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 1-3 of 3 results.

A371826 a(n) = Sum_{k=0..floor(n/2)} n^k * binomial(2*n-k,n-2*k).

Original entry on oeis.org

1, 2, 8, 35, 170, 872, 4740, 26994, 161006, 1001009, 6476976, 43480373, 302250196, 2170406149, 16070240276, 122453910495, 958755921686, 7701233828576, 63381318474768, 533793776053926, 4595440308780620, 40400161269188412, 362367733795887848
Offset: 0

Views

Author

Seiichi Manyama, Apr 07 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\2, n^k*binomial(2*n-k, n-2*k));

Formula

a(n) = [x^n] 1/((1-x-n*x^2) * (1-x)^n).
a(n) ~ exp(3*sqrt(n)/2) * n^(n/2) / 2. - Vaclav Kotesovec, Apr 07 2024

A371825 a(n) = Sum_{k=0..n} n^k * binomial(2*n,n-k).

Original entry on oeis.org

1, 3, 18, 146, 1510, 19302, 296520, 5339924, 110447046, 2581169510, 67274981356, 1934941601628, 60878718397276, 2080009638726684, 76694241037743300, 3035502492679964520, 128364744764608411718, 5776084128332328033798, 275565308510875579650348
Offset: 0

Views

Author

Seiichi Manyama, Apr 07 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, n^k*binomial(2*n, n-k));

Formula

a(n) = [x^n] 1/((1-(n+1)*x) * (1-x)^n).
a(n) ~ exp(2) * n^n. - Vaclav Kotesovec, Apr 07 2024

A371837 a(n) = Sum_{k=0..floor(n/3)} n^k * binomial(2*n-3*k-1,n-3*k).

Original entry on oeis.org

1, 1, 3, 13, 51, 201, 834, 3529, 15075, 65431, 288278, 1285263, 5799470, 26492103, 122432628, 572291385, 2705760291, 12937116213, 62542367166, 305668511259, 1510080076410, 7539381024297, 38034307340076, 193835252945487, 997724306958606, 5185731234177001
Offset: 0

Views

Author

Seiichi Manyama, Apr 08 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Join[{1}, Table[Sum[n^k*Binomial[2*n-3*k-1,n-1], {k, 0, n/3}], {n, 1, 25}]] (* Vaclav Kotesovec, Apr 08 2024 *)
  • PARI
    a(n) = sum(k=0, n\3, n^k*binomial(2*n-3*k-1, n-3*k));

Formula

a(n) = [x^n] 1/((1-n*x^3) * (1-x)^n).
a(n) ~ exp(n^(2/3) + n^(1/3)/2 + 1/3) * n^(n/3) / 3. - Vaclav Kotesovec, Apr 08 2024
Showing 1-3 of 3 results.