OFFSET
0,2
COMMENTS
The sequence of Apéry numbers A005258 defined by A005258(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k) 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.
FORMULA
Examples of supercongruences:
a(11) - a(1)= 2114316676856525533 - 7 = 2*(3^2)*7*(11^3)*12607281056471 == 0 (mod 11^3).
a(10) - a(0) = 29640623864436949 - 1 = (2^2)*3*(11^3)*504001*3682109 == 0 (mod 11^3).
From Vaclav Kotesovec, Sep 29 2024: (Start)
Recurrence: n^4*(171*n^2 - 551*n + 446)*a(n) = 7*(2052*n^6 - 10716*n^5 + 21806*n^4 - 22301*n^3 + 12415*n^2 - 3630*n + 440)*a(n-1) + (27189*n^6 - 196365*n^5 + 577739*n^4 - 882118*n^3 + 733309*n^2 - 313390*n + 53944)*a(n-2) + (n-2)^4*(171*n^2 - 209*n + 66)*a(n-3).
a(n) ~ (5/12 + sqrt(37)*cos(arccos(433/(74*sqrt(37)))/3)/6) * (28 + 6*sqrt(93)*cos(arccos((199*sqrt(3/31))/62)/3))^n / (Pi^2*n^2). (End)
MAPLE
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Sep 24 2024
STATUS
approved