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 A328414 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has no solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.. 2

%I

%S 7,12,13,17,19,25,28,31,34,37,38,43,47,49,52,57,59,61,62,67,71,73,76,

%T 77,79,80,84,85,91,92,93,94,97,100,101,103,104,107,108,109,112,117,

%U 118,121,122,124,127,129,133,137,139,142,143,144,148,149,151,152,157,160,161,163,164

%N Numbers k such that (Z/mZ)* = C_2 X C_(2k) has no solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m..

%C Indices of 0 in A328410, A328411 and A328412.

%C By definition, if there is no such m that psi(m) = 2k, psi = A002322, then m is a term of this sequence.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>

%e 12 is a term: if there exists m such that (Z/mZ)* = C_2 X C_24 = C_2 X C_8 X C_3, then m must have a factor q such that q is an odd prime power and phi(q) = 8 or phi(q) = 24, phi = A000010, which is impossible.

%e 80 is a term: if there exists m such that (Z/mZ)* = C_2 X C_80 = C_2 X C_16 X C_5, then m must have a factor q such that q is an odd prime power and phi(q) = 80 or phi(q) = 16, which is impossible.

%o (PARI) isA328414(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))==r/2, return(0)); if(k==N+1, return(1)))

%o for(n=1, 200, if(isA328414(n), print1(n, ", ")))

%Y Cf. A328410, A328411, A328412. Complement of A328413.

%Y Cf. also A000010, A002322, A005277, A079695.

%K nonn

%O 1,1

%A _Jianing Song_, Oct 14 2019

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Last modified July 27 17:05 EDT 2021. Contains 346308 sequences. (Running on oeis4.)