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a(n) = Sum_{k=0..n} (3*n-2*k)!/((n-k)!^3*k!)*(-3)^k, n >= 0.
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%I #27 Mar 17 2023 11:29:46

%S 1,3,27,303,3771,49653,677979,9496791,135572859,1963940073,

%T 28783474677,425872190241,6350923156059,95341185353781,

%U 1439433069482547,21839152342265703,332769145298428539,5089688869615075521,78108038975852093889,1202268428203687094493,18555675891246972931221

%N a(n) = Sum_{k=0..n} (3*n-2*k)!/((n-k)!^3*k!)*(-3)^k, n >= 0.

%C Diagonal of rational function 1/(1 - (x + y + z - 3*x*y*z)).

%H Seiichi Manyama, <a href="/A318108/b318108.txt">Table of n, a(n) for n = 0..829</a> (terms 0..100 from Gheorghe Coserea)

%F G.f. y=A(x) satisfies: 0 = x*(6*x - 1)*(27*x^3 + 27*x^2 - 18*x + 1)*y'' + (486*x^4 + 216*x^3 - 189*x^2 + 36*x - 1)*y' + 3*(3*x + 1)*(18*x^2 - 6*x + 1)*y.

%F Recurrence: n^2*(3*n - 4)*a(n) = 3*(3*n - 2)*(6*n^2 - 10*n + 3)*a(n-1) - 9*(9*n^3 - 30*n^2 + 29*n - 6)*a(n-2) - 27*(n-2)^2*(3*n - 1)*a(n-3). - _Vaclav Kotesovec_, Mar 01 2019

%F From _Peter Bala_, Mar 16 2023: (Start)

%F a(n) = Sum_{k = 0..n} (-3)^(n-k)*binomial(n,k)*binomial(n + 2*k,n)* binomial(2*k,k).

%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)

%e A(x) = 1 + 3*x + 27*x^2 + 303*x^3 + 3771*x^4 + 49653*x^5 + 677979*x^6 + ...

%t Table[Sum[(3n-2k)!/(((n-k)!)^3 k!) (-3)^k,{k,0,n}],{n,0,20}] (* _Harvey P. Dale_, Mar 01 2019 *)

%o (PARI)

%o a(n) = sum(k=0, n, (3*n-2*k)!/((n-k)!^3*k!)*(-3)^k);

%o vector(21, n, a(n-1))

%Y Cf. A000172, A124435, A318107, A318109.

%K nonn,easy

%O 0,2

%A _Gheorghe Coserea_, Sep 20 2018