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A350383
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a(n) = [x^n] 1/(1 + x + x^2)^n.
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4
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1, -1, 1, 2, -15, 49, -98, 48, 561, -2860, 8151, -12948, -9282, 149226, -594320, 1428952, -1448655, -5538975, 37450900, -122995950, 239589735, -37528755, -1886983020, 8939152560, -24579514050, 35197176924, 51580335366, -541312482256, 2033695030128, -4624358661240
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OFFSET
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0,4
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,3*k).
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
a(n) = (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1).
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.
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)
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MAPLE
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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
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MATHEMATICA
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a[n_] := Coefficient[Series[1/(1 + x + x^2)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)
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PROG
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(PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1+k, k)*binomial(n, 3*k));
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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