|
%I #38 Sep 08 2022 08:46:06
%S 1,3,55,1428,43263,1430715,50067108,1822766520,68328754959,
%T 2619631042665,102240109897695,4048514844039120,162250238001816900,
%U 6568517413771094628,268225186597703313816,11034966795189838872624,456949965738717944767791
%N a(n) = binomial(6*n, 2*n) / (4*n + 1).
%C This is the case l=4, 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 First bisection of A001764.
%H 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 a(n) = A047749(4*n-2) for n>0.
%F From _Ilya Gutkovskiy_, Jun 21 2018: (Start)
%F G.f.: 4F3(1/6,1/3,2/3,5/6; 1/2,3/4,5/4; 729*x/16).
%F a(n) ~ 3^(6*n+1/2)/(sqrt(Pi)*2^(4*n+7/2)*n^(3/2)). (End)
%t Table[Binomial[6 n, 2 n]/(4 n + 1), {n, 0, 20}]
%o (Magma) l:=4; 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 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), this sequence (l=4, k=2), A235536 (l=6, k=2), A187357 (l=3, k=3), A235535 (l=6, k=3).
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Jan 12 2014
|
