A328412 - OEIS

A328412

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

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 a-file 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