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%I #23 Mar 17 2023 10:47:00
%S 1,4,46,652,10186,168304,2884456,50723824,909192538,16538659384,
%T 304391739796,5655971294824,105929883322576,1997228410630912,
%U 37871584674309376,721672204654077952,13811327854028171098,265324110145941691912,5114208160758838538044,98874597697991698311832,1916741738060370782929036
%N a(n) = Sum_{k=0..n} (3*n-2*k)!/((n-k)!^3*k!)*(-2)^k.
%C Diagonal of rational function 1/(1 - (x + y + z - 2*x*y*z)).
%H Seiichi Manyama, <a href="/A318109/b318109.txt">Table of n, a(n) for n = 0..766</a> (terms 0..100 from Gheorghe Coserea)
%F G.f. y=A(x) satisfies: 0 = x*(x - 1)*(4*x - 1)*(8*x^2 + 20*x - 1)*y'' + (96*x^4 + 64*x^3 - 120*x^2 + 42*x - 1)*y' + 4*(2*x + 1)*(4*x^2 - 2*x + 1)*y.
%F From _Peter Bala_, Mar 16 2023: (Start)
%F n^2*(3*n - 4)*a(n) = (3*n - 2)*(21*n^2 - 35*n + 10)*a(n-1) - 4*(9*n^3 - 30*n^2 + 29*n - 6)*a(n-2) - 8*(3*n - 1)*(n - 2)^2*a(n-3) with a(0) = 1, a(1) = 4 and a(2) = 46.
%F Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. (End)
%F a(n) ~ (1 + sqrt(3))^(3*n + 1) / (2*Pi*sqrt(3)*n). - _Vaclav Kotesovec_, Mar 17 2023
%e A(x) = 1 + 4*x + 46*x^2 + 652*x^3 + 10186*x^4 + 168304*x^5 + 2884456*x^6 + ...
%o (PARI)
%o a(n) = sum(k=0, n, (3*n-2*k)!/((n-k)!^3*k!)*(-2)^k);
%o vector(21, n, a(n-1))
%Y Cf. A000172, A124435, A318107, A318108.
%K nonn,easy
%O 0,2
%A _Gheorghe Coserea_, Sep 20 2018
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