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.

A238578 Expansion of -(-4*x^4 + sqrt(-4*x^2-4*x+1) * (2*x^3+x^2-2*x) -12*x^3-7*x^2+2*x) / (sqrt(-4*x^2-4*x+1) * (4*x^3+8*x^2+3*x-1) - 4*x^3-8*x^2-3*x+1).

Original entry on oeis.org

0, 1, 3, 11, 45, 191, 833, 3695, 16593, 75199, 343233, 1575551, 7265921, 33637631, 156234497, 727681791, 3397475585, 15896054783, 74512968705, 349859309567, 1645121398785, 7746058698751, 36516283891713, 172332643868671, 814108326764545, 3849410342715391
Offset: 0

Views

Author

Vladimir Kruchinin, Mar 01 2014

Keywords

Crossrefs

Cf. A052709.

Programs

  • Mathematica
    Table[Sum[Binomial[n - 1, k - 1] * Sum[Binomial[k, n - k - i] * Binomial[k + i - 1, k - 1], {i, 0, n - k}], {k, n}], {n, 0, 20}] (* Wesley Ivan Hurt, Mar 02 2014 *)
    CoefficientList[Series[-(-4*x^4 + Sqrt[-4*x^2-4*x+1]*(2*x^3+x^2-2*x) -12*x^3-7*x^2+2*x)/(Sqrt[-4*x^2-4*x+1]*(4*x^3+8*x^2+3*x-1) - 4*x^3-8*x^2-3*x+1), {x, 0, 50}], x] (* G. C. Greubel, Jun 01 2017 *)
  • Maxima
    a(n):= sum((sum(binomial(k,n-k-i)*binomial(k+i-1,k-1), i,0,n-k)) *binomial(n-1,k-1), k,1,n);
    
  • PARI
    my(x='x+O('x^50)); concat([0], Vec(-(-4*x^4 + sqrt(-4*x^2-4*x+1)*(2*x^3+x^2-2*x) -12*x^3-7*x^2+2*x)/(sqrt(-4*x^2-4*x+1)*(4*x^3+8*x^2+3*x-1) - 4*x^3-8*x^2-3*x+1))) \\ G. C. Greubel, Jun 01 2017
    
  • PARI
    for(n=0,25, print1(sum(k=1,n, binomial(n-1,k-1)*sum(i=0,n-k, binomial(k,n-k-i)*binomial(k+i-1,k-1))), ", ")) \\ G. C. Greubel, Jun 01 2017

Formula

a(n) = Sum_{k=1..n} Sum_{i=0..(n-k)} C(k,n-k-i)*C(k+i-1,k-1)*C(n-1,k-1).
G.f.: A(x) = x*(F(x)-x)*F'(x)/F(x)^2, where F(x) = (1-sqrt(-4*x^2-4*x+1))/(2*x+2), F(x) is the g.f. of A052709.
D-finite with recurrence: (for n>5): (n-5)*(n-1)*a(n) = (3*n^2 - 20*n + 23)*a(n-1) + 2*(n-2)*(4*n-19)*a(n-2) + 4*(n-4)*(n-3)*a(n-3). - Vaclav Kotesovec, Mar 03 2014
a(n) ~ (2 + 2*sqrt(2))^n / (2^(5/4) * sqrt(1+sqrt(2)) * sqrt(Pi*n)). - Vaclav Kotesovec, Mar 03 2014