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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 #8 Jul 12 2023 11:04:12

%S 1,5,163,14409,2511251,730485013,320259339415,197591579213969,

%T 163325387776051459,174310058440646865021,233402385203650889753429,

%U 383208210107883180333696265,757120215942256247847040802463,1772210276849283299764079883683173

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

%C First subdiagonal of A364113.

%F Conjectures:

%F 1) the supercongruences a(p) == 2*p + 3 (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. A364113, A364116.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 08 2023