%I A066947
%S A066947 3,13,27,31,55,57,175,109,127,133,391,183,231,447,607,307,439,381,895,
%T A066947 811,535,553,2463,751,735,973,1623,871,1791,993,2335,1875,1231,1855,
%U A066947 3079,1407,1527,2575,5631,1723,3247,1893,3751,3519,2215,2257,8511,2745
%N A066947 Number of elements of order 2 in GL(2,Z_n).
%H A066947 Alec Mihailovs, <a href="http://www.shepherd.edu/mathweb/solution16.html">
               Problem 16 Solution</a>
%H A066947 Alec Mihailovs, <a href="http://webpages.shepherd.edu/amihailo/AbstractAlgebra.html">
               Abstract Algebra with Maple</a>
%H A066947 Alec Mihailovs, <a href="http://www.mapleapps.com/powertools/abstractalgebra/
               html/Chapter5.html">Chapter 5. Cyclic Groups</a>
%F A066947 If n = 2^m*p^a...q^b where p, ..., q are the odd prime divisors of n, 
               then a(n)=c(m)*(p^{2a}+p^{2a-1}+2)...(q^{2b}+q^{2b-1}+2) - 1 where 
               c(0) = 1, c(1) = 4, c(2) = 28 and c(m) = 9*4^{m-1}+ 32 for m > 2. 
               The integer function f(n) = a(n)+1 is multiplicative, i.e. f(m*n)=f(m)*f(n) 
               for coprime m and n. - Alec Mihailovs (alec(AT)mihailovs.com), Mar 
               24 2002
%e A066947 E.g. a(3000) = (a(8)+1)(a(3)+1)(a(125)+1)-1=(9*4^2+2)*(3^2+3+2)*(5^6+5^5+2)-1 
               = 46204927 because 3000=2^3*3*5^3.
%p A066947 Ord2inGL2 := proc(n::posint) local i,j,m,c; if n=1 then return 0 end 
               if; m := ifactors(n)[2]; c := 1; j := 1; if (m[1,1]=2) then j := 
               2; if m[1,2]=1 then c := 4 elif m[1,2]=2 then c := 28 else c := 9*4^(m[1,
               2]-1)+32 end if end if; c := c*mul((m[i,1]+1)*m[i,1]^(2*m[i,2]-1)+2,
               i=j..nops(m))-1 end;
%Y A066947 Sequence in context: A120074 A056706 A052454 this_sequence A031011 A099062 
               A002304
%Y A066947 Adjacent sequences: A066944 A066945 A066946 this_sequence A066948 A066949 
               A066950
%K A066947 nice,easy,nonn
%O A066947 2,1
%A A066947 Alec Mihailovs (alec(AT)mihailovs.com), Jan 24 2002 and Mar 24, 2002
%E A066947 More terms from Alec Mihailovs (alec(AT)mihailovs.com), Mar 24 2002