%I #23 Aug 03 2023 01:43:09
%S 2,-47,1142,3793,-4094806,371557891,-13021558306,-1374157073639,
%T 281067953420114,-22220280272696387,-51611579093593498,
%U 257837341935815261683,-35155217354672369625958,1761633462267526777842223,202464167122130621896038062
%N a(n) = (2/n^3)*( Sum_{k=0..n-1} (-1)^k*(3*k^2+3*k+1)*binomial(n-1,k)^3*binomial(n+k,k)^3 ).
%C Conjecture: Let n be any positive integer. For m = 0, 2, 4, ..., we have Sum_{k=0..n-1} (3k^2+3k+1)*(binomial(n-1,k)*binomial(n+k,k))^m == 0 (mod n^3); for m = 1, 3, 5, ... we have 2*Sum_{k=0..n-1} (-1)^k*(3k^2+3k+1)*(binomial(n-1,k)*binomial(n+k,k))^m == 0 (mod n^3).
%C The Zeilberger algorithm could yield a complicated fifth-order recurrence for a(n).
%C The author proved the conjecture in the latest version of arXiv:1408.5381. - _Zhi-Wei Sun_, Sep 14 2014
%H Zhi-Wei Sun, <a href="/A246543/b246543.txt">Table of n, a(n) for n = 1..70</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], 2014-2018.
%e a(2) = -47 since (2/2^3)*( Sum_{k=0..1} (-1)^k*(3k^2+3k+1)*binomial(1,k)^3*binomial(2+k,k)^3 ) = (1/4)*(1-7*3^3) = -47.
%t a[n_] := Sum[(3 k^2 + 3 k + 1) (-1)^k (Binomial[n - 1, k] Binomial[n + k, k])^3, {k, 0, n - 1}] 2/n^3
%t Table[a[n], {n, 1, 14}]
%Y Cf. A246512, A246542.
%K sign
%O 1,1
%A _Zhi-Wei Sun_, Aug 29 2014