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%I #9 Jul 02 2021 03:49:25
%S 1,3,3,3,4,3,3,15,3,3,8,3,6,19,3,3,12,35,3,39,3,3,12,3,8,51,3,55,20,3,
%T 3,49,8,3,24,3,3,63,24,3,28,3,27,87,3,15,32,95,3,77,3,3,16,3,3,111,3,
%U 115,28,119,12,123,51,3,44,3,8,95,3,3,48,143,16,129
%N Number of distinct residues of x^n (mod 2n+1), x=0..2n.
%C a(n) = 3 if 2n+1 prime because the corresponding residues are 0, 1 and 2n (mod 2n+1).
%H Amiram Eldar, <a href="/A198020/b198020.txt">Table of n, a(n) for n = 0..10000</a>
%e a(7) = 15 because x^7 == 0, 1, …,14 (mod 15) => 15 distinct residues.
%t Table[Length[Union[PowerMod[Range[0, 2*n], n, 2*n+1]]], {n,0, 100}]
%Y Cf. A195637, A197929.
%K nonn
%O 0,2
%A _Michel Lagneau_, Oct 20 2011
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