A244039 a(n) = 2^(2*n-1) * ( binomial(3*n/2,n) + binomial((3*n-1)/2,n) ).

1, 5, 39, 338, 3075, 28770, 274134, 2645844, 25781283, 253068530, 2498678754, 24788450076, 246889978062, 2467197059124, 24725226928140, 248396412496488, 2500825206700323, 25225687837101330, 254877697946626410, 2579123090162503500, 26133512970919973850, 265126176290618366460

Offset: 0

Author

N. J. A. Sloane, Jun 28 2014

Keywords

• nonn

References

• R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• I. M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656 [math.CO], 2014. See f_3(n).

Crossrefs

Cf. A045741, A091527, A244038, A244469.

Programs

Magma

[Round(2^(2*n-1)*( Gamma(3*n/2+1)/Gamma(n/2+1) + Gamma((3*n+1)/2)/Gamma((n+1)/2) )/Factorial(n)): n in [0..25]]; // G. C. Greubel, Aug 20 2019

Maple

a := n -> 2^(2*n-1)*(binomial(3*n/2,n) + binomial((3*n-1)/2,n));
seq(a(n), n=0..25);

Mathematica

Table[2^(2n-1)*(Binomial[3n/2, n] + Binomial[(3n-1)/2, n]), {n, 0, 25}] (* Vincenzo Librandi, Jun 29 2014 *)

PARI

a(n) = 2^(2*n-1)*(binomial(3*n/2, n) + binomial((3*n-1)/2, n));
vector(25, n, n--; a(n)) \\ G. C. Greubel, Aug 20 2019

Sage

[2^(2*n-1)*(binomial(3*n/2, n) + binomial((3*n-1)/2, n)) for n in (0..25)] # G. C. Greubel, Aug 20 2019

Formula

From Peter Bala, Mar 04 2022: (Start)

a(n) = [x^n] ( (1 + 2*x)^3/(1 + x) )^n. Cf. A091527.

a(n) = Sum_{k = 0..n} (-1)^k * 2^(n-k) * binomial(3*n,n-k) * binomial(n+k-1,k).

n*(n-1)*(6*n-11)*a(n) = - 18*(n-1)*a(n-1) + 12*(3*n-4)*(3*n-5)*(6*n-5)*a(n-2) with a(0) = 1 and a(1) = 5.

The o.g.f. A(x) = 1 + 5*x + 39*x^2 + 338*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + 2*x)^3/(1 + x)) and hence is an algebraic function over the field of rational functions Q(x) by Stanley 1999, Theorem 6.33, p. 197.

Calculation gives (1 - 108*x^2)*A(x)^3 - (1 + 9*x)*A(x) = x.

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)

a(n) = 2^n*binomial(3*n, n)*hypergeom([-n, n], [2*n + 1], 1/2). - Peter Luschny, Mar 07 2022

From Seiichi Manyama, Aug 08 2025: (Start)

a(n) = Sum_{k=0..n} binomial(3*n,k) * binomial(2*n-k,n-k).

a(n) = [x^n] (1+x)^(3*n)/(1-x)^(n+1).

a(n) = [x^n] 1/((1-x)^n * (1-2*x)^(n+1)).

a(n) = Sum_{k=0..n} 2^k * binomial(n+k,k) * binomial(2*n-k-1,n-k). (End)