Search: id:A066947 Results 1-1 of 1 results found. %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, Problem 16 Solution %H A066947 Alec Mihailovs, Abstract Algebra with Maple %H A066947 Alec Mihailovs, Chapter 5. Cyclic Groups %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 Search completed in 0.001 seconds