OFFSET
0,2
COMMENTS
More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2) with S(0)=1, |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
FORMULA
G.f.: sqrt( (1-6*x - sqrt(1-12*x+16*x^2))/10 )/x.
a(n) = Fibonacci(2*n+2) * binomial(2*n,n)/(n+1).
a(n) = Fibonacci(n+1) * Lucas(n+1) * binomial(2*n,n)/(n+1), where Lucas(n+1) = 2*Fibonacci(n) + Fibonacci(n+1) = A000032(n+1).
n*(n+1)*a(n) -6*n*(2*n-1)*a(n-1) +4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/16^n = 4*sqrt(1-2/sqrt(5)). - Amiram Eldar, May 06 2023
EXAMPLE
G.f.: A(x) = 1 + 3*x + 16*x^2 + 105*x^3 + 770*x^4 + 6048*x^5 + 49764*x^6 +...
such that the coefficients equal the term-wise products:
A = [1*1, 3*1, 8*2, 21*5, 55*14, 144*42, 377*132, 987*429, 2584*1430, ...].
Related expansions.
A(x)^2 = 1 + 6*x + 41*x^2 + 306*x^3 + 2426*x^4 + 20076*x^5 + 171481*x^6 +...
A(x)^3 = 1 + 9*x + 75*x^2 + 630*x^3 + 5400*x^4 + 47223*x^5 + 420277*x^6 +...
Incidentally, note that (2*n+1) divides [x^n] A(x)^3:
A^3 = [1*1, 3*3, 5*15, 7*90, 9*600, 11*4293, 13*32329, 15*253110, ...].
MATHEMATICA
Table[Fibonacci[2*n+2]*Binomial[2*n, n]/(n+1), {n, 0, 25}] (* Vincenzo Librandi, Aug 28 2012 *)
PROG
(PARI) {a(n)=fibonacci(2*n+2)*binomial(2*n, n)/(n+1)}
(PARI) {a(n)=fibonacci(n+1)*(2*fibonacci(n)+fibonacci(n+1))*binomial(2*n, n)/(n+1)}
(PARI) {a(n)=polcoeff( sqrt( (1-6*x - sqrt(1-12*x+16*x^2 +x^3*O(x^n)))/10 )/x, n)}
for(n=0, 21, print1(a(n), ", "))
(Magma) [Fibonacci(2*n+2)*Binomial(2*n, n)/(n+1): n in [0..22]] // Vincenzo Librandi, Aug 28 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Aug 27 2012
STATUS
approved