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A302710
a(n) = trinomial(2*n, 4) = (1/6)*n*(2*n - 1)*(2*n^2 + 7*n - 3).
0
0, 1, 19, 90, 266, 615, 1221, 2184, 3620, 5661, 8455, 12166, 16974, 23075, 30681, 40020, 51336, 64889, 80955, 99826, 121810, 147231, 176429, 209760, 247596, 290325, 338351, 392094, 451990, 518491, 592065, 673196, 762384, 860145, 967011, 1083530, 1210266, 1347799, 1496725, 1657656, 1831220
OFFSET
0,3
COMMENTS
The irregular triangle of trinomial coefficients is given in A027907 with the Comtet reference.
REFERENCES
L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)
FORMULA
a(n) = A027907(2*n, 4), n >= 0. a(n) = A027907(2*n, 4*(n-1)), for n >= 1 (symmetry).
a(n) = binomial(2*n, 2) + (2*n)*binomial(2*n-1, 2) + binomial(2*n, 4)(from the trinomial definition) = (1/6)*n*(2*n - 1)*(2*n^2 + 7*n - 3).
G.f.: x*(1 + 14*x + 5*x^2 - 4*x^3)/(1 - x)^5.
a(n) = (1/Pi)*Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n)*R(4*(n-2), x), n >= 0, with the R polynomial coefficients given in A127672. Note that R(-n, x) = R(n, x) [Comtet, p. 77, the integral formula for q = 3, n -> 2*n, k = 4, rewritten with x = 2*cos(phi)]. For the odd numbered rows see A302709.
MATHEMATICA
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 19, 90, 266}, 41] (* Vincenzo Librandi, Feb 28 2018 *)
PROG
(Magma) [(1/6)*n*(2*n-1)*(2*n^2+7*n-3): n in [0..40]]; // Vincenzo Librandi, Apr 28 2018
(PARI) a(n) = n*(2*n-1)*(2*n^2+7*n-3)/6; \\ Altug Alkan, May 01 2018
(PARI) a(n) = polcoeff((1 + x + x^2)^(2*n), 4); \\ Michel Marcus, May 04 2018
CROSSREFS
Cf. A014105 (k=2), A027907, A131423 (k=3), A127672, A302709.
Sequence in context: A224097 A088574 A096031 * A157875 A039409 A038653
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Apr 27 2018
STATUS
approved