OFFSET
0,2
COMMENTS
Cf. A005258(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2.
The sequence of Apéry numbers A005258 satisfies the pair of supercongruences
1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005258(n*p^r - 1) == A005258(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.
Constant terms in powers of the Laurent polynomial (d + 1) * (c + d + 1) * (b + c + d + 1) * (a + c + 1) * (a + b + c + d + 1) / (a*b*c*d). - F. Chapoton, May 08 2026
FORMULA
a(n) ~ 3^(3*n/2) / (Pi^2 * n^2 * 2^(9*n+6) * sin(Pi/9)^(9*n+4)). - Vaclav Kotesovec, May 08 2026
EXAMPLE
Examples of supercongruences:
(1a) a(13) - a(1) = 26126080107002322789528197 - 11 = 2*3*(7^3)*11*(13^3)*101* 5200975019533651 == 0 (mod 13^3);
(1b) a(13 - 1) - a(1 - 1) = 192832695330486199076011 - 1 = 2*3*5*(13^3)*107*457* 1366649*43779661 == 0 (mod 13^3).
(2a) a(5^2) - a(5) = 1806832316206138774124773283611717478292644188023761 - 416148761 = (2^3)*(3^3)*(5^8)*953*22470418607359387809254496551076506667197 = 0 (mod 5^8);
(2b) a(5^2 - 1) - a(5 - 1) = 12360782465873293068286987426696670402529985934251 - 3996751 = (2^2)*3*(5^6)*11779*1268093*9320371*634752259*746015256807592939 == 0 (mod 5^6).
MAPLE
MATHEMATICA
Table[Sum[(-1)^(n+k)*Binomial[n, k]*Binomial[n+k, k]^2 * HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 08 2026 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Sep 24 2024
STATUS
approved
