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.

A374487 Expansion of 1/(1 - 2*x - 7*x^2)^(3/2).

Original entry on oeis.org

1, 3, 18, 70, 315, 1281, 5348, 21708, 88245, 355135, 1425270, 5692050, 22666735, 89986365, 356400840, 1408459928, 5555679849, 21877337979, 86020384730, 337769595870, 1324677499299, 5189411915897, 20308936981932, 79406140870500, 310206869770525, 1210898719869111
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Module[{x}, CoefficientList[Series[1/(1 - (7*x + 2)*x)^(3/2), {x, 0, 25}], x]] (* Paolo Xausa, Aug 25 2025 *)
  • PARI
    a(n) = binomial(n+2, 2)*sum(k=0, n\2, 2^k*binomial(n, 2*k)*binomial(2*k, k)/(k+1));

Formula

a(0) = 1, a(1) = 3; a(n) = ((2*n+1)*a(n-1) + 7*(n+1)*a(n-2))/n.
a(n) = binomial(n+2,2) * A025235(n).
From Seiichi Manyama, Aug 20 2025: (Start)
a(n) = ((n+2)/2) * Sum_{k=0..floor(n/2)} 2^k * binomial(n+1,n-2*k) * binomial(2*k+1,k).
a(n) = Sum_{k=0..n} (1/2)^k * (7/2)^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(k,n-k). (End)
a(n) ~ sqrt(n) * (1 + 2*sqrt(2))^(n + 3/2) / (2^(11/4) * sqrt(Pi)). - Vaclav Kotesovec, Aug 21 2025

A374488 Expansion of 1/(1 - 2*x - 11*x^2)^(3/2).

Original entry on oeis.org

1, 3, 24, 100, 555, 2541, 12628, 59004, 281655, 1315765, 6171132, 28692456, 133315273, 616780815, 2848833120, 13124483344, 60364983987, 277142478921, 1270586298520, 5817063737100, 26600252408961, 121501917998263, 554429553154044, 2527595449990500
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Module[{x}, CoefficientList[Series[1/(1 - (11*x + 2)*x)^(3/2), {x, 0, 25}], x]] (* Paolo Xausa, Aug 25 2025 *)
  • PARI
    a(n) = binomial(n+2, 2)*sum(k=0, n\2, 3^k*binomial(n, 2*k)*binomial(2*k, k)/(k+1));

Formula

a(0) = 1, a(1) = 3; a(n) = ((2*n+1)*a(n-1) + 11*(n+1)*a(n-2))/n.
a(n) = binomial(n+2,2) * A025237(n).
From Seiichi Manyama, Aug 20 2025: (Start)
a(n) = ((n+2)/2) * Sum_{k=0..floor(n/2)} 3^k * binomial(n+1,n-2*k) * binomial(2*k+1,k).
a(n) = Sum_{k=0..n} (1/2)^k * (11/2)^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(k,n-k). (End)
a(n) ~ sqrt(n) * (1 + 2*sqrt(3))^(n + 3/2) / (4 * 3^(3/4) * sqrt(Pi)). - Vaclav Kotesovec, Aug 21 2025

A374508 Expansion of 1/(1 - 2*x + 5*x^2)^(5/2).

Original entry on oeis.org

1, 5, 5, -35, -140, -84, 840, 2640, 495, -16445, -41041, 11375, 282100, 559300, -474300, -4399260, -6807225, 11062275, 63677075, 73363675, -208411280, -865816600, -665544100, 3475847700, 11129861925, 4130560161, -53332660395, -135538728395, 9634906640
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2024

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:= Pochhammer[n+1, 4]*Hypergeometric2F1[(1-n)/2, -n/2, 3, -4]/4!; Array[a,29,0] (* Stefano Spezia, Jul 10 2024 *)
  • PARI
    a(n) = binomial(n+4, 2)/6*sum(k=0, n\2, (-1)^k*binomial(n+2, n-2*k)*binomial(2*k+2, k));

Formula

a(0) = 1, a(1) = 5; a(n) = ((2*n+3)*a(n-1) - 5*(n+3)*a(n-2))/n.
a(n) = (binomial(n+4,2)/6) * Sum_{k=0..floor(n/2)} (-1)^k * binomial(n+2,n-2*k) * binomial(2*k+2,k).
a(n) = Pochhammer(n+1, 4)*hypergeom([(1-n)/2, -n/2], [3], -4)/4!. - Stefano Spezia, Jul 10 2024
a(n) = (-1)^n * Sum_{k=0..n} 2^k * (5/2)^(n-k) * binomial(-5/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 23 2025

A374509 Expansion of 1/(1 - 2*x + 5*x^2)^(7/2).

Original entry on oeis.org

1, 7, 14, -42, -294, -462, 1386, 7722, 9009, -37037, -160160, -123760, 835380, 2848860, 1046520, -16550520, -45140865, 3533145, 296447690, 648593330, -393463070, -4895709390, -8489647530, 10975099590, 75528298755, 100311659721, -230350834728, -1097798696456
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2024

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:= Pochhammer[n+1, 6]*Hypergeometric2F1[(1-n)/2, -n/2, 4, -4]/6!; Array[a,28,0] (* Stefano Spezia, Jul 10 2024 *)
  • PARI
    a(n) = binomial(n+6, 3)/20*sum(k=0, n\2, (-1)^k*binomial(n+3, n-2*k)*binomial(2*k+3, k));

Formula

a(0) = 1, a(1) = 7; a(n) = ((2*n+5)*a(n-1) - 5*(n+5)*a(n-2))/n.
a(n) = (binomial(n+6,3)/20) * Sum_{k=0..floor(n/2)} (-1)^k * binomial(n+3,n-2*k) * binomial(2*k+3,k).
a(n) = Pochhammer(n+1, 6)*hypergeom([(1-n)/2, -n/2], [4], -4)/6!. - Stefano Spezia, Jul 10 2024
a(n) = (-1)^n * Sum_{k=0..n} 2^k * (5/2)^(n-k) * binomial(-7/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 23 2025
Showing 1-4 of 4 results.