A007054 Super ballot numbers: 6(2n)!/(n!(n+2)!).
3, 2, 3, 6, 14, 36, 99, 286, 858, 2652, 8398, 27132, 89148, 297160, 1002915, 3421710, 11785890, 40940460, 143291610, 504932340, 1790214660, 6382504440, 22870640910, 82334307276, 297670187844, 1080432533656, 3935861372604, 14386251913656, 52749590350072
Offset: 0
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Emily Allen and Irina Gheorghiciuc, A Weighted Interpretation for the Super Catalan Numbers, J. Int. Seq. 17 (2014) # 14.10.7.
- David Callan, A combinatorial interpretation for a super-Catalan recurrence, arXiv:math/0408117 [math.CO], 2004.
- David Callan, A Combinatorial Interpretation for a Super-Catalan Recurrence, Journal of Integer Sequences, Vol. 8 (2005), Article 05.1.8.
- David Callan, A variant of Touchard's Catalan number identity, arXiv preprint arXiv:1204.5704 [math.CO], 2012. - From _N. J. A. Sloane_, Oct 10 2012
- Ira M. Gessel, Letter to N. J. A. Sloane, Jul. 1992
- Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194
- Ira M. Gessel, Rational Functions With Nonnegative Integer Coefficients, 50th Séminaire Lotharingien de Combinatoire, 2003.
- Ira M. Gessel and Guoce Xin, A Combinatorial Interpretation of the Numbers 6(2n)!/n!(n+2)!, Journal of Integer Sequences, Vol. 8 (2005), Article 05.2.3.
- Ira M. Gessel and Guoce Xin, A Combinatorial Interpretation of The Numbers 6(2n)! /n! (n+2)!, arXiv:math/0401300v2 [math.CO], 2004.
- Nicholas Pippenger and Kristin Schleich, Topological characteristics of random triangulated surfaces (section 7), Random Structures Algorithms 28 (2006) 247-288; arXiv:gr-qc/0306049, 2003.
- G. Schaeffer, A combinatorial interpretation of super-Catalan numbers of order two, (2001).
- Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 96.
Programs
-
Magma
[6*Factorial(2*n)/(Factorial(n)*Factorial(n+2)): n in [0..30]]; // Vincenzo Librandi, Aug 20 2011
-
Maple
seq(3*(2*n)!/(n!)^2/binomial(n+2,n), n=0..22); # Zerinvary Lajos, Jun 28 2007 A007054 := n -> 6*4^n*GAMMA(1/2+n)/(sqrt(Pi)*GAMMA(3+n)): seq(A007054(n),n=0..28); # Peter Luschny, Dec 14 2015
-
Mathematica
Table[6(2n)!/(n!(n+2)!),{n,0,30}] (* or *) CoefficientList[Series[ (-1+Sqrt[1-4*x]+(6-4*Sqrt[1-4*x])*x)/(2*x^2),{x,0,30}],x] (* Harvey P. Dale, Oct 05 2011 *)
-
PARI
a(n)=6*(2*n)!/(n!*(n+2)!); /* Joerg Arndt, Sep 01 2012 */
-
Sage
def A007054(n): return (-4)^(2 + n)*binomial(3/2, 2 + n)/2 print([A007054(n) for n in range(29)]) # Peter Luschny, Nov 04 2021
Formula
G.f.: c(x)*(4-c(x)), where c(x) = g.f. for Catalan numbers A000108; Convolution of Catalan numbers with negative Catalan numbers but -C(0)=-1 replaced by 3. - Wolfdieter Lang
E.g.f. in Maple notation: exp(2*x)*(4*x*(BesselI(0, 2*x)-BesselI(1, 2*x))-BesselI(1, 2*x))/x. Integral representation as n-th moment of a positive function on [0, 4], in Maple notation: a(n)=int(x^n*(4-x)^(3/2)/x^(1/2), x=0..4)/(2*Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Oct 10 2001
E.g.f.: Sum_{n>=0} a(n)*x^(2*n) = 3*BesselI(2, 2x).
a(n) = A000108(n)*6/(n+2). - Philippe Deléham, Oct 30 2007
G.f.: ((6-4*sqrt(1-4*x))*x+sqrt(1-4*x)-1)/(2*x^2) - Harvey P. Dale, Oct 05 2011
D-finite with recurrence (n+2)*a(n) +2*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Dec 03 2012
G.f.: 1/(x^2*G(0)) + 3/x - (1/2)/x^2, where G(k) = 1 + 1/(1 - 2*x*(2*k+3)/(2*x*(2*k+3) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: 3/x - 1/(2*x^2) + G(0)/(4*x^2), where G(k) = 1 + 1/(1 - 2*x*(2*k-3)/(2*x*(2*k-3) + (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 18 2013
0 = a(n)*(+16*a(n+1) - 14*a(n+2)) + a(n+1)*(+6*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 18 2014
A002421(n+2) = 2*a(n) for all n in Z. - Michael Somos, Sep 18 2014
a(n) = 3*(2*n)!*[x^(2*n)]hypergeometric([],[3],x^2). - Peter Luschny, Feb 01 2015
a(n) = 6*4^n*Gamma(1/2+n)/(sqrt(Pi)*Gamma(3+n)). - Peter Luschny, Dec 14 2015
a(n) = (-4)^(2 + n)*binomial(3/2, 2 + n)/2. - Peter Luschny, Nov 04 2021
From Amiram Eldar, May 16 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + 20*Pi/(81*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/25 - 8*log(phi)/(25*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n-1) = 3*A000984(n)/((2*n-1)*(n+1)). - R. J. Mathar, Jul 12 2024
Extensions
Corrected and extended by Vincenzo Librandi, Aug 20 2011
Comments