A302709 a(n) = Trinomial(2*n+1, 4) = (1/6)*n*(2*n + 1)*(2*n^2 + 9*n + 1), n >= 0.

0, 6, 45, 161, 414, 880, 1651, 2835, 4556, 6954, 10185, 14421, 19850, 26676, 35119, 45415, 57816, 72590, 90021, 110409, 134070, 161336, 192555, 228091, 268324, 313650, 364481, 421245, 484386, 554364, 631655, 716751, 810160, 912406, 1024029, 1145585, 1277646, 1420800, 1575651, 1742819, 1922940

Offset: 0

Author

Wolfdieter Lang, Apr 19 2018

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A027907, A000384 (k=2), A030440 (k=3), A127672.

Programs

Magma

[(1/6)*n*(2*n+1)*(2*n^2+9*n+1): n in [0..50]]; // Vincenzo Librandi, Apr 20 2018

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 *)

PARI

a(n) = n*(2*n+1)*(2*n^2+9*n+1)/6; \\ Altug Alkan, Apr 20 2018

Formula

a(n) = A027907(2*n+1, 4), n >= 0. a(n) = A027907(2*n+1, 2*(2*n-1)), for n >= 1 (symmetry).

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)].