A361716 a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)^2*binomial(n-1,k).

0, 1, -3, -8, 45, 126, -840, -2400, 17325, 50050, -378378, -1100736, 8576568, 25069968, -199536480, -585307008, 4732755885, 13919870250, -113936715750, -335813478000, 2775498395670, 8194328596740, -68263497731520, -201822515032320

Offset: 0

Author

Peter Bala, Mar 23 2023

Keywords

• sign
• easy

Comments

Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and positive integers n and k.

Links

• Paolo Xausa, Table of n, a(n) for n = 0..1000
• Wikipedia, Dixon's identity.

Crossrefs

Cf. A006480, A245086, A361710, A361711.

Programs

Maple

seq(add((-1)^k*binomial(n,k)^2*binomial(n-1,k), k = 0..n-1), n = 0..20);

Mathematica

A361716[n_]:=Sum[(-1)^k*Binomial[n,k]^2Binomial[n-1,k],{k,0,n-1}];Array[A361716,30,0] (* Paolo Xausa, Oct 06 2023 *)

PARI

a(n) = sum(k = 0, n-1, (-1)^k*binomial(n,k)^2*binomial(n-1,k)); \\ Michel Marcus, Mar 26 2023

Python

from math import comb
def A361716(n): return (sum(comb(n,k)**3*k if k&1 else -comb(n,k)**3*k for k in range(n+1)))//(n if n&1 else -n) if n else 0 # Chai Wah Wu, Mar 27 2023

Formula

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

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

a(2*n) = (1/2)*A245086(2*n) = (1/2)*(-1)^n*A006480(n) for n >= 1.

a(2*n+1) = A361710(2*n+1) = A361711(2*n+1).

a(n) = hypergeom([1 - n, - n, - n], [1, 1], 1) for n >= 1.

P-recursive: n^2*(n-1)*(6*n^2-20*n+17)*a(n) = -( 6*(3*n^2-6*n+2)*(n-1)*a(n-1) + (3*n-6)*(3*n-5)*(3*n-4)*(6*n^2-8*n+3)*a(n-2) ).

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

a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n+k,n)*binomial(2*n-k-1,n). - Peter Bala, Sep 13 2023