%I #11 Nov 17 2019 01:46:12
%S 1,2,10,68,586,5252,49204,475400,4723786,47937812,494786260,
%T 5177188040,54794164660,585565913480,6309889976680,68484312535568,
%U 747985368753226,8214968193003860,90669516557975524,1005156080857529768,11187435500257898836,124964856185950621832
%N a(n) = Sum_{k=0..n} C(n,k)^2*T(k)*T(n-k), where T(k) = A002426(k) is the coefficient of x^k in the expansion of (x^2+x+1)^k.
%C The author introduced this sequence in arXiv:1911.05456 and made the following conjecture.
%C Conjecture: Let p be an odd prime and let S = Sum_{k=0..p-1}a(k)/(-4)^k. If p == 1 (mod 12) and p = x^2 + 9*y^2 with x and y integers, then S == 4*x^2-2*p (mod p^2). If p == 5 (mod 12) and p = x^2 + y^2 with x == y (mod 3), then S == 4*x*y (mod p^2). If p == 3 (mod 4), then S == 0 (mod p^2).
%C Note that if p > 3 is a prime, then a(p-1) == Sum_{k=0..p-1} T(k)*T(p-1-k) == Legendre(p/3)*Sum_{k=0..p-1}T(k)^2/(-3)^k == 1 (mod p) by (1.7) and (2.3) of the author's 2014 paper in Sci. China Math.
%H Seiichi Manyama, <a href="/A329475/b329475.txt">Table of n, a(n) for n = 0..931</a> (terms 0..150 from Zhi-Wei Sun)
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1007/s11425-014-4809-z">Congruences involving generalized central trinomial coefficients</a>, Sci. China Math. 57(2014), no.7, 1375-1400.
%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/978-1-4939-1601-6_18">On sums related to central binomial and trinomial coefficients</a>, in: M. B. Nathanson (ed.), Combinatorial and Additive Number Theory: CANT 2011 and 2012, Springer Proc. in Math. & Stat., Vol. 101, Springer, New York, 2014, pp. 257-312. Also available from <a href="https://arxiv.org/abs/1101.0600v25"> arXiv:1101.0600 [math.NT]</a>, 2011-2014.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1911.05456">Characterizing rational Ramanujan-type series for 1/Pi via congruences</a>, arXiv:1911.05456 [math.NT], 2019.
%F a(n) ~ (3/2)*12^n/(n*Pi)^(3/2) as n tends to the infinity.
%e a(1) = 2 since Sum_{k=0,1} C(1,k)^2*T(k)*T(1-k) = C(1,0)^2*T(0)*T(1) + C(1,1)^2*T(1)*T(0) = 2*T(0)*T(1) = 2*1*1 = 2.
%t T[0]=1; T[1]=1; T[n_]:=T[n]=((2n-1)T[n-1]+3*(n-1)*T[n-2])/n;
%t a[n_]:=a[n]=Sum[Binomial[n,k]^2*T[k]*T[n-k],{k,0,n}];
%t Table[a[n],{n,0,21}]
%Y Cf. A002426, A002895.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Nov 13 2019