

A339946


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


3



2, 24, 812, 52920, 5635002, 889789866, 195289709624, 56872979140536, 21222308525755790, 9874215185197183524, 5604584032515576621372, 3811820779676364251891562, 3060364611485092496329558842, 2863915888926428097267223280790, 3090075825959616714726175633059312
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OFFSET

1,1


COMMENTS

We conjecture that the following supercongruences hold for any prime p >= 5: a(n*p^k) == a(n*p^(k1)) ( 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



MAPLE

with(orthopoly):
a:= n>coeftayl(1/P(n, (1x)/(1+x)), x = 0, n):
seq(a(n), n = 1..20);


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



