Number of finite subgroups of GO3.

[John Conway]

   The number of groups of order n , counted up to conjugacy, is given by:

n = 1 2 3 4 5 6 7 8 9 ... 
    1 3 1 5 1 5 1 7 1 5 1 8 1 5 1 7 1 5 1 7 1 5 1 10 1 5 1 7 1 5 1 7 1 5 1

namely it has the period 1 5 1 7  except that

   g(2) = 3, g(4) = 5, g(12) = 8, g(24) = 10, g(48) = g(60) = g(120) = 8.

   Neil - did I give you this one already?

   In SO3 ge get a similar sequence - let me work it out - the groups are

  nn,  of order n
 22n, of order 2n
 332,432,532 of orders 12,24,60

but we have the coincidence  22 = 221 indicated below:

  11  22  33  44  55  66
     =221    222     223

so the sequence is

 1 1 1 2 1 2 1 2 1 2 1 3 1 2 1 2 1 2 1 2 1 2 1 3 1 2 1 ...

the only other exception to the  1 2  period being  f(60) = 3.

  Regards,  JHC