a(0)=1, for n > 0: a(n) = Sum_{j=0..n-1} Sum_{i=0..j} a(i)*a(j-i). G.f.: A(x) = 1 + x*A(x)^2/(1-x). - Mario Catalani (mario.catalani(AT)unito.it), Jun 19 2003
a(n) = Sum_{i=ceiling((n-1)/2)..n-1} (3^(2i+1-n)*binomial(n, i)*binomial(i, n-i-1))/n. -
Emeric Deutsch, Jul 23 2002
a(n) = Sum_{k=1..n} binomial(2k, k)*binomial(n-1, k-1)/(k+1), i.e., binomial transform of the Catalan numbers 1, 2, 5, 14, 42, ... (
A000108). a(n) = Sum_{k=0..floor((n-1)/2)} 3^(n-1-2*k)*binomial(2k, k)*binomial(n-1, 2k)/(k+1). -
Emeric Deutsch, Aug 05 2002
D-finite with recurrence: a(1)=1, a(n) = (3(2n-1)*a(n-1)-5(n-2)*a(n-2))/(n+1) for n > 1. -
Emeric Deutsch, Dec 18 2002
a(n) is asymptotic to c*5^n/n^(3/2) with c=0.63.... -
Benoit Cloitre, Jun 23 2003
In closed form, c = (1/2)*sqrt(5/Pi) = 0.63078313050504... -
Vaclav Kotesovec, Oct 04 2012
Reversion of Sum_{n>0} a(n)x^n = -Sum_{n>0}
A001906(n)(-x)^n.
G.f. A(x) satisfies xA(x)^2 + (1-x)(1-A(x)) = 0.
G.f.: (1 - x - sqrt(1 - 6x + 5x^2))/(2x). For n > 1, a(n) = 3*a(n-1) + Sum_{k=1..n-2} a(k)*a(n-k-1). -
John W. Layman, Feb 22 2001
The Hankel transform of this sequence gives
A001519 = 1, 2, 5, 13, 34, 89, ... E.g., Det([1, 1, 3, 10, 36; 1, 3, 10, 36, 137; 3, 10, 36, 137, 543; 10, 36, 137, 543, 2219; 36, 137, 543, 2219, 9285 ])= 34. -
Philippe Deléham, Jan 25 2004
a(n+1) = Sum_{k=0..n} 2^(n-k)*M(k)*binomial(n,k), where M(k) =
A001006(k) is the k-th Motzkin number (from here it follows that a(n+1) and M(n) have the same parity). -
Emeric Deutsch, May 10 2007
G.f.: 1/(1-x/(1-x-x/(1-x/(1-x-x/(1-x/(1-x-x/(1-... (continued fraction). -
Paul Barry, May 16 2009
G.f.: (1-x)/(1-2x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-.... (continued fraction). -
Paul Barry, Oct 17 2009
G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1-x) (continued fraction); more generally g.f. C(x/(1-x)) where C(x) is the g.f. for the Catalan numbers (
A000108). -
Joerg Arndt, Mar 18 2011
a(n) = -5^(1/2)/(10*(n+1)) * (5*hypergeom([1/2, n], [1], 4/5) -3*hypergeom([1/2, n+1], [1], 4/5)) (for n>0). -
Mark van Hoeij, Nov 12 2009
For n >= 1, a(n) = (1/(2*Pi))*Integral_{x=1..5} x^(n-1)*sqrt((x-1)*(5-x)) dx. -
Groux Roland, Mar 16 2011
a(n) = upper left term in M^(n-1), M = an infinite square production matrix as follows (with 3,2,2,2,... as the main diagonal):
3, 1, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 1, 2, 1, 0, 0, ...
1, 1, 1, 2, 1, 0, ...
1, 1, 1, 1, 2, 0, ...
...
Alternatively, let M = the previous matrix but change the 3 to a 2. Then a(n) = sum of top row terms of M^(n-1). (End)
a(n) = hypergeometric([1-n,3/2],[3],-4), for n>0. -
Peter Luschny, Aug 15 2012
a(n) = GegenbauerC(n-1, -n, -3/2)/n for n >= 1. -
Peter Luschny, May 09 2016
E.g.f.: 1 + Integral (exp(3*x) * BesselI(1,2*x) / x) dx. -
Ilya Gutkovskiy, Jun 01 2020
G.f.: 1 + x/G(0) with G(k) = (1 - 3*x - x^2/G(k+1)) (continued fraction). -
Nikolaos Pantelidis, Dec 12 2022
G.f.: 1 + x/(1 - x) * c(x/(1 - x))^2 = 1 + x/(1 - 5*x) * c(-x/(1 - 5*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers
A000108.
a(n+1) = Sum_{k = 0..n} binomial(n, k)*Catalan(k+1).
a(n+1) = hypergeom([-n, 3/2], [3], -4).
a(n+1) = 5^n * Sum_{k = 0..n} (-5)^(-k)*binomial(n, k)*Catalan(k+1).
a(n+1) = 5^n * hypergeom([-n, 3/2], [3], 4/5). (End)
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