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%I #49 Apr 19 2023 09:54:30
%S 1,-1,1,2,-15,49,-98,48,561,-2860,8151,-12948,-9282,149226,-594320,
%T 1428952,-1448655,-5538975,37450900,-122995950,239589735,-37528755,
%U -1886983020,8939152560,-24579514050,35197176924,51580335366,-541312482256,2033695030128,-4624358661240
%N a(n) = [x^n] 1/(1 + x + x^2)^n.
%H Seiichi Manyama, <a href="/A350383/b350383.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,3*k).
%F Recurrence: 3*(n-1)*n*(4*n - 7)*a(n) = -2*(n-1)*(28*n^2 - 63*n + 27)*a(n-1) - 3*(3*n - 5)*(3*n - 4)*(4*n - 3)*a(n-2). - _Vaclav Kotesovec_, Mar 18 2023
%F From _Peter Bala_, Apr 15 2023: (Start)
%F a(n) = (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1).
%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. Cf. A228960.
%F More generally, let k be a positive integer, m an integer and let f(x) = g(x)/h(x), where g(x) and h(x) are both finite products of cyclotomic polynomials. Then we conjecture that the same supercongruences hold, except for a finite number of primes p depending on f(x), for the sequence {a_(k,m,f)(n): n >= 0} defined by a_(k,m,f)(n) = [x^(k*n)] f(x)^(m*n). (End)
%p a := n -> (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1): seq(simplify(a(n)), n = 0..30); # _Peter Bala_, Apr 17 2023
%t a[n_] := Coefficient[Series[1/(1 + x + x^2)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* _Amiram Eldar_, Dec 29 2021 *)
%o (PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1+k, k)*binomial(n, 3*k));
%Y Cf. A002426, A027907, A110556, A165817, A228960, A350406, A350407.
%K sign,easy
%O 0,4
%A _Seiichi Manyama_, Dec 29 2021