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.

A365724 G.f. satisfies A(x) = 1 + x^3*A(x)^3*(1 + x*A(x)).

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 3, 7, 4, 12, 45, 55, 77, 286, 546, 728, 1960, 4760, 7548, 15504, 39729, 75582, 140448, 336490, 723327, 1366200, 2992990, 6758895, 13522275, 28094040, 63183315, 133231800, 273896532, 600805296, 1305229332, 2720740792, 5843241088, 12797739672
Offset: 0

Views

Author

Seiichi Manyama, Sep 17 2023

Keywords

Crossrefs

Programs

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

Formula

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

A365725 G.f. satisfies A(x) = 1 + x^3*A(x)^4*(1 + x*A(x)).

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 4, 9, 5, 22, 78, 91, 175, 680, 1224, 1938, 6270, 14630, 24794, 63756, 166980, 322920, 720720, 1900080, 4125888, 8803008, 22151360, 51778804, 111882100, 267682272, 645736432, 1442390092, 3346519020, 8094247798, 18657762006, 42890295734
Offset: 0

Views

Author

Seiichi Manyama, Sep 17 2023

Keywords

Crossrefs

Programs

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

Formula

a(n) = Sum_{k=0..floor(n/3)} binomial(k,n-3*k) * binomial(n+k+1,k) / (n+k+1).
G.f.: (1/x) * Series_Reversion( x*(1 - x^3*(1 + x)) ). - Seiichi Manyama, Sep 24 2024

A365726 G.f. satisfies A(x) = 1 + x^3*A(x)^5*(1 + x*A(x)).

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 5, 11, 6, 35, 120, 136, 336, 1330, 2310, 4301, 15456, 35100, 64701, 193662, 508921, 1023000, 2643432, 7298984, 16196682, 38795055, 105939288, 254015541, 596987183, 1575487320, 3959803694, 9418896773, 24081344034, 61781452530, 150293865540
Offset: 0

Views

Author

Seiichi Manyama, Sep 17 2023

Keywords

Crossrefs

Programs

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

Formula

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