n/3)} (2*k+1) * binomial(2*k,k) * binomial(n-2*k+1,k+1).
Showing 1-2 of 2 results.