%I #18 Jul 09 2023 12:14:33
%S 1,2,14,128,1310,14252,161168,1872096,22179102,266766500,3247293764,
%T 39914850560,494587904720,6170138404640,77420709800000,
%U 976308769560128,12365391374849310,157214288994620820,2005631418267291740
%N a(n) = Sum_{k = floor((n+1)/2)..n} (-1)^(n+k)*binomial(n,k)*binomial(n+k-1,k)*binomial(2*k,n).
%C Strehl's first identity for the Franel numbers A000172 is A000172(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(2*k,n). Here we modify the right-hand side of Strehl's identity and consider the sequence defined by a(n) = (-1)^n * Sum_{k = 0..n} binomial(n,k)*binomial(-n,k)*binomial(2*k,n) = Sum_{k = 0..n} (-1)^(n+k)* binomial(n,k)*binomial(n+k-1,k)*binomial(2*k,n).
%C The Franel numbers satisfy the supercongruences A000172(n*p^r) == A000172(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.
%H Eric W. Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/StrehlIdentities.html">Strehl identities</a>
%F a(2*n) = (-1)^n*(2/3)*(3*n)!/n!^3 * hypergeom([3*n, n + 1/2, -n],[n + 1, 1/2], 1) for n >= 1.
%F a(2*n+1) = 2*binomial(4*n+1, 2*n+1)*binomial(4*n+1, 2*n) * hypergeom([-n, -(2*n + 1), -(2*n + 1)/2], [-(4*n + 1), -(4*n + 1)/2], 1).
%F P-recursive: 2*(n^2)*(n - 1)*(5*n^2 - 16*n + 13)*a(n) = (n - 1)*(145*n^4 - 609*n^3 + 868*n^2 - 480*n + 96)*a(n-1) - 3*(n - 2)*(3*n - 4)*(3*n - 5)*(5*n^2 - 6*n + 2)*a(n-2) with a(0) = 1 and a(1) = 2.
%e Examples of supercongruences:
%e a(7) - a(1) = 1872096 - 2 = 2*(7^3)*2729 == 0 (mod 7^3).
%e a(2*11) - a(2) = 54602077661833355122560 - 14 = 2*7*(11^3)*182893*16021604008633 == 0 (mod 11^3).
%e a(5^2) - a(5) = 118334929857938631776326752 - 14252 = (2^2)*(5^6)*7*19*701* 3126449*6495490213 == 0 (mod 5^6).
%p seq(add((-1)^(n+k)*binomial(n,k)*binomial(n+k-1,k)*binomial(2*k,n), k = 0..n), n = 0..20);
%Y Cf. A000172, A363984.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Jul 01 2023