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.

A367713 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * (1 + x) * A(x^2))).

Original entry on oeis.org

1, 2, 4, 9, 20, 46, 105, 243, 560, 1296, 2994, 6928, 16019, 37060, 85713, 198282, 458636, 1060932, 2454070, 5676750, 13131210, 30374892, 70262196, 162528916, 375957183, 869654746, 2011661506, 4653323265, 10763942479, 24898869052, 57595401116, 133228161565
Offset: 0

Views

Author

Seiichi Manyama, Nov 28 2023

Keywords

Crossrefs

Programs

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

Formula

a(n) = 1 + Sum_{k=0..n-1} a(floor(k/2)) * a(n-1-k).

A367715 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * (1 + x + x^2 + x^3) * A(x^4))).

Original entry on oeis.org

1, 2, 4, 8, 16, 33, 68, 140, 288, 594, 1225, 2526, 5208, 10740, 22148, 45673, 94184, 194224, 400524, 825950, 1703249, 3512395, 7243168, 14936668, 30801992, 63519044, 130987274, 270118452, 557031032, 1148694482, 2368807011, 4884890405, 10073490200
Offset: 0

Views

Author

Seiichi Manyama, Nov 28 2023

Keywords

Crossrefs

Programs

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

Formula

a(n) = 1 + Sum_{k=0..n-1} a(floor(k/4)) * a(n-1-k).

A367717 G.f. A(x) satisfies A(x) = 1 / ((1 + x) * (1 - x * (1 + x + x^2) * A(x^3))).

Original entry on oeis.org

1, 0, 2, 2, 5, 8, 16, 30, 57, 108, 206, 390, 741, 1407, 2670, 5068, 9622, 18262, 34666, 65806, 124911, 237109, 450092, 854368, 1621784, 3078519, 5843709, 11092672, 21056400, 39969753, 75871567, 144021302, 273384733, 518945611, 985075356, 1869894158, 3549478993
Offset: 0

Views

Author

Seiichi Manyama, Nov 28 2023

Keywords

Crossrefs

Programs

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

Formula

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