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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 (list; graph; refs; listen; history; text; internal format)
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
MAPLE
with(orthopoly):
a:= n->coeftayl(1/P(n, (1-x)/(1+x)), x = 0, n):
seq(a(n), n = 1..20);
CROSSREFS
Sequence in context: A012128 A099704 A265879 * A172492 A322895 A264559
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Dec 23 2020
STATUS
approved

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Last modified April 17 02:32 EDT 2024. Contains 371756 sequences. (Running on oeis4.)