A245086 Central values of the n-th discrete Chebyshev polynomials of order 2n.

1, 0, -6, 0, 90, 0, -1680, 0, 34650, 0, -756756, 0, 17153136, 0, -399072960, 0, 9465511770, 0, -227873431500, 0, 5550996791340, 0, -136526995463040, 0, 3384731762521200, 0, -84478098072866400, 0, 2120572665910728000, 0, -53494979785374631680, 0

Offset: 0

Author

Nikita Gogin, Jul 11 2014

Keywords

• sign
• easy

Comments

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).

References

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

Links

• 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

Crossrefs

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

Programs

Mathematica

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

Formula

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