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

1,1

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}.

Links

Table of n, a(n) for n=1..67.

Wikipedia, Multiplicative group of integers modulo n

Example

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.

Prog

(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)))

Crossrefs

Cf. A062373, A328411 (largest m).

Sequence in context: A082867 A075713 A274290 * A089025 A088977 A070043

Adjacent sequences: A328407 A328408 A328409 * A328411 A328412 A328413

Keyword

nonn

Author

Jianing Song, Oct 14 2019

Status

approved