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a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n-1 k).
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%I #7 Sep 29 2024 09:20:18

%S 1,5,289,32909,4846001,824723005,154077084505,30725590395197,

%T 6429986799134257,1396511202452212733,312375607924873231289,

%U 71567605248444374973205,16725718218774077760354953,3974882968204513021199112653,958241214962583413382405975337,233879553339714596653197104362909

%N a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n-1 k).

%C The sequence of Apéry numbers A005259 defined by A005259(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2 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 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.

%e Examples of supercongruences:

%e a(11) - a(1) = 71567605248444374973205 - 5 = (2^4)*(5^2)*(11^3)*19*56040893*126246629 == 0 (mod 11^3).

%e a(10) - a(0) = 312375607924873231289 - 1 = (2^3)*(11^3)*17*1725679541724893 == 0 (mod 11^3).

%p A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):

%p a(n) := add(binomial(n, k)^2*binomial(n+k, k)^2*A108625(n-1, k), k = 0..n):

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

%Y Cf. A005259, A376458 - A376466.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Sep 25 2024