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a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, n-k).
1

%I #11 Sep 29 2024 09:19:13

%S 1,-1,-17,143,751,-20251,30871,2584847,-21586193,-251907751,

%T 5176221733,5498864117,-913327142441,5540080670669,120825094592983,

%U -1860921180719857,-8346832617144593,401702184476719649,-1403893237226212151,-64680833271083055607,743195619082337134501,6754996433001423371159,-192371016736634220839987,139058974519768723621493,36163089652079749214625751,-298797649039016749340832751

%N a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, 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 pair of supercongruences

%C 1) 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 and

%C 2) A005259(n*p^r - 1) == A005259(n*p^(r-1) - 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 the same pair of supercongruences. Some examples are given below.

%F P-recursive: n^3*(n - 1)*(946*n^2 - 3053*n + 2475)*a(n) = -2*(n - 1)*(3784*n^5 - 17888*n^4 + 31787*n^3 - 26726*n^2 + 11051*n - 1824)*a(n-1) - 2*(104060*n^6 - 752070*n^5 + 2212238*n^4 - 3374927*n^3 + 2802671*n^2 - 1196821*n + 205920)*a(n-2) - 2*(n - 2)^3*(2*n - 3)*(946*n^2 - 1161*n + 368)*a(n-3) with a(0) = 1, a(1) = -1 and a(2) = -17.

%e Examples of supercongruences:

%e (1a) a(11) - a(1) = 5498864117 - (-1) = 2*(3^3)*(11^3)*76507 == 0 (mod 11^3);

%e (1b) a(10) - a(0) = 5176221733 - 1 = (2^2)*(3^5)*(11^3)*4001 == 0 mod 11^3).

%e (2a) a(5^2) - a(5) = -298797649039016749340832751 - (-20251) = - (2^2)*3*(5^6)*

%e (11^2)*47*89*1683049*1870707593 == 0 (mod 5^6);

%e (2b) a(5^2 - 1) - a(5 - 1) = 36163089652079749214625751 - 751 = (2^3)*3*(5^6)*7*11*

%e 17*101*729412564491671 == 0 (mod 5^6).

%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, n-k), k = 0..n):

%p seq(a(n), n = 0..25);

%Y Cf. A005259, A108625, A376458, A376460 - A376466.

%K sign,easy

%O 0,3

%A _Peter Bala_, Sep 24 2024