

A246130


Binomial(2n,n)2 mod n.


6



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
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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 nth 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 primetesting 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



FORMULA

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


EXAMPLE

a(7)=0 because cb(7)2 = binomial(14,7) 2 = 34322 = 490*7. Check also that cn(7) = 3432/8 = 429 and 4292 = 61*7 so that (cn(7)2) mod 7 = 0.


PROG

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


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



