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 1-2 of 2 results.

A372004 G.f. A(x) satisfies A(x) = ( 1 + 9*x*A(x)*(1 + x*A(x)) )^(1/3).

Original entry on oeis.org

1, 3, 3, 0, 9, 0, -63, 189, 0, -1944, 6399, 0, -72009, 245430, 0, -2921832, 10184130, 0, -125775585, 445134690, 0, -5641620192, 20188568790, 0, -260832419406, 941254831539, 0, -12342425759136, 44833549152825, 0, -594857401230510, 2172276845159733, 0
Offset: 0

Views

Author

Seiichi Manyama, Apr 15 2024

Keywords

Crossrefs

Programs

  • Maple
    A372004 := proc(n)
        add(9^k*binomial((n+1)/3,k)*binomial(k,n-k),k=0..n)/(n+1) ;
    end proc:
    seq(A372004(n),n=0..60) ; # R. J. Mathar, Apr 22 2024
  • PARI
    a(n) = sum(k=0, n, 9^k*binomial(n/3+1/3, k)*binomial(k, n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} 9^k * binomial(n/3+1/3,k) * binomial(k,n-k).
a(3*n+2) = 0 for n > 0.
a(n) = 9^n*binomial((n+1)/3, n)*hypergeom([(1-n)/2, -n/2], [2*(2-n)/3], 4/9)/(n+1). - Stefano Spezia, Apr 18 2024
D-finite with recurrence n*(n-1)*(2*n-11)*a(n) -108*(n-5)*(n-3)^2*a(n-3) -135*(n-5)*(n-8)*(2*n-5)*a(n-6)=0. - R. J. Mathar, Apr 22 2024

A372005 G.f. A(x) satisfies A(x) = ( 1 + 16*x*A(x)*(1 + x*A(x)) )^(1/4).

Original entry on oeis.org

1, 4, -4, 0, 136, -1152, 5152, 0, -230560, 2267136, -11355008, 0, 594412800, -6184304640, 32458736640, 0, -1828185954816, 19583341166592, -105435193825280, 0, 6195266435870720, -67554137604096000, 369569533686562816, 0, -22322916873246359552, 246346071588005216256
Offset: 0

Views

Author

Seiichi Manyama, Apr 15 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, 16^k*binomial(n/4+1/4, k)*binomial(k, n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} 16^k * binomial(n/4+1/4,k) * binomial(k,n-k).
a(4*n+3) = 0 for n >= 0.
a(n) = 16^n*binomial((n+1)/4, n)*hypergeom([(1-n)/2, -n/2], [(5-3*n)/4], 1/4)/(n+1). - Stefano Spezia, Apr 18 2024
Showing 1-2 of 2 results.