OFFSET
0,3
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.
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(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
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 *)
PROG
(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
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Peter Bala, Mar 23 2023
STATUS
approved