A339946 - OEIS

%I #20 Jul 10 2023 08:20:03

%S 2,24,812,52920,5635002,889789866,195289709624,56872979140536,

%T 21222308525755790,9874215185197183524,5604584032515576621372,

%U 3811820779676364251891562,3060364611485092496329558842,2863915888926428097267223280790,3090075825959616714726175633059312

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

%C We conjecture that the following supercongruences hold for any prime p >= 5: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k.

%C It is known that other sequences, which are related to the Legendre polynomials in a similar manner to this one, satisfy these congruences. Examples include A103882 and the two kinds of Apéry numbers A005258 and A005259: it can be shown that A103882(n) = [x^n] Legendre_P(n,(1 + x)/(1 - x)), while A005258(n) = [x^n] 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)) and A005259(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^2.

%C Calculation suggests that, for the present sequence, we have stronger congruences for prime p >= 7, namely a(p) == a(1) ( mod p^5 ) (checked up to p = 61).

%C a(p) == 2 (mod p^5) verified for primes p with 7 <= p <= 401. - _Robert Israel_, Dec 29 2020

%C For m a positive integer, define a_m(n) = [x^(m*n)] 1/Legendre_P(n,(1 - x)/(1 + x)). We conjecture that the supercongruence a_m(p) == a_m(1) ( mod p^5 ) holds for all primes p >= 7. - _Peter Bala_, Mar 10 2022

%H Robert Israel, <a href="/A339946/b339946.txt">Table of n, a(n) for n = 1..220</a>

%p with(orthopoly):

%p a:= n->coeftayl(1/P(n,(1-x)/(1+x)), x = 0, n):

%p seq(a(n), n = 1..20);

%Y Cf. A005258, A005259, A099601, A103882.

%K nonn,easy

%O 1,1

%A _Peter Bala_, Dec 23 2020