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A328412 Number of solutions to (Z/mZ)* = C_2 X C_(2n), where (Z/mZ)* is the multiplicative group of integers modulo m. 7


%S 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,

%T 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,

%U 7,4,0,4,0,3,3,0,0,12,0,0,4,2,3,0,0,3,4,2,3,9,0,0

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

%C It is sufficient to check all numbers in the range [A049283(4n), A057635(4n)] for m if 4n is a totient number.

%C Conjecture: every number occurs in this sequence. That is to say, A328416(n) > 0 for every n.

%C Conjecture: this sequence is unbounded. That is to say, A328417 and A328418 are infinite.

%H Jianing Song, <a href="/A328412/b328412.txt">Table of n, a(n) for n = 1..10000</a>

%H Jianing Song, <a href="/A328412/a328412.txt">Solutions to (Z/mZ)* = C_2 X C_(2n), n <= 5000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>

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

%o (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

%Y Cf. A328413 (numbers k such that a(k) > 0), A328414 (indices of 0), A328415 (indices of 1).

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

%Y Cf. A328417, A328418 (records in this sequence).

%Y Cf. also A049823, A057635.

%K nonn

%O 1,1

%A _Jianing Song_, Oct 14 2019

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Last modified June 18 04:41 EDT 2021. Contains 345098 sequences. (Running on oeis4.)