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A235536 a(n) = binomial(8*n, 2*n) / (6*n + 1).

%I A235536 #34 Sep 08 2022 08:46:06
%S A235536 1,4,140,7084,420732,27343888,1882933364,134993766600,9969937491420,
%T A235536 753310723010608,57956002331347120,4524678117939182220,
%U A235536 357557785658996609700,28545588568201512137904,2298872717007844035521848,186533392975795702301759056
%N A235536 a(n) = binomial(8*n, 2*n) / (6*n + 1).
%C A235536 This is the case l=6, k=2 of binomial((l+k)*n,k*n)/((l*n+1)/gcd(k,l*n+1)), see Theorem 1.1 in Zhi-Wei Sun's paper.
%C A235536 First bisection of A002293.
%C A235536 Also, the sequence is between A002296 and A235535.
%H A235536 Zhi-Wei Sun, <a href="https://doi.org/10.1017/S1446788712000171">On Divisibility Of Binomial Coefficients</a>, Journal of the Australian Mathematical Society 93 (2012), p. 189-201.
%F A235536 a(n) = A124753(6*n).
%F A235536 From _Ilya Gutkovskiy_, Jun 21 2018: (Start)
%F A235536 G.f.: 6F5(1/8,1/4,3/8,5/8,3/4,7/8; 1/3,1/2,2/3,5/6,7/6; 65536*x/729).
%F A235536 a(n) ~ 2^(16*n-1)/(sqrt(Pi)*3^(6*n+3/2)*n^(3/2)). (End)
%t A235536 Table[Binomial[8 n, 2 n]/(6 n + 1), {n, 0, 20}]
%o A235536 (Magma) l:=6; k:=2; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* where l is divisible by all the prime factors of k */
%Y A235536 Cf. similar sequences generated by binomial((l+k)*n,k*n)/(l*n+1), where l is divisible by all the factors of k: A000108 (l=1, k=1), A001764 (l=2, k=1), A002293 (l=3, k=1), A002294 (l=4, k=1), A002295 (l=5, k=1), A002296 (l=6, k=1), A007556 (l=7, k=1), A062994 (l=8, k=1), A059968 (l=9, k=1), A230388 (l=10, k=1), A048990 (l=2, k=2), A235534 (l=4, k=2), this sequence (l=6, k=2), A187357 (l=3, k=3), A235535 (l=6, k=3).
%K A235536 nonn,easy
%O A235536 0,2
%A A235536 _Bruno Berselli_, Jan 12 2014