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%I #17 Aug 21 2014 20:31:38
%S 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,
%T 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,
%U 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
%N Binomial(2n,n)-2 mod n.
%C 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.
%H Stanislav Sykora, <a href="/A246130/b246130.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CentralBinomialCoefficient.html">Central Binomial Coefficient</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem">Wolstenholme's theorem</a>
%F For any prime p, a(p)=0.
%e 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.
%o (PARI) a(n) = (binomial(2*n,n)-2)%n
%Y Cf. A000108, A000984, A128311, A246131 (pseudoprimes of a(n)), A246132 (e=2), A246133 (e=3), A246134 (e=4).
%K nonn
%O 1,6
%A _Stanislav Sykora_, Aug 16 2014
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