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%I #47 Mar 23 2023 07:57:57
%S 1,6,110,2562,66222,1815506,51697802,1511679210,45076309166,
%T 1364497268946,41800229045610,1292986222651646,40317756506959050,
%U 1265712901796074842,39965073938276694002,1268208750951634765562,40419340092267053380782,1293151592990764737265490
%N The number of integer sequences of length d = 2n+1 such that the sum of the terms is 0 and the sum of the absolute values of the terms is d-1.
%C Let b(n) = S(d,n) be the coordination sequence of the lattice A_d. Then this sequence is a(n) = S(2n,n). See Conway-Sloane. The sequence is defined by Couveignes et al.
%H R. H. Hardin and Colin Barker, <a href="/A156554/b156554.txt">Table of n, a(n) for n = 0..300</a> (terms up to n=26 from R. H. Hardin)
%H J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (<a href="http://neilsloane.com/doc/Me220.pdf">pdf</a>).
%H J.-M. Couveignes, T. Ezome, and R. Lercier, <a href="https://arxiv.org/abs/0810.2853">Elliptic periods and primality proving</a>, arXiv:0810.2853 [math.NT], 2008-2009.
%F a(n) = S(2n,n) where S(d,n) = Sum_{k=0..d} C(d,k)^2*C(n-k+d-1,d-1) from formula (22) in Conway-Sloane.
%F a(n) ~ (1 + sqrt(2))^(4*n + 1/2) / (2^(5/4) * Pi * n). - _Vaclav Kotesovec_, Apr 10 2018
%F From _Peter Bala_, Dec 19 2020: (Start)
%F a(n) = Sum_{k = 0..n} C(2*n,n-k)^2 * C(2*n+k-1,k).
%F a(n) = Sum_{k = 1..n} C(2*n, k)*C(2*n+k, k)*C(n-1,k-1) for n >= 1.
%F a(n) = [x^n] P(2*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A103882.
%F a(n) = C(2*n,n)^2 * hypergeom([-n, -n, 2*n], [n+1, n+1], 1).
%F n^2*(2*n - 1)^2*(24*n^3 - 105*n^2 + 152*n - 73)*a(n) = (3264*n^7 - 20808*n^6 + 53900*n^5 - 73159*n^4 + 55963*n^3 - 24107*n^2 + 5436*n - 504)*a(n - 1) - (2*n - 1)*(2*n - 3)*(n - 2)^2*(24*n^3 - 33*n^2 + 14*n - 2)*a(n - 2).
%F Conjectural: for any prime p >= 5, a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k.
%F More generally, if r and s are positive integers, we conjecture that the same supercongruences hold for the sequence defined by [x^(r*n)] P(s*n,(1 + x)/(1 - x)). (End)
%F Even more generally, we conjecture that the same supercongruences hold for the sequence defined by [x^(r*n)] (1 + x)^(A*n) * (1 - x)^(B*n) * P(s*n,(1 + x)/(1 - x)), where A and B are integers. - _Peter Bala_, Mar 17 2023
%e The a(1) = 6 sequences are (1,-1,0), (-1,1,0), (1,0,-1), (-1,0,1), (0,1,-1) and (0,-1,1).
%p S:=proc(d,n) add(binomial(d,k)^2*binomial(n-k+d-1,d-1),k=0..d); end proc; a:=n->S(2*n,n);
%t Table[ Binomial[-1 + 3 n, -1 + 2 n] HypergeometricPFQ[{-2 n, -2 n, -n}, {1, 1 - 3 n}, 1], {n, 0, 10}] (* _Eric W. Weisstein_, Feb 10 2009 *)
%o (PARI) S(d, n) = sum(k=0, d, binomial(d,k)^2*binomial(n-k+d-1, d-1));
%o concat(1, vector(20, n, S(2*n,n))) \\ _Colin Barker_, Dec 24 2015
%Y a(n) = A103881(2n, n), A103882.
%K easy,nice,nonn
%O 0,2
%A _W. Edwin Clark_, Feb 09 2009
%E Formula incorrectly copied from A143699 removed by _R. J. Mathar_, Mar 11 2010
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