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A364117 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^(n+1) for n >= 0.

%I A364117 #8 Jul 12 2023 11:04:12
%S A364117 1,5,163,14409,2511251,730485013,320259339415,197591579213969,
%T A364117 163325387776051459,174310058440646865021,233402385203650889753429,
%U A364117 383208210107883180333696265,757120215942256247847040802463,1772210276849283299764079883683173
%N A364117 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^(n+1) for n >= 0.
%C A364117 First subdiagonal of A364113.
%F A364117 Conjectures:
%F A364117 1) the supercongruences a(p) == 2*p + 3 (mod p^3) hold for all primes p >= 5 (checked up to p = 101).
%F A364117 2) the supercongruences a(p - 1) == 1 (mod p^4) hold for all primes p >= 3 (checked up to p = 101).
%F A364117 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 A364117 a(n) := coeff(series( 1/(1-x)* LegendreP(n, (1+x)/(1-x))^(n+1), x, 21), x, n):
%p A364117 seq(a(n), n = 0..20);
%Y A364117 Cf. A364113, A364116.
%K A364117 nonn,easy
%O A364117 0,2
%A A364117 _Peter Bala_, Jul 08 2023