A328413 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 14, 15, 16, 18, 20, 21, 22, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 39, 40, 41, 42, 44, 45, 46, 48, 50, 51, 53, 54, 55, 56, 58, 60, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 78, 81, 82, 83, 86, 87, 88, 89, 90, 95, 96, 98, 99, 102, 105, 106, 110, 111

Offset: 1

Author

Jianing Song, Oct 14 2019

Keywords

• nonn

Comments

For n > 1, it is easy to see A114871(n)/2 is a term of this sequence. The smallest term here not of the form A114871(k)/2 is 24: 48 is not of the form (p-1)*p^k for any prime p, but (Z/mZ)* = C_2 X C_48 has solutions m = 119, 153, 238, 306.

Examples

			(Z/mZ)* = C_2 X C_2 has solutions m = 8, 12; (Z/mZ)* = C_2 X C_4 has solutions m = 15, 16, 20, 30; (Z/mZ)* = C_2 X C_6 has solutions m = 21, 28, 36, 42; (Z/mZ)* = C_2 X C_8 has solutions m = 32; (Z/mZ)* = C_2 X C_10 has solutions m = 33, 44, 66; (Z/mZ)* = C_2 X C_12 has solutions m = 35, 39, 45, 52, 70, 78, 90. So 1, 2, 3, 4, 5, 6 are all terms.
		

Crossrefs

Cf. A328412. Complement of A328414.

Cf. also A114871.

Programs

PARI

isA328413(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(1)); if(k==N+1, return(0)))
for(n=1, 100, if(isA328413(n), print1(n, ", ")))