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%I #18 Aug 13 2015 21:20:56
%S 4,2,6,4,2,2,2,2,4,2,2,6,2,2,2,4,2,2,2,2,2,2,2,2,4,2,6,2,2,2,2,2,2,2,
%T 2,4,2,2,2,2,2,2,2,2,2,2,2,6,4,2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2,2,
%U 2,2,2,2,2,2,6,2,2,2,2,2,4,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2,2,2
%N Number of units in the imaginary quadratic field Q(sqrt(-n)).
%C Sequence of n such that a(n)=2 gives A092206; a(n)=4 gives A000290; a(n)=6 gives A033428. - _Marc LeBrun_, Apr 12 2006
%H Nathaniel Johnston, <a href="/A092205/b092205.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Unit.html">Unit</a>
%F a(A005117(n)) = A236213(n). - _Jonathan Sondow_, Jan 29 2014
%e For n=1, the units are +/-1, +/-i, so a(1) = 4.
%e For n=3, the units are +/-1, +/-w, +/-w^2, where w is a cube root of unity, so a(3) = 6. [Corrected by _Jonathan Sondow_, Jan 29 2014]
%p A092205 := proc(n) if(type(sqrt(n),integer))then return 4: elif(n mod 3 = 0 and type(sqrt(n/3),integer))then return 6: else return 2: fi: end: seq(A092205(n),n=1..105); # _Nathaniel Johnston_, Jun 26 2011
%t a[n_] := Which[ IntegerQ[ Sqrt[n] ], 4, Mod[n, 3] == 0 && IntegerQ[ Sqrt[n/3] ], 6, True, 2]; Table[a[n], {n, 1, 105}] (* _Jean-François Alcover_, Oct 30 2012, after _Nathaniel Johnston_ *)
%o (PARI) a(n)=if(issquare(n),return(4));if(n%3==0&&issquare(n/3),6,2) \\ _Charles R Greathouse IV_, Oct 30 2012
%Y Cf. A092206, A000290, A033482, A236213.
%K nonn,easy
%O 1,1
%A _Eric W. Weisstein_, Feb 24 2004
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