

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



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


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 (p1)*p^k for any prime p, but (Z/mZ)* = C_2 X C_48 has solutions m = 119, 153, 238, 306.


LINKS

Table of n, a(n) for n=1..71.


EXAMPLE

(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.


PROG

(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, ", ")))


CROSSREFS

Cf. A328412. Complement of A328414.
Cf. also A114871.
Sequence in context: A220506 A045546 A050026 * A174788 A129143 A039225
Adjacent sequences: A328410 A328411 A328412 * A328414 A328415 A328416


KEYWORD

nonn


AUTHOR

Jianing Song, Oct 14 2019


STATUS

approved



