OFFSET
1,4
COMMENTS
When e=2, the numbers binomial(2n, n)-2 mod n^e are 0 whenever n is a prime (see A246130 for introductory comments). This follows from Wolstenholme's theorem or, in a simpler way, from the identity binomial(2n, n)-2 = sum_{k=1..(n-1)} binomial(n,k)^2, in which every RHS term is divisible by n^2 whenever n is a prime. No composite number n for which a(n)=0 was found up to n=431500; nevertheless, the existence of such a composite is likely (personal opinion, based on the combinatorial nature of the problem).
LINKS
Stanislav Sykora, Table of n, a(n) for n = 1..10000
Wikipedia, Wolstenholme's theorem
FORMULA
For any prime p, a(p)=0.
EXAMPLE
a(7) = (binomial(14,7)-2) mod 7^2 = (3432-2) mod 49 = 70*49 mod 49 = 0.
PROG
(PARI) a(n) = (binomial(2*n, n)-2)%n^2
CROSSREFS
KEYWORD
nonn
AUTHOR
Stanislav Sykora, Aug 16 2014
STATUS
approved