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.

A367830 E.g.f. A(x) satisfies A(x) = (1 + (exp(x) - 1) * A(2*x)) / (1 - x).

Original entry on oeis.org

1, 2, 13, 208, 7817, 681626, 136872113, 62739300968, 64993463748977, 150619722938940622, 773428868899900772345, 8724654696222415759129388, 214574098061440421518595200025, 11429824974654804201081062775335234, 1311103770238649103823410558613476172193
Offset: 0

Views

Author

Seiichi Manyama, Dec 02 2023

Keywords

Crossrefs

Programs

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

Formula

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

A367829 E.g.f. A(x) satisfies A(x) = (1 - log(1 - x) * A(3*x)) / (1 - x).

Original entry on oeis.org

1, 2, 17, 530, 60332, 24882484, 36501847110, 186651759218364, 3267898148335418280, 193010228785740170125728, 37993098362777240856612204096, 24678625994736515097158433120107040, 52461378922253347510159057679901573120528
Offset: 0

Views

Author

Seiichi Manyama, Dec 02 2023

Keywords

Crossrefs

Programs

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

Formula

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

A367925 Expansion of e.g.f. 1/(4 - x - 3*exp(x)).

Original entry on oeis.org

1, 4, 35, 459, 8025, 175383, 4599507, 140728437, 4920898317, 193579534155, 8461200381111, 406815231899409, 21337866382711521, 1212458502624643719, 74193773349948903483, 4864422156647044661949, 340191752483516373189621, 25278147388666498256368323
Offset: 0

Views

Author

Seiichi Manyama, Dec 05 2023

Keywords

Crossrefs

Programs

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

Formula

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