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.

A365177 E.g.f. satisfies A(x) = 1 + x*A(x)^4*exp(x*A(x)^3).

Original entry on oeis.org

1, 1, 10, 201, 6220, 261465, 13925286, 898994383, 68240292856, 5956670911041, 587896878021130, 64738492669538391, 7869297152389747284, 1046629627952327990545, 151192146681811716344878, 23573456446401808474471455, 3945806733850334447131941616
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2023

Keywords

Crossrefs

Programs

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

Formula

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

A365176 E.g.f. satisfies A(x) = 1 + x*A(x)^4*exp(x*A(x)^2).

Original entry on oeis.org

1, 1, 10, 195, 5836, 236925, 12177966, 758458603, 55528414264, 4674208189977, 444823048027450, 47227542351423951, 5534636939373353604, 709653811287800826421, 98825110036657191358822, 14853654178825132742729715, 2396666529204491489278153456
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2023

Keywords

Crossrefs

Programs

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

Formula

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

A382044 E.g.f. A(x) satisfies A(x) = 1 + x*A(x)^4*exp(2*x*A(x)).

Original entry on oeis.org

1, 1, 12, 252, 8096, 352120, 19372512, 1290832480, 101078857728, 9098805892608, 925857411706880, 105098610198360064, 13167689873652178944, 1804954814456584081408, 268702350796640969736192, 43172786067215188056023040, 7446421094705349321120677888, 1372319952106065844255081037824
Offset: 0

Views

Author

Seiichi Manyama, Mar 13 2025

Keywords

Crossrefs

Programs

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

Formula

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