%I #8 Oct 18 2019 17:07:01

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

%N 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.

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

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

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>

%e 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.

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

%Y Cf. A328412, A328418.

%K nonn,hard,more

%O 1,2

%A _Jianing Song_, Oct 14 2019