A049823 -id:A049823 - cp's OEIS frontend

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

Offset: 1

Jianing Song, Oct 14 2019

nonn

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.

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

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

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.

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

cp's OEIS Frontend

A328412 Number of solutions to (Z/mZ)* = C_2 X C_(2n), where (Z/mZ)* is the multiplicative group of integers modulo m.

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