A246130 - OEIS

A246130

Binomial(2n,n)-2 mod n.

0, 0, 0, 0, 0, 4, 0, 4, 0, 4, 0, 2, 0, 4, 13, 4, 0, 4, 0, 18, 4, 4, 0, 10, 0, 4, 18, 26, 0, 2, 0, 4, 7, 4, 5, 14, 0, 4, 18, 18, 0, 40, 0, 2, 43, 4, 0, 10, 0, 4, 1, 42, 0, 4, 30, 30, 37, 4, 0, 34, 0, 4, 10, 4, 3, 64, 0, 34, 64, 38, 0, 34, 0, 4, 43, 30, 75, 64, 0, 18, 18, 4, 0, 26, 63, 4, 76, 86, 0, 38, 89, 22, 18, 4, 3, 58, 0

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,6

COMMENTS

By Wolstenholme's theorem, when n>3 is prime and cb(n) is the central binomial coefficient A000984(n), then cb(n)-2 is divisible by n^3. This implies that it is also divisible by n^e for e=1,2 and 3, but not necessarily for e=4. It follows also that cn(n)-2, with cn(n)=cb(n)/(n+1) being the n-th Catalan number A000108(n), is divisible by any prime n. In fact, for any n>0, cn(n)-2 = (n+1)cb(n)-2 implies (cn(n)-2) mod n = (cb(n)-2) mod n = a(n). The sequence a(n) is of interest as a prime-testing sequence similar to Fermat's, albeit not a practical one until/unless an efficient algorithm to compute moduli of binomial coefficients is found. For more info, see A246131 through A246134.

LINKS

Stanislav Sykora, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Central Binomial Coefficient

Wikipedia, Wolstenholme's theorem

FORMULA

For any prime p, a(p)=0.

EXAMPLE

a(7)=0 because cb(7)-2 = binomial(14,7) -2 = 3432-2 = 490*7. Check also that cn(7) = 3432/8 = 429 and 429-2 = 61*7 so that (cn(7)-2) mod 7 = 0.

PROG

(PARI) a(n) = (binomial(2*n, n)-2)%n

CROSSREFS

Cf. A000108, A000984, A128311, A246131 (pseudoprimes of a(n)), A246132 (e=2), A246133 (e=3), A246134 (e=4).