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A364299
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a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-1) for n >= 0.
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2
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1, 1, 19, 721, 49251, 5370751, 859748023, 190320431953, 55743765411043, 20884452115700251, 9745388924112505269, 5543574376457462884111, 3776677001062829977964007, 3036161801705682492174749691, 2844274879825369072829081331519
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OFFSET
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0,3
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COMMENTS
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Compare with the Apéry numbers A005258, which are related to the Legendre polynomials by A005258(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x)).
A005258 satisfies the supercongruences
1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
and the shifted supercongruences
2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
for all primes p >= 5 and positive integers n and r.
We conjecture that the present sequence also satisfies the supercongruences 1) and 2).
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LINKS
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FORMULA
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Conjectures:
1) 13*a(p) - 7*a(p-1) == 6 (mod p^5) for all primes p >= 3 (checked up to p = 101).
2) for r >= 2, 13*a(p^r) - 7*a(p^r - 1) == 13*a(p^(r-1)) - 7*a(p^(r-1) - 1) (mod p^(3*r+3)) for all primes p >= 5.
3) a(p)^13 == a(p-1)^7 (mod p^5) for all primes p >= 3 (checked up to p = 101).
4) for r >= 2, a(p^r)^13 * a(p^(r-1) - 1)^7 == a(p^(r-1))^13 * a(p^r - 1)^7 (mod p^(3*r+3)) for all primes p >= 5.
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MAPLE
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a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-1), x, 21), x, n):
seq(a(n), n = 0..20);
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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