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.

A365225 G.f. satisfies A(x) = 1 + x*A(x)^5 / (1 + x*A(x)^2).

Original entry on oeis.org

1, 1, 4, 24, 169, 1301, 10605, 89963, 785943, 7023148, 63892489, 589771350, 5509967214, 52001860377, 495048989686, 4748144843341, 45838627944500, 445072967642096, 4343508043479012, 42581707009501604, 419158119684986781, 4141270208611084284
Offset: 0

Views

Author

Seiichi Manyama, Aug 27 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(2*n+3*k+1, k)*binomial(n-1, n-k)/(2*n+3*k+1));

Formula

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

A365224 G.f. satisfies A(x) = 1 + x*A(x)^4 / (1 + x*A(x)^5).

Original entry on oeis.org

1, 1, 3, 10, 30, 56, -167, -2813, -21515, -126135, -601812, -2179039, -3455504, 32238155, 430944400, 3334419890, 20083350422, 97094186751, 338485665435, 274332822425, -8491831747320, -97735154210032, -732963337489636, -4341176221239330
Offset: 0

Views

Author

Seiichi Manyama, Aug 27 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(5*n-k+1, k)*binomial(n-1, n-k)/(5*n-k+1));

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(5*n-k+1,k) * binomial(n-1,n-k)/(5*n-k+1).

A365223 G.f. satisfies A(x) = 1 + x*A(x)^3 / (1 + x*A(x)^4).

Original entry on oeis.org

1, 1, 2, 3, -3, -50, -244, -714, -530, 8522, 63548, 259473, 535647, -1321437, -19094684, -103022071, -322370363, -142186810, 5537336460, 41081448638, 170484444654, 332739198585, -1241023311708, -15677607031084, -83737193010368, -255608722098225, -12706843586158
Offset: 0

Views

Author

Seiichi Manyama, Aug 27 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(4*n-k+1, k)*binomial(n-1, n-k)/(4*n-k+1));

Formula

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