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

1,1

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}.

Links

Table of n, a(n) for n=1..64.

Wikipedia, Multiplicative group of integers modulo n

Example

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.

Prog

(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)))

Crossrefs

Cf. A062373, A328410 (largest m).

Cf. also A049283, A057635.

Sequence in context: A325802 A326019 A334804 * A333840 A145470 A108278

Adjacent sequences: A328408 A328409 A328410 * A328412 A328413 A328414

Keyword

nonn

Author

Jianing Song, Oct 14 2019

Status

approved