%I #14 Sep 29 2024 09:20:28
%S 1,1,-7,1,569,-3749,-45151,806737,1052729,-130060889,740060243,
%T 16076432923,-238772815711,-1050791121197,49401000432497,
%U -171944622257999,-7658491447803847,87632552103603679,768037618172427023,-22023427875902878553,19183786570616924819,4030690809877385503081,-33792039667279104716677,-520860578851790657166869
%N a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, n-k).
%C Compare with the following identity relating the sequence of Apéry numbers to the table of crystal ball sequences for the A_n lattices: A005259(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k), which can be verified by using the MulZeil procedure in Zeilberger's MultiZeilberger Maple package to find a recurrence for the double sum on the right-hand side of the identity.
%C The sequence of Apéry numbers A005259 satisfies the supercongruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
%C We conjecture that the present sequence satisfies same supercongruences and in addition satisfies the stronger congruences a(p) == a(1) (mod p^5) for all primes p >= 7 (checked up to p = 199) and a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for all primes p >= 5 and integers r >= 2.
%F a(n) = Sum_{0 <= i <= k <= n} (-1)^k * binomial(n, k) * binomial(2*n-k, n-k) * binomial(n-1, i)^2 * binomial(n+k-i-1, k-i).
%F P-recursive: (2*n - 3)*n^3*(n - 1)^2*(473*n^5 - 4988*n^4 + 20888*n^3 - 43462*n^2 + 45019*n - 18634)*a(n) = - 2*(n - 1)^2*(3784*n^9 - 51256*n^8 + 303801*n^7 - 1037327*n^6 + 2252744*n^5 - 3220636*n^4 + 3006247*n^3 - 1739455*n^2 + 555714*n - 75024)*a(n-1) - 2*(n - 2)*(2*n - 1)*(52030*n^9 - 756800*n^8 + 4787337*n^7 - 17271387*n^6 + 39143817*n^5 - 57806236*n^4 + 55708921*n^3 - 33926177*n^2 + 11955879*n - 1890360)*a(n-2) - 2*(n - 2)*(n - 3)^3*(2*n - 1)*(2*n - 3)*(473*n^5 - 2623*n^4 + 5666*n^3 - 5996*n^2 + 3172*n - 704)*a(n-3) with a(0) = 1, a(1) = -1 and a(2) = 7.
%e Examples of supercongruences:
%e a(7) - a(1) = 806737 - 1 = (2^4)*3*(7^5) == 0 (mod 7^5).
%e a(11) - a(1) = 16076432923 - 1 = 2*3*(11^5)*127*131 == 0 (mod 11^5).
%e a(5^2) - a(5) = 22511570786292886382808751 - (-3749) = (2^2)*(3^2)*(5^9)*67*97*
%e 7741*49223*129289 == 0 (mod 5^9).
%p A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
%p a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, n-k), k = 0..n):
%p seq(a(n), n = 0..25);
%Y Cf. A005259, A108625, A183204, A352653, A376459 - A376466.
%K sign,easy
%O 0,3
%A _Peter Bala_, Sep 23 2024