OFFSET
1,1
COMMENTS
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.
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.
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).
a(p) == 2 (mod p^5) verified for primes p with 7 <= p <= 401. - Robert Israel, Dec 29 2020
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
LINKS
Robert Israel, Table of n, a(n) for n = 1..220
MAPLE
with(orthopoly):
a:= n->coeftayl(1/P(n, (1-x)/(1+x)), x = 0, n):
seq(a(n), n = 1..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Dec 23 2020
STATUS
approved