0,3

In the general case the n-th discrete Chebyshev polynomial of order N is D(N,n;x) = Sum_{i = 0..n} (-1)^i*C(n,i)*C(N-x,n-i)*C(x,i). For N = 2*n , x = n, one gets a(n) = D(2n,n;n) = Sum_{i = 0..n} (-1)^i*C(n,i)^3 that equals (due to Dixon's formula) 0 for odd n and (-1)^m*(3m)!/(m!)^3 for n = 2*m. (Riordan, 1968) So, a(2*m) = (-1)^m*A006480(m).

John Riordan, Combinatorial Identities, John Willey&Sons Inc., 1968.

Nikita Gogin and Mika Hirvensalo, On the Generating Function of Discrete Chebyshev Polynomials, TUCS Technical Reports 819, Turku Centre for Computer Science, 2007.

Nikita Gogin and Mika Hirvensalo, Recurrent Constructions of the MacWilliams and Chebyshev Matrices, Fundamenta Informaticae, vol.116, issue 1-4, January 2012, pages 93-110.

H. W. Gould, Tables of Combinatorial Identities, Edited by J. Quaintance.

Darij Grinberg, Introduction to Modern Algebra (UMN Spring 2019 Math 4281 Notes), University of Minnesota (2019).

Y. Sun and X. Wang, A new proof of a curious identity, Math Gazette, Vol. 91, No. 520 (Mar 2007) 105-107.

Wikipedia, Dixon's identity

a(n) is a coefficient at t^n in (1-t^2)^n*P(0,-(2*n+1);n;(1+t^2)/(1-t^2)), where P(a,b;k;x) is the k-th Jacobi polynomial (Gogin and Hirvensalo, 2007).

G.f.: Hypergeometric2F1[1/3,2/3,1,-27*x^2].

a(2*m+1) = 0, a(2*m) = (-1)^m*A006480(m).

From Peter Bala, Aug 04 2016: (Start)

a(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(n + k,n) (Sun and Wang).

a(n) = Sum_{k = 0..n} (-1)^(n + k)*binomial(n + k, n - k)*binomial(2*k, k)*binomial(2*n - k, n) (Gould, Vol.5, 9.23).

a(n) = -1/(n + 1)^3 * A273630(n+1). (End)

From Peter Bala, Mar 22 2022: (Start)

a(n) = - (3*(3*n-2)*(3*n-4)/n^2)*a(n-2).

a(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A002894(n) = binomial(2*n,n)^2 = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)). Cf. A103882. (End)

From Peter Bala, Jul 23 2023: (Start)

a(n) = [x^n] G(x)^(3*n), where the power series G(x) = 1 - x^2 + 2*x^4 - 14*x^6 + 127*x^8 - 1364*x^10 + ... appears to have integer coefficients.

exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^3, where the power series F(x) = 1 - x^2 + 8*x^4 - 101*x^6 + 1569*x^8 - 27445*x^10 + ..., appears to have integer coefficients. See A229452.

Row 1 of A364303. (End)

a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(n+k, k)^2 * binomial(3*n+1, n-k). Cf. A183204.- Peter Bala, Sep 20 2024

Table[Coefficient[Simplify[JacobiP[n, 0, -(2*n+1), (1+t^2)/(1-t^2)]*(1-t^2)^n], t, n], {n, 0, 20}]

(Python)

from math import factorial

def A245086(n): return 0 if n&1 else (-1 if (m:=n>>1)&1 else 1)*factorial(3*m)//factorial(m)**3 # Chai Wah Wu, Oct 04 2022

Cf. A000172, A002894, A005260, A006480, A103882, A183204, A273630, A364303.

Sequence in context: A199044 A156488 A057399 * A365909 A145223 A365979

Adjacent sequences: A245083 A245084 A245085 * A245087 A245088 A245089

sign,easy

Nikita Gogin, Jul 11 2014

approved