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%I #8 Jul 20 2023 10:09:50
%S 1,1,73,10805,3100001,1479318759,1062573281785,1073267499046525,
%T 1451614640844881665,2534009926232394596267,5548110762587726241026801,
%U 14890865228866506199602545427,48084585660733078332263158771313,183923731031112887024255817209295155,822427361894711201025101782425695273529
%N a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-n) for n >= 0.
%C Main diagonal of A364298 (with extra initial term 1). Compare with A364116.
%C Compare with the two types of Apéry numbers A005258 and A005259, which are related to the Legendre polynomials by A005258(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x)) and A005259(n) = [x^k] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^2.
%C A005258 is the main diagonal of A108625 and A005259 is the main diagonal of A143007.
%F Conjectures:
%F 1) a(p) == 2*p - 1 (mod p^4) for all primes p >= 5 (checked up to p = 101).
%F More generally, the supercongruence a(p^k) == 2*p^k - 1 (mod p^(3+k)) may hold for all primes p >= 5 and all k >= 1.
%F 2) a(p-1) == 1 (mod p^3) for all primes p except p = 3 (checked up to p = 101).
%F More generally, the supercongruence a(p^k - p^(k-1)) == 1 (mod p^(2+k)) may hold for all primes p >= 5 and all k >= 1.
%p a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-n), x, 21), x, n):
%p seq(a(n), n = 0..20);
%Y Cf. A005258, A005259, A108625, A143007, A364116, A364298, A364302.
%K nonn,easy
%O 0,3
%A _Peter Bala_, Jul 18 2023
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