login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

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

%I #20 Mar 26 2023 10:27:20

%S 1,1,-8,5,126,-168,-2400,4125,50050,-98098,-1100736,2339064,25069968,

%T -56279520,-585307008,1367240589,13919870250,-33510798750,

%U -335813478000,827780223270,8194328596740,-20587404077760,-201822515032320,515067876905400,5009403008531376,-12953308371172848

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

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

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Dixon%27s_identity">Dixon's identity</a>.

%F 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.

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

%F a(2*n+1) = A361710(2*n+1) = A361716(2*n+1).

%F a(n) = hypergeom([1 -n, -1 - n, -1 - n], [1, 1], 1).

%F 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.

%e Examples of supercongruences:

%e a(11) - a(1) = - (11^3)*827 == 0 (mod 11^3);

%e a(13) - a(1) = (13^3)*11411 == 0 (mod 13^3);

%e a(23) - a(1) = -(23^3)*16587697463 == 0 (mod 23^3);

%e a(5^2) - a(5) = 2*(3^2)*(5^6)*7*6791*374681 == 0 (mod 5^6).

%p 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:

%p seq(a(n), n = 1..25);

%o (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

%Y Cf. A006480, A361710, A361716.

%K sign,easy

%O 1,3

%A _Peter Bala_, Mar 21 2023