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%I #32 Sep 02 2014 10:42:25
%S -1,2,37,324,2403,16582,109961,711176,4521607,28394442,176648877,
%T 1090974156,6698429931,40930511502,249105886737,1510954029072,
%U 9138180665871,55128765109906,331851698655797,1993747216968788
%N a(n) = (sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)*C(n+k,k))/n^2, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).
%C Conjecture: Let n be any positive integer.
%C (i) For any m = 0,1,2,..., we have sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^m*C(-n-1,k)^m == 0 (mod n^2). Also, we may replace 4*k^3-1 by 2*k^2+2*k+1 or 2*k^3+k^2+k.
%C (ii) For any s,t = 0,1,2,..., both sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t and sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t are multiples of n.
%C The author proved the conjecture in the latest version of arXiv:1408.5381, thus a(n) is always an integer. - _Zhi-Wei Sun_, Sep 01 2014
%H Zhi-Wei Sun, <a href="/A243101/b243101.txt">Table of n, a(n) for n = 1..200</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, 2014.
%F Recurrence (obtained via the Zeilberger algorithm):
%F n*(2*n+3)*(n^2+3*n+1)*a(n) - 2*(2*n^2+4*n+1)*(3*n^2+6*n+2)*a(n+1) + (n+2)*(2*n+1)*(n^2+n-1)*a(n+2) = 0.
%F a(n) ~ (3+2*sqrt(2))^n * sqrt(n) / (2^(5/4) * sqrt(Pi)). - _Vaclav Kotesovec_, Sep 02 2014
%e a(2) = 2 since (sum_{k=0,1}(4*k^3-1)*C(1,k)*C(2+k,k))/2^2 = (-1 + 3*3)/4 = 2.
%p A243101:=n->add((4*k^3-1)*binomial(n-1,k)*binomial(n+k,k), k=0..n-1)/n^2: seq(A243101(n), n=1..20); # _Wesley Ivan Hurt_, Sep 01 2014
%t a[n_]:=Sum[(4k^3-1)*Binomial[n-1,k]Binomial[n+k,k],{k,0,n-1}]/n^2
%t Table[a[n],{n,1,20}]
%Y Cf. A246542, A246543.
%K sign
%O 1,2
%A _Zhi-Wei Sun_, Aug 29 2014
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