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A246542 a(n) = 2/n^2*( sum_{k=0..n-1} (2*k+1)*C(n-1,k)^2*C(n+k,k) ), where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

%I A246542 #22 Feb 11 2015 06:34:53
%S A246542 2,5,22,132,918,6981,56390,475796,4149286,37133043,339307098,
%T A246542 3154030050,29741815998,283896719073,2738445478758,26656533873204,
%U A246542 261561469613190,2584718580416919,25703179602581234,257046296680889600,2583719988283365322,26090463844931102715,264570229302222957162,2693187696469413499902,27511970457139362253598
%N A246542 a(n) = 2/n^2*( sum_{k=0..n-1} (2*k+1)*C(n-1,k)^2*C(n+k,k) ), where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).
%C A246542 The following general conjecture implies that a(n) is always an integer.
%C A246542 Conjecture: Let a and b be nonnegative integers, and let n be a positive integer. If a == b (mod 2), then sum_{k=0}^{n-1}(2k+1)*C(n-1,k)^a*C(-n-1,k)^b == 0 (mod n^2); if a - b is odd, then 2*sum_{k=0}^{n-1}(-1)^k*(2k+1)C(n-1,k)^a*C(-n-1,k)^b == 0 (mod n^2).
%C A246542 An extension of the conjecture was proved in the latest version of arXiv:1408.5381. - _Zhi-Wei Sun_, Sep 14 2014
%H A246542 Zhi-Wei Sun, <a href="/A246542/b246542.txt">Table of n, a(n) for n = 1..150</a>
%H A246542 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 A246542 Recurrence (obtained via the Zeilberger algorithm):
%F A246542 -n*(n-2)*(20*n^2+50*n+29)*a(n) - (220*n^4+550*n^3+369*n^2+59*n-10)*a(n+1)
%F A246542 + (n+2)^2*(20*n^2+10*n-1)*a(n+2) = 0.
%e A246542 a(2) = 5 since 2/2^2*( sum_{k=0,1} (2k+1)*C(1,k)^2*C(2+k,k) ) = 1/2*(1+3*3) = 5.
%t A246542 a[n_] := Sum[(2 k + 1)Binomial[n - 1, k]^2 Binomial[n + k, k], {k, 0, n - 1}] 2/n^2
%t A246542 Table[a[n], {n, 1, 25}]
%Y A246542 Cf. A180364, A246460, A246511, A246543.
%K A246542 nonn
%O A246542 1,1
%A A246542 _Zhi-Wei Sun_, Aug 29 2014