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A246130 Binomial(2n,n)-2 mod n. 7
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).

Sequence in context: A147988 A230031 A019920 * A010675 A035673 A035638

Adjacent sequences:  A246127 A246128 A246129 * A246131 A246132 A246133

KEYWORD

nonn

AUTHOR

Stanislav Sykora, Aug 16 2014

STATUS

approved

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Last modified August 21 04:27 EDT 2017. Contains 290858 sequences.