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A364300 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-2) for n >= 0.

%I A364300 #7 Jul 20 2023 10:09:34
%S A364300 1,3,73,3747,329001,44127003,8405999785,2160445363107,720972846685225,
%T A364300 303256387595475003,157007652309393485073,98141188253799911132091,
%U A364300 72882030213423405890701449,63436168183711463443127520699,63968150042375034921379294100073,73985402858435691329113991048739747
%N A364300 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-2) for n >= 0.
%C A364300 Row 2 of A364298.
%C A364300 Compare with the Apéry numbers A005259, which are related to the Legendre polynomials by A005259(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^2.
%C A364300 A005259 satisfies the supercongruences
%C A364300 1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
%C A364300 and the shifted supercongruences
%C A364300 2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
%C A364300 for all primes p >= 5 and positive integers n and r.
%C A364300 We conjecture that the present sequence also satisfies the supercongruences 1) and 2).
%F A364300 Conjectures:
%F A364300 1) 17*a(p) - 11*a(p-1) == 40 (mod p^5) for all primes p >= 7 (checked up to p = 101).
%F A364300 2) for r >= 2, 17*a(p^r) - 11*a(p^r - 1) == 17*a(p^(r-1)) - 11*a(p^(r-1) - 1) (mod p^(3*r+3)) for all primes p >= 5.
%F A364300 3) a(p)^(3*17) == a(1)^(3*17) * a(p-1)^11 (mod p^5) for all primes p except p = 5 (checked up to p = 101).
%F A364300 4) for r >= 2, a(p^r)^(3*17) * a(p^(r-1) - 1)^11 == a(p^(r-1))^(3*17) * a(p^r - 1)^11 (mod p^(3*r+3)) for all primes p >= 5.
%p A364300 a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-2), x, 21), x, n):
%p A364300 seq(a(n), n = 0..20);
%Y A364300 Cf. A005259, A364298, A364299, A364301, A364302.
%K A364300 nonn,easy
%O A364300 0,2
%A A364300 _Peter Bala_, Jul 18 2023