OFFSET
0,3
COMMENTS
Compare with the identity Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k) = (1/2) * binomial(2*n, n) = (1/2) * A000984(n) for n >= 1.
The central binomial coefficients satisfy the supercongruence (1/2) * binomial(2*p, p) == 1 (mod p^3) for all primes p >= 5 (Wolstenholme's theorem).
We conjecture that for the present sequence the stronger supercongruence a(p) == 1 (mod p^5) holds for all primes p >= 7.
Further, we conjecture that for r >= 2 and prime p >= 5, a(p^r) == a(p^(r-1)) (mod p^(3*r+3)).
More generally, for a positive integer m, define a sequence {b_m(n) : n >= 0} by setting b_m(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k)^(2*m+1). We conjecture that the supercongruence b_m(p) == 1 (mod p^(2*m+3)) holds for all primes p >= 2*m + 5, and for r >= 2, the supercongruence b_m(p^r) == b_m(p^(r-1)) (mod p^(3*r+2*m+1)) also holds for all primes p >= 2*m + 5.
LINKS
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).
FORMULA
a(n) ~ 2^(6*n-1) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Aug 08 2024
EXAMPLE
Examples of supercongruences:
a(7) - a(1) = 20012582304 - 1 = (7^5)*389*3061 == 0 (mod 7^5);
a(17) - a(1) = 6354063452012073380091209250 - 1 = (17^5)*6007*15725921* 47373254231 == 0 (mod 17^5).
MAPLE
seq(add( (-1)^(n+k+1)*binomial(2*n, k)^3, k = 0..n-1), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Aug 04 2024
STATUS
approved