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.

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

Offset: 0

Author

Jianing Song, Oct 14 2019

Keywords

• nonn
• hard
• more

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.

Examples

			(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.
		

Links

• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A328412.

Programs

PARI

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