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%I #8 Jul 20 2023 10:08:59
%S 1,1,19,721,49251,5370751,859748023,190320431953,55743765411043,
%T 20884452115700251,9745388924112505269,5543574376457462884111,
%U 3776677001062829977964007,3036161801705682492174749691,2844274879825369072829081331519
%N a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-1) for n >= 0.
%C Row 1 of A364298.
%C Compare with the Apéry numbers A005258, which are related to the Legendre polynomials by A005258(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x)).
%C A005258 satisfies the supercongruences
%C 1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
%C and the shifted supercongruences
%C 2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
%C for all primes p >= 5 and positive integers n and r.
%C We conjecture that the present sequence also satisfies the supercongruences 1) and 2).
%F Conjectures:
%F 1) 13*a(p) - 7*a(p-1) == 6 (mod p^5) for all primes p >= 3 (checked up to p = 101).
%F 2) for r >= 2, 13*a(p^r) - 7*a(p^r - 1) == 13*a(p^(r-1)) - 7*a(p^(r-1) - 1) (mod p^(3*r+3)) for all primes p >= 5.
%F 3) a(p)^13 == a(p-1)^7 (mod p^5) for all primes p >= 3 (checked up to p = 101).
%F 4) for r >= 2, a(p^r)^13 * a(p^(r-1) - 1)^7 == a(p^(r-1))^13 * a(p^r - 1)^7 (mod p^(3*r+3)) for all primes p >= 5.
%p a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-1), x, 21), x, n):
%p seq(a(n), n = 0..20);
%Y Cf. A005258, A364298, A364300, A364301, A364302.
%K nonn,easy
%O 0,3
%A _Peter Bala_, Jul 18 2023
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