|
%I #54 Feb 24 2023 02:30:37
%S 5,231,14586,1062347,84021990,7012604550,607892634420,54200780036595,
%T 4938927219474990,457909109348466930,43057935618181929900,
%U 4096531994713828810686,393617202432246696493436,38142088615983865845923052,3723160004902167033863327592
%N a(n) = binomial(6*n, 3*n)*binomial(3*n, n)/(2*(2*n+1)*binomial(2*n, n)).
%C During April 26-28, 2010, _Zhi-Wei Sun_ introduced this new sequence and proved that a(n) = binomial(6n,3n)*binomial(3n,n)/(2*(2n+1)*binomial(2n,n)) is a positive integer for every n=1,2,3,... He also observed that a(n) is odd if and only if n is a power of two, and that 3a(n)=0 (mod 2n+3). By Stirling's formula, we have lim_n (8n*sqrt(n*Pi)a(n)/108^n) = 1. It is interesting to find a combinatorial interpretation or recursion for the sequence.
%C From _Tatiana Hessami Pilehrood_, Dec 01 2015: (Start)
%C _Zhi-Wei Sun_ formulated two conjectures concerning a(n) (see Conjectures 1.1 and 1.2 in Z.-W. Sun, "Products and sums divisible by central binomial coefficients" and Conjecture A89 in "Open conjectures on congruences"). The first conjecture states that Sum_{n=1..p-1} a(n)/(108^n) is congruent to 0 or -1 modulo a prime p > 3 depending on whether p is congruent to +-1 or +-5 modulo 12, respectively.
%C The second conjecture asks about an exact formula for a companion sequence of a(n). Both conjectures as well as many numerical congruences involving a(n) and (2n+1)a(n) were solved by Kh. Hessami Pilehrood and T. Hessami Pilehrood, see the link below. (End)
%H Indranil Ghosh, <a href="/A176898/b176898.txt">Table of n, a(n) for n = 1..450</a>
%H Kh. Hessami Pilehrood and T. Hessami Pilehrood, <a href="http://arxiv.org/abs/1504.07944">Jacobi polynomials and congruences involving some higher-order Catalan numbers and binomial coefficients</a>, preprint, arXiv:1504.07944 [math.NT], 2015.
%H K. H. Pilehrood and T. H. Pilehrood, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Pilehrood/pile5.html">Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients</a>, J. Int. Seq. 18 (2015) 15.11.7
%H M. R. Sepanski, <a href="https://doi.org/10.37236/3350">On Divisibility of Convolutions of Central Binomial Coefficients</a>, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1004.4623">Products and sums divisible by central binomial coefficients</a>, preprint, arXiv:1004.4623 [math.NT], 2010.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0911.5665">Open conjectures on congruences</a>, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
%H Brian Y. Sun, J. X. Meng, <a href="https://arxiv.org/abs/1606.08153">Proof of a Conjecture of Z.-W. Sun on Trigonometric Series</a>, arXiv preprint arXiv:1606.08153 [math.CO], 2016.
%F G.f.: (1-6*s)/((12*s-1)*(8*s-2)) - 1/2, where x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2 = 0. - _Mark van Hoeij_, May 06 2013
%F a(n) ~ 2^(2*n-3) * 3^(3*n) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Jan 09 2023
%F From _Peter Bala_, Feb 21 2023: (Start)
%F a(n+1) = 6*(6*n + 1)*(6*n + 5)/((n + 1)*(2*n + 3))*a(n).
%F a(n) = (2^(2*n-1)) * Product_{1 <= i <= j <= 2*n-1} (2*i + j + 2)/(2*i + j - 1). Cf. A006013. (End)
%e For n=2 we have a(2) = binomial(12,6)*binomial(6,2)/(2*(2*2+1)*binomial(4,2)) = 231.
%p ogf := eval((1-6*s)/((12*s-1)*(8*s-2)) - 1/2, s=RootOf(x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2,s));
%p series(ogf,x=0,30); # _Mark van Hoeij_, May 06 2013
%t S[n_]:=Binomial[6n,3n]Binomial[3n,n]/(2(2n+1)Binomial[2n,n]) Table[S[n],{n,1,50}]
%o (Magma) [Binomial(6*n, 3*n)*Binomial(3*n, n)/(2*(2*n+1)*Binomial(2*n, n)): n in [1..15]]; // _Vincenzo Librandi_, Dec 02 2015
%o (PARI) a(n) = binomial(6*n, 3*n) * binomial(3*n, n) / (2*(2*n+1) * binomial(2*n, n)); \\ _Indranil Ghosh_, Mar 05 2017
%o (Python)
%o import math
%o f=math.factorial
%o def C(n,r): return f(n)/f(r)/f(n-r)
%o def A176898(n): return C(6*n, 3*n) * C(3*n, n) / (2*(2*n+1) * C(2*n, n)) # _Indranil Ghosh_, Mar 05 2017
%Y Cf. A000984, A006013, A173774, A176285, A176477.
%K nonn,easy
%O 1,1
%A _Zhi-Wei Sun_, Apr 28 2010
| 