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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
7, 12, 13, 17, 19, 25, 28, 31, 34, 37, 38, 43, 47, 49, 52, 57, 59, 61, 62, 67, 71, 73, 76, 77, 79, 80, 84, 85, 91, 92, 93, 94, 97, 100, 101, 103, 104, 107, 108, 109, 112, 117, 118, 121, 122, 124, 127, 129, 133, 137, 139, 142, 143, 144, 148, 149, 151, 152, 157, 160, 161, 163, 164
OFFSET
1,1
COMMENTS
Indices of 0 in A328410, A328411 and A328412.
By definition, if there is no such m that psi(m) = 2k, psi = A002322, then m is a term of this sequence.
EXAMPLE
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.
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.
PROG
(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)[2])==r/2, return(0)); if(k==N+1, return(1)))
for(n=1, 200, if(isA328414(n), print1(n, ", ")))
CROSSREFS
Cf. A328410, A328411, A328412. Complement of A328413.
Sequence in context: A070420 A223423 A274334 * A083681 A178660 A048653
KEYWORD
nonn
AUTHOR
Jianing Song, Oct 14 2019
STATUS
approved

  NODES
orte 1
see 1
Story 1