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a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)*A108625(n, k).
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%I #11 Sep 29 2024 08:53:45

%S 1,7,199,8359,423751,23973757,1457477659,93222362023,6192061490503,

%T 423467225669257,29640623864436949,2114316676856525533,

%U 153205735737013589371,11249571292274407086001,835446050985954811273699,62655191693751473885672359,4739286937335653870416516423,361196527809084536022005145121

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

%C The sequence of Apéry numbers A005258 defined by A005258(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k) satisfies the pair of supercongruences

%C 1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r

%C and

%C 2) A005258(n*p^r - 1) == A005258(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 Examples of supercongruences:

%F a(11) - a(1)= 2114316676856525533 - 7 = 2*(3^2)*7*(11^3)*12607281056471 == 0 (mod 11^3).

%F a(10) - a(0) = 29640623864436949 - 1 = (2^2)*3*(11^3)*504001*3682109 == 0 (mod 11^3).

%F From _Vaclav Kotesovec_, Sep 29 2024: (Start)

%F Recurrence: n^4*(171*n^2 - 551*n + 446)*a(n) = 7*(2052*n^6 - 10716*n^5 + 21806*n^4 - 22301*n^3 + 12415*n^2 - 3630*n + 440)*a(n-1) + (27189*n^6 - 196365*n^5 + 577739*n^4 - 882118*n^3 + 733309*n^2 - 313390*n + 53944)*a(n-2) + (n-2)^4*(171*n^2 - 209*n + 66)*a(n-3).

%F a(n) ~ (5/12 + sqrt(37)*cos(arccos(433/(74*sqrt(37)))/3)/6) * (28 + 6*sqrt(93)*cos(arccos((199*sqrt(3/31))/62)/3))^n / (Pi^2*n^2). (End)

%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)*A108625(n, k), k = 0..n):

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

%Y Cf. A005258, A376458 - A376466.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Sep 24 2024