

A246132


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


6



0, 0, 0, 4, 0, 22, 0, 4, 18, 54, 0, 122, 0, 102, 43, 68, 0, 274, 0, 18, 361, 246, 0, 538, 250, 342, 504, 166, 0, 722, 0, 580, 865, 582, 5, 50, 0, 726, 18, 818, 0, 1510, 0, 310, 493, 1062, 0, 538, 1029, 2254, 2041, 406, 0, 922, 855, 1206, 379, 1686, 0, 3454, 0, 1926, 3538, 580, 3123, 922, 0, 4114, 547, 1298, 0, 4930, 0, 2742, 2518, 790, 3309, 2950, 0
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OFFSET

1,4


COMMENTS

When e=2, the numbers binomial(2n, n)2 mod n^e are 0 whenever n is a prime (see A246130 for introductory comments). This follows from Wolstenholme's theorem or, in a simpler way, from the identity binomial(2n, n)2 = sum_{k=1..(n1)} binomial(n,k)^2, in which every RHS term is divisible by n^2 whenever n is a prime. No composite number n for which a(n)=0 was found up to n=431500; nevertheless, the existence of such a composite is likely (personal opinion, based on the combinatorial nature of the problem).


LINKS



FORMULA

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


EXAMPLE

a(7) = (binomial(14,7)2) mod 7^2 = (34322) mod 49 = 70*49 mod 49 = 0.


PROG

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


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



