%N Number of solutions to (Z/mZ)* = C_2 X C_(2n), where (Z/mZ)* is the multiplicative group of integers modulo m.
%C It is sufficient to check all numbers in the range [A049283(4n), A057635(4n)] for m if 4n is a totient number.
%C Conjecture: every number occurs in this sequence. That is to say, A328416(n) > 0 for every n.
%C Conjecture: this sequence is unbounded. That is to say, A328417 and A328418 are infinite.
%H Jianing Song, <a href="/A328412/b328412.txt">Table of n, a(n) for n = 1..10000</a>
%H Jianing Song, <a href="/A328412/a328412.txt">Solutions to (Z/mZ)* = C_2 X C_(2n), n <= 5000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>
%e See the a-file for the solutions to (Z/mZ)* = C_2 X C_(2n) for n <= 5000.
%o (PARI) a(n) = my(i=0, r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); for(k=r+1, N, if(eulerphi(k)==r && lcm(znstar(k))==r/2, i++)); i
%Y Cf. A328413 (numbers k such that a(k) > 0), A328414 (indices of 0), A328415 (indices of 1).
%Y Cf. A328416 (smallest k such that a(k) = n).
%Y Cf. A328417, A328418 (records in this sequence).
%Y Cf. also A049823, A057635.
%A _Jianing Song_, Oct 14 2019