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A259613
a(n) = binomial(6*n,2*n)/3, n>0, a(0)=1.
1
1, 5, 165, 6188, 245157, 10015005, 417225900, 17620076360, 751616304549, 32308782859535, 1397281501935165, 60727722660586800, 2650087220696342700, 116043807643289338428, 5096278545356362962504, 224377658168860057076688
OFFSET
0,2
LINKS
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.
FORMULA
G.f.: A(x) = 1 + (x*B(x)')/(B(x)) where B(x) = 2 * (1 + x*B(x)^2)^2 / (1 - 2*x*B(x)^2 + sqrt(1-8*x*B(x)^2)).
a(n) ~ 3^(6*n-1/2) / (sqrt(Pi*n) * 2^(4*n+3/2)). - Vaclav Kotesovec, Jul 01 2015
a(n) = A025174(2*n), n>0. - R. J. Mathar, Jun 07 2016
From Peter Bala, Jun 08 2024: (Start)
a(n) = (9/2)*(6*n-1)*(6*n-5)*(3*n-1)*(3*n-2)/((4*n-1)*(4*n-3)*(2*n-1)*n) * a(n-1) with a(0) = 1 and a(1) = 5.
Right-hand side of the identity (1/3)*Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(5*n-k, 2*n-k) = (1/3)*binomial(6*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(5*n-k, 2*n-k) = binomial(4*n, 2*n). (End)
From Karol A. Penson, Jan 26 2025: (Start)
G.f. for 3*a(n),a(0)=1, denoted A, expressible entirely by radicals: A = A1 + A2 with
A1 = ((4*sqrt(4 - 27*sqrt(z)) + 12*i*sqrt(3)*z^(1/4))^(1/3) + (4*sqrt(4 - 27*sqrt(z)) - 12*i*sqrt(3)*z^(1/4))^(1/3))/(4*sqrt(4 - 27*sqrt(z))), and
A2 = (1/(4*sqrt(4 + 27*sqrt(z)) + 12*sqrt(3)*z^(1/4))^(1/3) + 1/(4*sqrt(4 + 27*sqrt(z)) - 12*sqrt(3)*z^(1/4))^(1/3))/sqrt(4 + 27*sqrt(z)),
where i = sqrt(-1), the imaginary unit. (End)
MATHEMATICA
Join[{1}, Table[Binomial[6 n, 2 n]/3, {n, 30}]] (* Vincenzo Librandi, Jul 01 2015 *)
PROG
(PARI) vector(20, n, n--; if (n==0, 1, binomial(6*n, 2*n)/3)) \\ Michel Marcus, Jul 01 2015
(Magma) [1] cat [Binomial(6*n, 2*n)/3: n in [1..20]]; // Vincenzo Librandi, Jul 01 2015
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Vladimir Kruchinin, Jun 30 2015
STATUS
approved