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 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 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified June 23 11:24 EDT 2021. Contains 345397 sequences. (Running on oeis4.)