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 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. 2
 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 (list; graph; refs; listen; history; text; internal format)
 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 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))==r/2, return(k)); if(k==r, return(0))) CROSSREFS Cf. A062373, A328410 (largest m). Cf. also A049823, A057635. Sequence in context: A093507 A325802 A326019 * A145470 A108278 A298077 Adjacent sequences:  A328408 A328409 A328410 * A328412 A328413 A328414 KEYWORD nonn AUTHOR Jianing Song, Oct 14 2019 STATUS approved

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Last modified January 20 23:20 EST 2020. Contains 331104 sequences. (Running on oeis4.)