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2^(p-1) modulo p^3, where p = prime(n).
2

%I #6 Apr 02 2016 21:40:01

%S 2,4,16,64,1024,1899,1667,1502,8856,10122,14602,20573,27840,10321,

%T 92638,86179,35283,54291,126363,211865,313171,338516,114209,317375,

%U 598297,702961,822971,1089047,684521,928748,421641,911761,739253,912258,2634023,829293,505855

%N 2^(p-1) modulo p^3, where p = prime(n).

%C H. S. Vandiver showed that a(n) = 1 if and only if sum{k=1, p-2}(1/k) == 0 (mod p^2), where k runs over the odd numbers up to p-2 (cf. Dickson, 1917, p. 187).

%C Clearly, if a(n) = 1, then p is a Wieferich prime, i.e., a term of A001220.

%H Felix Fröhlich, <a href="/A271234/b271234.txt">Table of n, a(n) for n = 1..10000</a>

%H L. E. Dickson, <a href="http://dx.doi.org/10.2307/2007234">Fermat's Last Theorem and the Origin and Nature of the Theory of Algebraic Numbers</a>, Annals of Mathematics, Second Series, Vol. 18, No. 4 (1917), 161-187.

%o (PARI) a(n) = my(p=prime(n)); lift(Mod(2, p^3)^(p-1))

%Y Cf. A001220, A196202.

%K nonn

%O 1,1

%A _Felix Fröhlich_, Apr 02 2016