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

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).

Sequence in context: A306488 A230031 A019920 * A010675 A035673 A035638

Adjacent sequences: A246127 A246128 A246129 * A246131 A246132 A246133

Keyword

nonn

Author

Stanislav Sykora, Aug 16 2014

Status

approved