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, 17490, 7590, 29610, 99330, 98670, 62790, 4830, 193830, 78540, 1411410, 1594320, 5962110, 690690, 3148530, 8592990, 14804790, 1381380, 21411390, 20281170, 26193090, 62607090, 57687630, 26246220, 36606570, 28318290, 11741730, 25571910, 55645590, 16546530

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..45.

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: A371851 A199157 A376845 * A372275 A373809 A010508

Adjacent sequences: A328413 A328414 A328415 * A328417 A328418 A328419

Keyword

nonn,hard

Author

Jianing Song, Oct 14 2019

Extensions

a(17)-a(45) from Amiram Eldar, Mar 17 2026

Status

approved