OFFSET
0,3
COMMENTS
Compare with the identity Sum_{k = 0..n-1} binomial(n+k-1, 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).
For prime p, binomial(p+k-1, k) == 0 (mod p) for 1 <= k <= p-1. It follows that a(p) == 1 (mod p^3) for all primes p. We conjecture that, in fact, the stronger congruence 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} binomial(n+k-1, k)^(2*m+1). Then the congruence b_m(p) == 1 (mod p^(2*m+1)) clearly holds for all primes p. We conjecture that the stronger 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.
Essentially a duplicate of A112028.
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) = Sum_{k = 0..n-1} (-1)^k * binomial(-n, k)^3.
a(n) ~ 2^(6*n-3)/(7*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Aug 03 2024
EXAMPLE
Examples of supercongruences:
a(7) - a(1) = 897376152 - 1 = (7^5)*107*499 == 0 (mod 7^5)
a(11) - a(1) = 7186614533569296 - 1 = 5*(11^5)*8924644409 == 0 (mod 11^5).
MAPLE
seq(add( binomial(n+k-1, k)^3, k = 0..n-1), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Aug 03 2024
STATUS
approved