

A328412


Number of solutions to (Z/mZ)* = C_2 X C_(2n), where (Z/mZ)* is the multiplicative group of integers modulo m.


7



2, 4, 4, 1, 3, 7, 0, 4, 4, 5, 3, 0, 0, 3, 7, 1, 0, 7, 0, 3, 6, 2, 3, 4, 0, 3, 1, 0, 3, 11, 0, 1, 7, 0, 3, 3, 0, 0, 3, 2, 3, 8, 0, 3, 4, 2, 0, 3, 0, 6, 3, 0, 3, 5, 5, 3, 0, 2, 0, 4, 0, 0, 3, 1, 3, 4, 0, 3, 7, 4, 0, 4, 0, 3, 3, 0, 0, 12, 0, 0, 4, 2, 3, 0, 0, 3, 4, 2, 3, 9, 0, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

It is sufficient to check all numbers in the range [A049283(4n), A057635(4n)] for m if 4n is a totient number.
Conjecture: every number occurs in this sequence. That is to say, A328416(n) > 0 for every n.
Conjecture: this sequence is unbounded. That is to say, A328417 and A328418 are infinite.


LINKS



EXAMPLE

See the afile for the solutions to (Z/mZ)* = C_2 X C_(2n) for n <= 5000.


PROG

(PARI) a(n) = my(i=0, r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); for(k=r+1, N, if(eulerphi(k)==r && lcm(znstar(k)[2])==r/2, i++)); i


CROSSREFS

Cf. A328416 (smallest k such that a(k) = n).


KEYWORD

nonn


AUTHOR



STATUS

approved



