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.

A103971 Expansion of (1 - sqrt(1 - 4*x - 16*x^2))/(2*x).

Original entry on oeis.org

1, 5, 10, 45, 190, 930, 4660, 24445, 131190, 719830, 4013260, 22684370, 129661740, 748252580, 4353379560, 25508284445, 150392391590, 891549228430, 5310994644060, 31775749689670, 190860711108740, 1150473009844380
Offset: 0

Views

Author

Paul Barry, Feb 23 2005

Keywords

Comments

Image of c(x), the g.f. of the Catalan numbers A000108 under the mapping g(x) -> (1+4x)g(x(1+4x)). In general, the image of the Catalan numbers under the mapping g(x)->(1+i*x)g(x(1+i*x)) is given by a(n) = Sum_{k=0..n} i^(n-k)C(k)C(k+1,n-k).
More generally, the sequence C for which C(0)=a, C(1)=b and C(n+1) = sum(C(k)*C(n-k),k=0..n) has the following g.f. f: f(z) = (1-sqrt(1-4*z*(a-(a^2-b)*z)))/(2*z). We obtain: C(n)=(sum(-1)^(p-1)*2^{n-p}a^{n-2*p-1}*(a^2-b)^p*((2*n-2*p-1)*...*5*3*1/(p!*(n-2*p+1)!)),p=0..floor((n+1)/2)). By following Comtet [Analyse Combinatoire Tomes 1 et 2, PUF, Paris 1970], we obtain also: (n+1)*C(n) - 2*a*(2*n-1)*C(n-1) + 4*(n-2)*(a^2-b)*C(n-2) = 0. - Richard Choulet, Dec 17 2009

Crossrefs

Programs

  • Maple
    n:=30:a(0):=1:a(1):=5: for k from 1 to n do a(k+1):=sum('a(p)*a(k-p)','p'=0..k):od:seq(a(k),k=0..n); # Richard Choulet, Dec 17 2009
  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-4x-16x^2])/(2x),{x,0,30}],x] (* Harvey P. Dale, Apr 02 2012 *)

Formula

G.f.: (1-sqrt(1-4*x*(1+4*x)))/(2*x).
a(n) = Sum_{k=0..n} 4^(n-k)*C(k)*C(k+1, n-k).
Another recurrence formula: (n+1)*a(n) = 2*(2*n-1)*a(n-1) + 16*(n-2)*a(n-2). - Richard Choulet, Dec 17 2009
a(n) ~ sqrt(10 + 2*sqrt(5))*(2 + 2*sqrt(5))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
Equivalently, a(n) ~ 5^(1/4) * 2^(2*n) * phi^(n + 1/2) / (sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021