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 A092205 Number of units in the imaginary quadratic field Q(sqrt(-n)). 5
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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 LINKS Nathaniel Johnston, Table of n, a(n) for n = 1..10000 Eric Weisstein's World of Mathematics, Unit FORMULA a(A005117(n)) = A236213(n). - Jonathan Sondow, Jan 29 2014 EXAMPLE For n=1, the units are +/-1, +/-i, so a(1) = 4. 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] MAPLE 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 MATHEMATICA 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 *) PROG (PARI) a(n)=if(issquare(n), return(4)); if(n%3==0&&issquare(n/3), 6, 2) \\ Charles R Greathouse IV, Oct 30 2012 CROSSREFS Cf. A092206, A000290, A033482, A236213. Sequence in context: A273667 A187109 A242599 * A272101 A059853 A136527 Adjacent sequences:  A092202 A092203 A092204 * A092206 A092207 A092208 KEYWORD nonn,easy AUTHOR Eric W. Weisstein, Feb 24 2004 STATUS approved

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Last modified May 31 02:51 EDT 2020. Contains 334747 sequences. (Running on oeis4.)