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 1-4 of 4 results.

A365752 Expansion of (1/x) * Series_Reversion( x*(1+x)*(1-x)^4 ).

Original entry on oeis.org

1, 3, 16, 103, 735, 5592, 44452, 364815, 3067558, 26290517, 228819168, 2016953848, 17968790029, 161536295244, 1463535347928, 13349907110367, 122499957767130, 1130001670577730, 10472708110616136, 97468774074103041, 910582642690819351
Offset: 0

Views

Author

Seiichi Manyama, Sep 18 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^k*binomial(n+k, k)*binomial(5*n-k+3, n-k))/(n+1);
    
  • SageMath
    def A365752(n):
        h = binomial(5*n + 3, n) * hypergeometric([-n, n + 1], [-5 * n - 3], -1) / (n + 1)
        return simplify(h)
    print([A365752(n) for n in range(21)])  # Peter Luschny, Sep 20 2023

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(n+k,k) * binomial(5*n-k+3,n-k).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(n+k,k) * binomial(4*n-2*k+2,n-2*k). - Seiichi Manyama, Jan 18 2024
a(n) = (1/(n+1)) * [x^n] 1/( (1+x) * (1-x)^4 )^(n+1). - Seiichi Manyama, Feb 16 2024

A365878 Expansion of (1/x) * Series_Reversion( x*(1+x)^3*(1-x)^4 ).

Original entry on oeis.org

1, 1, 5, 17, 83, 381, 1939, 9905, 52544, 282315, 1545130, 8552557, 47880020, 270401515, 1539288570, 8821594865, 50860072024, 294774097800, 1716506373521, 10037592274363, 58920231785426, 347051995986538, 2050627029532225, 12151336260368205
Offset: 0

Views

Author

Seiichi Manyama, Sep 21 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^k*binomial(3*n+k+2, k)*binomial(5*n-k+3, n-k))/(n+1);
    
  • SageMath
    def A365878(n):
        h = binomial(5*n + 3, n) * hypergeometric([-n, 3*(n + 1)], [-5 * n - 3], -1) / (n + 1)
        return simplify(h)
    print([A365878(n) for n in range(24)])  # Peter Luschny, Sep 21 2023

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(3*n+k+2,k) * binomial(5*n-k+3,n-k).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(3*n+k+2,k) * binomial(2*n-2*k,n-2*k). - Seiichi Manyama, Jan 18 2024
a(n) = (1/(n+1)) * [x^n] 1/( (1+x)^3 * (1-x)^4 )^(n+1). - Seiichi Manyama, Feb 16 2024

A365879 Expansion of (1/x) * Series_Reversion( x*(1+x)^3*(1-x)^5 ).

Original entry on oeis.org

1, 2, 10, 54, 332, 2162, 14734, 103630, 746857, 5486206, 40926152, 309202686, 2361065920, 18192978966, 141280871840, 1104603758526, 8687878404289, 68692224882620, 545681467048132, 4353096328518810, 34858239962177764, 280095777427596008
Offset: 0

Views

Author

Seiichi Manyama, Sep 21 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^k*binomial(3*n+k+2, k)*binomial(6*n-k+4, n-k))/(n+1);
    
  • SageMath
    def A365879(n):
        h = binomial(6*n + 4, n) * hypergeometric([-n, 3*(n + 1)], [-6 * n - 4], -1) / (n + 1)
        return simplify(h)
    print([A365879(n) for n in range(22)])  # Peter Luschny, Sep 21 2023

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(3*n+k+2,k) * binomial(6*n-k+4,n-k).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(3*n+k+2,k) * binomial(3*n-2*k+1,n-2*k). - Seiichi Manyama, Jan 18 2024
a(n) = (1/(n+1)) * [x^n] 1/( (1+x)^3 * (1-x)^5 )^(n+1). - Seiichi Manyama, Feb 16 2024

A370271 Coefficient of x^n in the expansion of 1/( (1-x)^3 * (1-x^2)^3 )^n.

Original entry on oeis.org

1, 3, 27, 246, 2379, 23628, 239058, 2450052, 25351755, 264270765, 2771024652, 29194911342, 308813298690, 3277454178144, 34883317836240, 372195546176496, 3979793738688075, 42635773396647054, 457529396858568837, 4917191231017846902, 52917857164300253004
Offset: 0

Views

Author

Seiichi Manyama, Feb 13 2024

Keywords

Crossrefs

Cf. A368079.

Programs

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

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(3*n+k-1,k) * binomial(4*n-2*k-1,n-2*k).
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-x)^3 * (1-x^2)^3 ). See A368079.
Showing 1-4 of 4 results.