|
%I #13 Aug 21 2014 20:27:59
%S 0,0,0,4,0,22,0,4,18,54,0,122,0,102,43,68,0,274,0,18,361,246,0,538,
%T 250,342,504,166,0,722,0,580,865,582,5,50,0,726,18,818,0,1510,0,310,
%U 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
%N Binomial(2n, n) - 2 mod n^2.
%C 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..(n-1)} 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).
%H Stanislav Sykora, <a href="/A246132/b246132.txt">Table of n, a(n) for n = 1..10000</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) = (binomial(14,7)-2) mod 7^2 = (3432-2) mod 49 = 70*49 mod 49 = 0.
%o (PARI) a(n) = (binomial(2*n,n)-2)%n^2
%Y Cf. A000984, A246130 (e=1), A246133 (e=3), A246134 (e=4).
%K nonn
%O 1,4
%A _Stanislav Sykora_, Aug 16 2014
|
