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.

A360885 G.f. satisfies A(x) = 1 + x * A(x * (1 + x^2)).

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 7, 16, 39, 93, 246, 671, 1884, 5578, 16887, 52854, 170649, 563703, 1914366, 6649798, 23610987, 85689987, 317054427, 1196183592, 4595744763, 17965311672, 71426213637, 288535755417, 1183807706841, 4929801601890, 20825803784129, 89210585925338
Offset: 0

Views

Author

Seiichi Manyama, Feb 25 2023

Keywords

Crossrefs

Programs

  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, (i-1)\3, binomial(i-1-2*j, j)*v[i-2*j])); v;

Formula

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

A360894 G.f. satisfies A(x) = 1 + x * A(x * (1 - x)).

Original entry on oeis.org

1, 1, 1, 0, -2, -1, 7, 0, -44, 69, 276, -1471, 675, 20407, -90560, -20552, 2141700, -10558223, -675239, 329376824, -2106253225, 2364924062, 67114942438, -621638176430, 1926931098484, 14768396756732, -236623058229675, 1371752460097440, 1098671590491324
Offset: 0

Views

Author

Seiichi Manyama, Feb 25 2023

Keywords

Crossrefs

Programs

  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, (i-1)\2, (-1)^j*binomial(i-1-j, j)*v[i-j])); v;

Formula

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

A360897 G.f. satisfies A(x) = 1 + x * A(x * (1 - x^3)).

Original entry on oeis.org

1, 1, 1, 1, 1, 0, -2, -5, -9, -8, 7, 48, 120, 161, -18, -798, -2486, -4088, -692, 19840, 71159, 130467, 31737, -688014, -2644266, -5066453, -866551, 31217375, 121457519, 231494879, -10834753, -1756652362, -6638239650, -12044755426, 5372265122, 117373545212
Offset: 0

Views

Author

Seiichi Manyama, Feb 25 2023

Keywords

Crossrefs

Programs

  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, (i-1)\4, (-1)^j*binomial(i-1-3*j, j)*v[i-3*j])); v;

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/4)} (-1)^k * binomial(n-1-3*k,k) * a(n-1-3*k).
Showing 1-3 of 3 results.