A273630 a(n) = Sum_{k = 0..n} (-1)^k*k^3*binomial(n,k)^3.

0, -1, 0, 162, 0, -11250, 0, 576240, 0, -25259850, 0, 1007242236, 0, -37685439792, 0, 1346871240000, 0, -46504059326010, 0, 1562983866658500, 0, -51407781284599740, 0, 1661123953798807680, 0, -52886433789393750000, 0, 1662782404368229351200

Offset: 0

Author

Peter Bala, Jul 17 2016

Keywords

• sign
• easy

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 = 1. See A273631 (case r = 2) and A245086 (case r = 0).

Links

• Peter Bala, A generalization of Dixon's identity
• J. Ward, 100 Years of Dixon's Identity, Irish Mathematical Society Bulletin 27, 46-54, 1991
• Wikipedia, Dixon's identity

Crossrefs

Cf. A006480, A245086, A273631.

Programs

Magma

[&+[(-1)^k*k^3 *Binomial(n, k)^3: k in [0..n]]: n in [0..70]]; // Vincenzo Librandi, Jul 23 2016

Maple

seq(add((-1)^k*k^3*binomial(n,k)^3, k = 0..n), n = 0..30);

Mathematica

Table[Sum[(-1)^k*k^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*k^3*binomial(n, k)^3) \\ Felix Fröhlich, Jul 22 2016

Python

from math import factorial
def A273630(n): return (1 if (m:=n>>1)&1 else -1)*n**3*factorial(3*m)//factorial(m)**3 if n&1 else 0 # Chai Wah Wu, Oct 04 2022

Formula

a(2*n) = 0; a(2*n + 1) = (-1)^(n+1)*(2*n + 1)^3*(3*n)!/n!^3.

a(2*n + 1) = -(2*n + 1)^3*A245086(2*n) = (-1)^(n+1)* (2*n + 1)^3*A006480(n).

a(n) = Sum_{k = 1..n} (-1)^k*multinomial(n, 1, k - 1, n - k)^3.

Recurrence: a(n) = -3*n^3*(3*n - 5)*(3*n - 7)/((n - 1)^2*(n - 2)^3) * a(n-2).