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a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-n-1) for n >= 0.
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%I #6 Jul 20 2023 10:10:07

%S 1,3,163,23623,6751251,3219777011,2313306332191,2337707082109071,

%T 3163417897474821763,5524913023443862515019,

%U 12101947272421487464092429,32493996621780038121738419591,104964758754905547830609842389527,401618040258524641485654323795309235

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

%C First subdiagonal of A364298.

%F Conjectures:

%F 1) the supercongruences a(p) == 2*p + 1 (mod p^3) hold for all primes p >= 5 (checked up to p = 101).

%F 2) the supercongruences a(p - 1) == 1 (mod p^4) hold for all primes p >= 3 (checked up to p = 101).

%F 3) more generally, the supercongruences a(p^k - 1) == 1 (mod p^(3+k)) may hold for all primes p >= 3 and all k >= 1.

%p a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-n-1), x, 21), x, n):

%p seq(a(n), n = 0..20);

%Y Cf. A364117, A364298, A364301.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 18 2023