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a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^n for n >= 0.
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%I #16 Jul 22 2023 21:18:59

%S 1,3,73,5623,908001,251831261,106898093065,64439674636863,

%T 52344140654486017,55113399257643294769,73004404532578627776801,

%U 118810038754810358401521065,233027150139808176596750408337,542098915811219991386976197616441

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

%C Main diagonal of A364113.

%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 >= 3 (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 >= 5 (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.

%F From _Vaclav Kotesovec_, Jul 10 2023: (Start)

%F a(n) ~ c * d^n * n^(2*n - 1/2), where d = 2.102423770105721036432437141524634595160013830317976222331887376263238499... (the same as for A033935) and c = 1.325068544739430738025458046917491360304162175529817456184402029433873399...

%F a(n) ~ A033935(n) * exp(2*n + 1) / (2*Pi*n).

%F a(n) ~ A033935(n) * exp(1) * n^(2*n) / n!^2. (End)

%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);

%t Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^n, {x,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Jul 09 2023 *)

%Y Cf. A005258, A005259, A108625, A143007, A364113, A364114, A364115, A364117, A364301.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 08 2023