OFFSET
1,3
COMMENTS
Conjecture: the supercongruence a(p^k) == a(p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and positive integer k.
LINKS
Wikipedia, Dixon's identity.
FORMULA
a(2*n) = (-1)^n * (1/6) * (2*n-3)/(2*n-1) * (3*n)!/n!^3 = (-1)^n * (1/6) * (2*n-3)/(2*n-1) * A006480(n) for n >= 1.
a(2*n+1) = (-1)^n * (3*n+1)/(2*n+1) * (3*n)!/n!^3 for n >= 1.
a(n) = hypergeom([1 -n, -1 - n, -1 - n], [1, 1], 1).
P-recursive: n^2*(n-2)*(3*n^2-14*n+17)*a(n) = -6*(6*n^3-24*n^2+29*n-9)*a(n-1) - 3*(n-3)*(3*n-4)*(3*n-5)*(3*n^2-8*n+6)*a(n-2) with a(1) = a(2) = 1.
EXAMPLE
Examples of supercongruences:
a(11) - a(1) = - (11^3)*827 == 0 (mod 11^3);
a(13) - a(1) = (13^3)*11411 == 0 (mod 13^3);
a(23) - a(1) = -(23^3)*16587697463 == 0 (mod 23^3);
a(5^2) - a(5) = 2*(3^2)*(5^6)*7*6791*374681 == 0 (mod 5^6).
MAPLE
a := proc(n) option remember; if n = 1 then 1 elif n = 2 then 1 else ( -6*(6*n^3-24*n^2+29*n-9)*a(n-1) - 3*(n-3)*(3*n-4)*(3*n-5)*(3*n^2-8*n+6)*a(n-2) )/( n^2*(n-2)*(3*n^2-14*n+17) ) end if; end:
seq(a(n), n = 1..25);
PROG
(PARI) a(n) = if (n==1, 1, sum(k = 0, n-2, (-1)^k * binomial(n, k)^2 * binomial(n-2, k))); \\ Michel Marcus, Mar 26 2023
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Peter Bala, Mar 21 2023
STATUS
approved