A328410 Smallest 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.

8, 15, 21, 32, 33, 35, 0, 51, 57, 55, 69, 0, 0, 87, 77, 128, 0, 95, 0, 123, 129, 115, 141, 119, 0, 159, 324, 0, 177, 143, 0, 256, 161, 0, 213, 219, 0, 0, 237, 187, 249, 203, 0, 267, 209, 235, 0, 291, 0, 303, 309, 0, 321, 327, 253, 339, 0, 295, 0, 287, 0, 0, 381, 512, 393, 299, 0

Offset: 1

Author

Jianing Song, Oct 14 2019

Keywords

• nonn

Comments

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/15Z)* = {+-1, +-2, +-4, +-8}, (Z/21Z)* = {+-1, +-5, +-4, +-20, +-16, +-17}.

Examples

			The solutions to (Z/mZ)* = C_2 X C_6 are m = 21, 28, 36 and 42, the smallest of which is 21, so a(3) = 21.
		

Links

• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A062373, A328411 (largest m).

Programs

PARI

a(n) = my(r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); for(k=r+1, N+1, if(eulerphi(k)==r && lcm(znstar(k)[2])==r/2, return(k)); if(k==N+1, return(0)))