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%I #8 Oct 18 2019 17:06:48
%S 7,4,1,5,2,10,21,6,42,90,150,30,78,210,2730,690,1050
%N Smallest k such that (Z/mZ)* = C_2 X C_(2k) has exactly n solutions for m, or 0 if no such k exists, where (Z/mZ)* is the multiplicative group of integers modulo m.
%C Conjecture: a(n) > 0 for all n. That is to say, every number occurs in A328412.
%C It seems that most terms are congruent to 2 modulo 4.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>
%e (Z/mZ)* = C_2 X C_42 has exactly 6 solutions m = 129, 147, 172, 196, 258, 294; for any k < 21, (Z/mZ)* = C_2 X C_(2k) has either fewer than or more than 6 solutions, so a(6) = 21.
%o (PARI) a(n) = my(k=1); while(A328412(k)!=n, k++); k \\ See A328412 for its program
%Y Cf. A328412.
%K nonn,hard,more
%O 0,1
%A _Jianing Song_, Oct 14 2019
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