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A246134 Binomial(2n, n) - 2 mod n^4.

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%I A246134 #13 Aug 21 2014 20:29:55
%S A246134 0,4,18,68,250,922,1029,580,2691,4754,2662,8474,4394,10294,2518,49732,
%T A246134 29478,65074,123462,128818,6535,93174,36501,12058,187750,162582,
%U A246134 297936,273782,536558,741422,59582,16964,118477,540434,132305,136130,1114366,1138598,2214594,2381618,1860867,2795686,1828661,1775622,2683618,1435710,1557345,3882778
%N A246134 Binomial(2n, n) - 2 mod n^4.
%C A246134 For e > 3, unlike the cases e=1,2,3, the numbers binomial(2n, n) - 2 mod n^e are not necessarily 0 for any n>1, be it prime or composite (see A246130 for introductory comments). Testing up to n=278000, the only number n>1 for which a(n)=0 is the first Wolstenholme prime 16843 (A088164), but no composite.
%H A246134 Stanislav Sykora, <a href="/A246134/b246134.txt">Table of n, a(n) for n = 1..10000</a>
%H A246134 R. J. McIntosh, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa71/aa7144.pdf">On the converse of Wolstenholme's theorem</a>, Acta Arithmetica 71 (4): 381-389, (1995)
%H A246134 Wikipedia, <a href="http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem">Wolstenholme's theorem</a>
%e A246134 a(7) = (binomial(14,7)-2) mod 7^4 = (3432-2) mod 2401 = 1029.
%o A246134 (PARI) a(n) = (binomial(2*n,n)-2)%n^4
%Y A246134 Cf. A000984, A088164, A246130 (e=1), A246132 (e=2), A246133 (e=3).
%K A246134 nonn
%O A246134 1,2
%A A246134 _Stanislav Sykora_, Aug 16 2014