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%I #12 Oct 18 2019 17:04:55
%S 12,30,42,32,66,90,0,102,114,150,138,0,0,174,198,128,0,270,0,246,294,
%T 230,282,306,0,318,324,0,354,450,0,256,414,0,426,438,0,0,474,374,498,
%U 522,0,534,594,470,0,582,0,750,618,0,642,810,726,678,0,590,0,738,0,0,762,512
%N Largest m such that (Z/mZ)* = C_2 X C_(2n), or 0 if no such m exists, 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 If (Z/mZ)* is isomorphic to C_2 X C_(2k) for some k, let x be any element in (Z/mZ)* such that the multiplicative order of x is 2k and that x != -1, then {-1, x} generates (Z/mZ)*. For example, (Z/16Z)* = {+-1, +-3, +-9, +-11}, (Z/32Z)* = {+-1, +-3, +-9, +-27, +-17, +-19, +-25, +-11}.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>
%e The solutions to (Z/mZ)* = C_2 X C_12 are m = 35, 39, 45, 52, 70, 78 and 90, the largest of which is 90, so a(6) = 90.
%o (PARI) a(n) = my(r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); forstep(k=N, r, -1, if(eulerphi(k)==r && lcm(znstar(k)[2])==r/2, return(k)); if(k==r, return(0)))
%Y Cf. A062373, A328410 (largest m).
%Y Cf. also A049823, A057635.
%K nonn
%O 1,1
%A _Jianing Song_, Oct 14 2019
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