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A259557 a(n) = binomial(4*n-1, 2*n).

%I A259557 #32 Jun 11 2024 19:47:43
%S A259557 1,3,35,462,6435,92378,1352078,20058300,300540195,4537567650,
%T A259557 68923264410,1052049481860,16123801841550,247959266474052,
%U A259557 3824345300380220,59132290782430712,916312070471295267,14226520737620288370
%N A259557 a(n) = binomial(4*n-1, 2*n).
%C A259557 Essentially the same as A100033.
%H A259557 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 A259557 G.f. A(x)=1+x*B(x)'/B(x), where B(x) is g.f. of A079489.
%F A259557 a(n) = A100033(n-1) for n>0.
%F A259557 D-finite with recurrence n*(2*n-1)*a(n) -2*(4*n-1)*(4*n-3)*a(n-1)=0. - _R. J. Mathar_, Jul 06 2015
%F A259557 a(n) = [x^(2*n)] 1/(1 - x)^(2*n). - _Ilya Gutkovskiy_, Oct 10 2017
%F A259557 From _Peter Bala_, Jun 11 2023: (Start)
%F A259557 a(n) = (1/2) * [x^n] ( (1 + x)^2/( 1 - x) )^(2*n) for n >= 1.
%F A259557 Right-hand side of the identity (1/2)*Sum_{k = 0..n} binomial(4*n,k)*binomial(3*n-k-1,n-k) = binomial(4*n-1,2*n) for n >= 1.
%F A259557 a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A119259(k)*x^k/k ). (End)
%F A259557 a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)*binomial(-3*n-k, 2*n-k) = Sum_{k = 0..2*n} (-1)^k*binomial(n+k-1, k)*binomial(5*n-1, 2*n-k). - _Peter Bala_, Jun 08 2024
%t A259557 Table[Binomial[4 n - 1, 2 n], {n, 0, 30}] (* _Vincenzo Librandi_, Jul 01 2015 *)
%o A259557 (PARI) vector(20, n, n--; binomial(4*n-1, 2*n)) \\ _Michel Marcus_, Jul 01 2015
%o A259557 (Magma) [Binomial(4*n-1, 2*n): n in [0..20]]; // _Vincenzo Librandi_, Jul 01 2015
%Y A259557 Cf. A001700, A079489, A100033.
%K A259557 nonn,easy
%O A259557 0,2
%A A259557 _Vladimir Kruchinin_, Jun 30 2015