A328411 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.

12, 30, 42, 32, 66, 90, 0, 102, 114, 150, 138, 0, 0, 174, 198, 128, 0, 270, 0, 246, 294, 230, 282, 306, 0, 318, 324, 0, 354, 450, 0, 256, 414, 0, 426, 438, 0, 0, 474, 374, 498, 522, 0, 534, 594, 470, 0, 582, 0, 750, 618, 0, 642, 810, 726, 678, 0, 590, 0, 738, 0, 0, 762, 512

Offset: 1

Author

Jianing Song, Oct 14 2019

Keywords

• nonn

Comments

It is sufficient to check all numbers in the range [A049283(4n), A057635(4n)] for m if 4n is a totient number.

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}.

Examples

			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.
		

Links

• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A062373, A328410 (largest m).

Cf. also A049283, A057635.

Programs

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)))