

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.


2



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
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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



