

A328416


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.


1



7, 4, 1, 5, 2, 10, 21, 6, 42, 90, 150, 30, 78, 210, 2730, 690, 1050
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OFFSET

0,1


COMMENTS

Conjecture: a(n) > 0 for all n. That is to say, every number occurs in A328412.
It seems that most terms are congruent to 2 modulo 4.


LINKS

Table of n, a(n) for n=0..16.
Wikipedia, Multiplicative group of integers modulo n


EXAMPLE

(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.


PROG

(PARI) a(n) = my(k=1); while(A328412(k)!=n, k++); k \\ See A328412 for its program


CROSSREFS

Cf. A328412.
Sequence in context: A225463 A010138 A199157 * A010508 A070403 A144468
Adjacent sequences: A328413 A328414 A328415 * A328417 A328418 A328419


KEYWORD

nonn,hard,more


AUTHOR

Jianing Song, Oct 14 2019


STATUS

approved



