

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
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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

Jianing Song, Table of n, a(n) for n = 1..10000
Jianing Song, Solutions to (Z/mZ)* = C_2 X C_(2n), n <= 5000
Wikipedia, Multiplicative group of integers modulo n


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. A328413 (numbers k such that a(k) > 0), A328414 (indices of 0), A328415 (indices of 1).
Cf. A328416 (smallest k such that a(k) = n).
Cf. A328417, A328418 (records in this sequence).
Cf. also A049823, A057635.
Sequence in context: A159778 A243278 A307059 * A079536 A058923 A107500
Adjacent sequences: A328409 A328410 A328411 * A328413 A328414 A328415


KEYWORD

nonn


AUTHOR

Jianing Song, Oct 14 2019


STATUS

approved



