%I #37 Apr 06 2017 21:23:44
%S 1,2,13,134,1783,27950,491335,9401390,192033565,4131488426,
%T 92723165533,2155279960586,51602299168639,1267128734047142,
%U 31803430252162579,813628992468938750,21168533016938471665,559044288633621863810,14962460440143262653685,405299365266569619086462
%N a(n) = (Sum_{k=0..n-1} C(n-1,k)^2*C(-n-1,k)^2/C(k+2,2))/n.
%C The author proved in arXiv:1408.5381 that a(n) is always an integer.
%C Note that Sum_{k=0..n-1} C(n-1,k)*C(-n-1,k)/C(k+2,2) = 0 for n > 1.
%C Conjecture: The sequence a(n+1)/a(n) (n > 0) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n > 1) is strictly decreasing to the limit 1.
%H Zhi-Wei Sun, <a href="/A246875/b246875.txt">Table of n, a(n) for n = 1..100</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and related divisibility results</a>, arXiv:1408.5381 [math.NT], 2014.
%F Recurrence (obtained via the Zeilberger algorithm): -n*(n-1)^2*(2*n+3)*a(n) + 4*(17*n^4+68*n^3+92*n^2+48*n+9)*a(n+1) - (n+2)*(n+3)^2*(2*n+1)*a(n+2) = 0.
%F a(n) ~ (17+12*sqrt(2))^n / (2^(5/4) * Pi^(3/2) * n^(9/2)). - _Vaclav Kotesovec_, Sep 07 2014
%F a(n) = 4F3(1-n,1-n,1+n,1+n;1,1,3;1)/n. - _Benedict W. J. Irwin_, Apr 04 2017
%e a(2) = 2 since (Sum_{k=0..1} C(2-1,k)^2*C(-2-1,k)^2/C(2+k,2))/2 = (1 + (-3)^2/3)/2 = 2.
%t a[n_]:=Sum[(Binomial[n-1,k]*Binomial[-n-1,k])^2/Binomial[k+2,2],{k,0,n-1}]/n
%t Table[a[n],{n,1,20}]
%t Table[HypergeometricPFQ[{1-n,1-n,1+n,1+n},{1,1,3},1]/n,{n,1,10}] (* _Benedict W. J. Irwin_, Apr 04 2017 *)
%o (PARI) a(n) = sum(k=0, n - 1, (binomial(n - 1, k) * binomial(-n - 1, k))^2/binomial(k + 2, 2))/n; \\ _Indranil Ghosh_, Apr 04 2017
%Y Cf. A243101, A246460, A246511, A246543, A246567.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Sep 07 2014