A235535 - OEIS

%I #49 Sep 08 2022 08:46:06

%S 1,12,1428,246675,50067108,11124755664,2619631042665,642312451217745,

%T 162250238001816900,41932353590942745504,11034966795189838872624,

%U 2946924270225408943665279,796607831560617902288322405,217550867863011281855594752680

%N a(n) = binomial(9*n, 3*n) / (6*n + 1).

%C This is the case l=6, k=3 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 Also, the sequence follows A002296 and A235536, namely binomial(7*n,n)/(6*n+1) and binomial(8*n,2*n)/(6*n+1); naturally, even binomial(10*n,4*n)/(6*n+1) is always integer.

%H Robert Israel, <a href="/A235535/b235535.txt">Table of n, a(n) for n = 0..403</a>

%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) = A001764(3*n) = A047749(6*n).

%F From _Ilya Gutkovskiy_, Jun 21 2018: (Start)

%F G.f.: 6F5(1/9,2/9,4/9,5/9,7/9,8/9; 1/3,1/2,2/3,5/6,7/6; 19683*x/64).

%F a(n) ~ 3^(9*n-1)/(sqrt(Pi)*4^(3*n+1)*n^(3/2)). (End)

%F D-finite with recurrence 8*(6*n + 5)*(2*n + 1)*(n + 1)*(3*n + 2)*(3*n + 1)*(6*n + 7)*a(n + 1) = 3*(9*n + 8)*(9*n + 7)*(9*n + 5)*(9*n + 4)*(9*n + 2)*(9*n + 1)*a(n). - _Robert Israel_, Feb 15 2021

%p seq(binomial(9*n,3*n)/(6*n+1), n=0..30); # _Robert Israel_, Feb 15 2021

%t Table[Binomial[9 n, 3 n]/(6 n + 1), {n, 0, 20}]

%o (Magma) l:=6; k:=3; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* here 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), A235534 (l=4, k=2), A235536 (l=6, k=2), A187357 (l=3, k=3), this sequence (l=6, k=3).

%K nonn,easy

%O 0,2

%A _Bruno Berselli_, Jan 12 2014