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%I #20 Dec 01 2014 00:22:48
%S 1,1,-2,-2,2,-3,-7,3,5,-11,-15,8,-18,-14,3,-12,19,-18,25,14,5,21,11,7,
%T -22,3,43,-40,-7,-53,54,23,11,-12,-7,41,6,-13,-66,71,-32,18,94,-20,
%U -79,7,-88,12,11,-73,3,29,-120,50,10,-60,-63,34,94,47,-113,131,-18,128,60,57,79,22,-45,-68,100,100,131,-171,56,-166,11,-153,-174,10
%N Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that L(2*n) == r (mod prime(n)), where L(k) denotes the Lucas number A000032(k).
%C Conjecture: a(n) is always nonzero, i.e., prime(n) never divides the Lucas number L(2*n).
%C We have verified this for all n = 1, ..., 2*10^6.
%C On Jul 26 2014, Bjorn Poonen (from MIT) found a counterexample with n = 14268177. - _Zhi-Wei Sun_, Jul 26 2014
%H Zhi-Wei Sun, <a href="/A245526/b245526.txt">Table of n, a(n) for n = 1..10000</a>
%e a(10) = -11 since L(2*10) = 15127 == -11 (mod prime(10)=29).
%t rMod[m_,n_]:=Mod[m,n,-(n-1)/2]
%t a[n_]:=rMod[LucasL[2n],Prime[n]]
%t Table[a[n],{n,1,80}]
%Y Cf. A000032, A000040, A245525.
%K sign
%O 1,3
%A _Zhi-Wei Sun_, Jul 25 2014
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