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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. 2

%I #6 Oct 18 2019 17:06:09

%S 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,

%T 35,36,39,40,41,42,44,45,46,48,50,51,53,54,55,56,58,60,63,64,65,66,68,

%U 69,70,72,74,75,78,81,82,83,86,87,88,89,90,95,96,98,99,102,105,106,110,111

%N 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.

%C 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.

%e (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.

%o (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)))

%o for(n=1, 100, if(isA328413(n), print1(n, ", ")))

%Y Cf. A328412. Complement of A328414.

%Y Cf. also A114871.

%K nonn

%O 1,2

%A _Jianing Song_, Oct 14 2019

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Last modified March 28 18:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)