%I #10 Jul 12 2023 11:03:49
%S 10,412,15076,643900,30440010,1541377330,81983235064,4524150828092,
%T 256902133600630,14924997512212912,883403610976880740,
%U 53105747607145638706,3234568078911042493578,199234128948556264779390,12391648147019445115584576,777286417688953098495554620
%N a(n) = 3*A364114(n) - 11*A364114(n-1).
%C It is conjectured that the sequence {A364114(n)} and the shifted sequence {A364114(n-1)} both satisfy the supercongruences A364114(p^r) == A364114(p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers r. Stronger supercongruences may hold for the present sequence, a linear combination of A364114(n) and A364114(n-1).
%C Conjectures: 1) the supercongruences a(p) == a(1) (mod p^5) hold for all primes p >= 7 (checked up to p = 101).
%C 2) for r >= 2, the supercongruences a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) hold for all primes p >= 7. Cf. A212334.
%C There is also a multiplicative version of this sequence. Define a sequence of rational numbers {b(n) : n >= 1} by b(n) = A364114(n)^21 / A364114(n-1)^11. Then we conjecture that the above pair of supercongruences also hold for the sequence {b(n)}.
%H Peter Bala, <a href="/A357956/a357956.pdf">Some conjectural supercongruences for the Apéry numbers</a>.
%p A364114 := n -> coeff(series( 1/(1-x)* LegendreP(n,(1+x)/(1-x))^3, x, 21), x, n):
%p seq(3*A364114(n) - 11*A364114(n-1), n = 1..20);
%Y Cf. A212334, A357506, A357507, A357568, A364114, A364119.
%K nonn,easy
%O 1,1
%A _Peter Bala_, Jul 12 2023