1,1

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}.

Jianing Song, Table of n, a(n) for n = 1..10000

Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Wikipedia, Multiplicative group of integers modulo n.

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.

(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)))

(PARI) a(n) = my(v=select(x->znstar(x)[2]==[2*n, 2], invphi(4*n))); if(v!=[], vecmax(v), 0) \\ Jianing Song, Apr 08 2026 using Max Alekseyev's invphi.gp

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

nonn,changed

Jianing Song, Oct 14 2019

approved