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%I #29 Jun 11 2024 19:48:03
%S 1,5,165,6188,245157,10015005,417225900,17620076360,751616304549,
%T 32308782859535,1397281501935165,60727722660586800,
%U 2650087220696342700,116043807643289338428,5096278545356362962504,224377658168860057076688
%N a(n) = binomial(6*n,2*n)/3, n>0, a(0)=1.
%H G. C. Greubel, <a href="/A259613/b259613.txt">Table of n, a(n) for n = 0..600</a>
%H V. V. Kruchinin and D. V. Kruchinin, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Kruchinin/kruch9.html">A Generating Function for the Diagonal T_{2n,n} in Triangles</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.
%F 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)).
%F a(n) ~ 3^(6*n-1/2) / (sqrt(Pi*n) * 2^(4*n+3/2)). - _Vaclav Kotesovec_, Jul 01 2015
%F a(n) = A025174(2*n), n>0. - _R. J. Mathar_, Jun 07 2016
%F From _Peter Bala_, Jun 08 2024: (Start)
%F 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.
%F 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)
%t Join[{1}, Table[Binomial[6 n, 2 n]/3, {n, 30}]] (* _Vincenzo Librandi_, Jul 01 2015 *)
%o (PARI) vector(20,n, n--; if (n==0, 1, binomial(6*n,2*n)/3)) \\ _Michel Marcus_, Jul 01 2015
%o (Magma) [1] cat [Binomial(6*n,2*n)/3: n in [1..20]]; // _Vincenzo Librandi_, Jul 01 2015
%Y Cf. A001448, A001450, A182959.
%K nonn,easy
%O 0,2
%A _Vladimir Kruchinin_, Jun 30 2015
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