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.

A366118 Expansion of (1/x) * Series_Reversion( x*(1+x+x^3)/(1+x)^4 ).

Original entry on oeis.org

1, 3, 12, 54, 259, 1292, 6616, 34507, 182395, 973691, 5236794, 28323936, 153842660, 838210232, 4577088300, 25029752243, 136984351418, 749861003157, 4103514165744, 22437771907824, 122529735856669, 667928556599614, 3632677249330084, 19701490439727844
Offset: 0

Views

Author

Seiichi Manyama, Sep 29 2023

Keywords

Comments

a(40) is negative.

Crossrefs

Programs

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

Formula

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

A366119 Expansion of (1/x) * Series_Reversion( x*(1+x+x^3)/(1+x)^5 ).

Original entry on oeis.org

1, 4, 22, 139, 950, 6831, 50899, 389402, 3040433, 24127641, 194015198, 1577355740, 12943616840, 107061667210, 891666351501, 7471125565836, 62932782325745, 532621542290355, 4526846632642489, 38621126867786485, 330635368752515710, 2839444812305017875
Offset: 0

Views

Author

Seiichi Manyama, Sep 29 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\3, (-1)^k*binomial(n+k, k)*binomial(4*n-k+4, n-3*k))/(n+1);

Formula

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

A366117 Expansion of (1/x) * Series_Reversion( x*(1+x+x^3)/(1+x)^3 ).

Original entry on oeis.org

1, 2, 5, 13, 33, 77, 147, 128, -726, -5870, -28918, -117501, -419748, -1330265, -3631065, -7585458, -5021553, 62555365, 485450778, 2473377984, 10459876208, 38765746168, 126610935508, 352046019398, 725506704096, 264040020875, -7980075940060, -58806846330257
Offset: 0

Views

Author

Seiichi Manyama, Sep 29 2023

Keywords

Crossrefs

Programs

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

Formula

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