cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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

Original entry on oeis.org

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

Views

Author

Nikita Gogin, Jul 11 2014

Keywords

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.

Crossrefs

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

A273631 a(n) = Sum_{k = 0..n} (-1)^k*binomial(k,2)^3*binomial(n,k)^3.

Original entry on oeis.org

0, 0, 1, 0, -1296, 0, 303750, 0, -36879360, 0, 3157481250, 0, -217564322976, 0, 12926105848656, 0, -689598074880000, 0, 33901459248661290, 0, -1562983866658500000, 0, 68423756889802253940, 0, -2870422192164339671040, 0, 116191495035298068750000
Offset: 0

Views

Author

Peter Bala, Jul 17 2016

Keywords

Comments

Let d(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)^3. Clearly, by symmetry of the binomial coefficients we have d(2*n + 1) = 0. Dixon's identity is the result d(2*n) = (-1)^n*(3*n)!/n!^3. A generalization is: for r a nonnegative integer there holds Sum_{k = 0..n} (-1)^k*binomial(k,r)^3*binomial(n,k)^3 = (-1)^r*binomial(n,r)^3*d(n - r). This is the case r = 2. See A273630 (case r = 1) and A245086 (case r = 0).

Crossrefs

Programs

  • Magma
    [&+[(-1)^k*Binomial(k,2)^3*Binomial(n,k)^3: k in [0..n]]: n in [0..70]]; // Vincenzo Librandi, Jul 23 2016
    
  • Maple
    seq(add((-1)^k*binomial(k,2)^3*binomial(n,k)^3, k = 0..n), n = 0..30);
  • Mathematica
    Table[Sum[(-1)^k*Binomial[k, 2]^3 Binomial[n, k]^3, {k, 0, n}], {n, 0, 27}] (* Michael De Vlieger, Jul 22 2016 *)
  • PARI
    a(n) = sum(k=0, n, (-1)^k*binomial(k, 2)^3*binomial(n, k)^3) \\ Felix Fröhlich, Jul 22 2016
    
  • Python
    from math import factorial
    def A273631(n): return 0 if n&1 or n == 0 else (-1 if (m:=n-1>>1)&1 else 1)*((m+1)*(n-1))**3*factorial(3*m)//factorial(m)**3 # Chai Wah Wu, Oct 04 2022

Formula

a(0) = 0 and a(2*n + 2) = (-1)^n*binomial(2*n + 2,2)^3*(3*n)!/n!^3 for n >= 0. a(2*n + 1) = 0.
a(2*n + 2) = (-1)^n*(n + 1)^3*(2*n + 1)^3 * A006480(n) for n >= 0.
a(n) = Sum_{k = 2..n} (-1)^k*multinomial(n, 2, k - 2, n - k)^3.
Recurrence: a(n) = -3*n^3*(n - 1)^3*(3*n - 8)*(3*n - 10)/((n - 2)^5*(n - 3)^3) * a(n-2).
Showing 1-2 of 2 results.