OFFSET
0,2
COMMENTS
The irregular triangle of trinomial coefficients is given in A027907. There the Comtet reference is given.
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.)
LINKS
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = binomial(2*n+1, 2) + (2*n+1)*binomial(2*n, 2) + binomial(2*n+1, 4) (from the trinomial definition) = (1/6)*n*(2*n + 1)*(2*n^2 + 9*n + 1).
G.f.: x*(6 + 15*x - 4*x^2 - x^3)/(1 - x)^5.
a(n) = (1/Pi)*Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(2*(2*n-3), 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+1, k = 4, rewritten with x = 2*cos(phi)].
MATHEMATICA
CoefficientList[Series[x (6 + 15 x - 4 x^2 - x^3) / (1 - x)^5, {x, 0, 40}], x] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 6, 45, 161, 414}, 45] (* Vincenzo Librandi, Apr 20 2018 *)
PROG
(Magma) [(1/6)*n*(2*n+1)*(2*n^2+9*n+1): n in [0..50]]; // Vincenzo Librandi, Apr 20 2018
(PARI) a(n) = n*(2*n+1)*(2*n^2+9*n+1)/6; \\ Altug Alkan, Apr 20 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Apr 19 2018
STATUS
approved