A328417 Numbers k such that A328412(k) sets a new record; numbers k such that (Z/mZ)* = C_2 X C_(2k) has more solutions for m than all k' < k, where (Z/mZ)* is the multiplicative group of integers modulo m.

1, 2, 6, 30, 78, 210, 690, 1050, 4830

Offset: 1

Author

Jianing Song, Oct 14 2019

Keywords

• nonn
• hard
• more

Comments

Conjecture: this sequence is infinite. That is to say, A328412 is unbounded.

It seems that a(n) == 2 (mod 4) for n > 1.

Examples

			For k = 30: (Z/mZ)* = C_2 X C_60 has 11 solutions, namely m = 143, 155, 175, 183, 225, 244, 286, 310, 350, 366, 450; for all k' < 30, (Z/mZ)* = C_2 X C_(2k') has fewer than 11 solutions. So 30 is a term.
		

Links

• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A328412, A328418.

Programs

PARI

my(t=0); for(k=1, 5000, if(A328412(k)>t, print1(k, ", "); t=A328412(k))) \\ See A328412 for its program