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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.
2
1, 2, 6, 30, 78, 210, 690, 1050, 4830
OFFSET
1,2
COMMENTS
Conjecture: this sequence is infinite. That is to say, A328412 is unbounded.
It seems that a(n) == 2 (mod 4) for n > 1.
EXAMPLE
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.
PROG
(PARI) my(t=0); for(k=1, 5000, if(A328412(k)>t, print1(k, ", "); t=A328412(k))) \\ See A328412 for its program
CROSSREFS
Sequence in context: A211889 A174276 A117849 * A290760 A088857 A099081
KEYWORD
nonn,hard,more
AUTHOR
Jianing Song, Oct 14 2019
STATUS
approved