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A294646 a(n) = (1/2)^(2*n) mod (2*n+1).

%I A294646 #21 Nov 07 2017 03:03:12
%S A294646 1,1,1,7,1,1,4,1,1,16,1,11,25,1,1,25,4,1,10,1,1,16,1,36,13,1,9,43,1,1,
%T A294646 16,61,1,52,1,1,64,60,1,79,1,16,22,1,64,70,44,1,70,1,1,16,1,1,28,1,59,
%U A294646 16,4,67,31,11,1,97,1,106,79,1,1,106,69,136,100,1,1,52,64,1,40,32,1,31,1,131,169
%N A294646 a(n) = (1/2)^(2*n) mod (2*n+1).
%C A294646 a(n) is the smallest k > 0 such that k*2^(2*n) == 1 (mod 2*n+1).
%C A294646 a(n)*A177023(n) == 1 (mod 2*n+1).
%C A294646 a(n)=1 iff 2*n+1 is in A015919.
%C A294646 1 <= a(n) <= 2*n, and is always coprime to 2*n+1.
%C A294646 Conjecture: a(n) is never 2 or 2*n or 2*n-2.
%C A294646 a(n) = 2*n-1 iff 2*n+1 is in A006521.
%H A294646 Robert Israel, <a href="/A294646/b294646.txt">Table of n, a(n) for n = 1..10000</a>
%e A294646 For n = 3, 2*n+1 = 7, (1/2)^6 == 4^6 == 1 (mod 7) so a(3)=1.
%p A294646 seq((1/2 mod (2*n+1)) &^(2*n) mod (2*n+1), n=1..200);
%o A294646 (PARI) a(n) = (1/2)^(2*n) % (2*n+1); \\ _Michel Marcus_, Nov 06 2017
%Y A294646 Cf. A006521, A015919, A177023.
%K A294646 nonn
%O A294646 1,4
%A A294646 _Robert Israel_ and _Thomas Ordowski_, Nov 05 2017