

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




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.


LINKS

Table of n, a(n) for n=1..9.
Wikipedia, Multiplicative group of integers modulo n


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

Cf. A328412, A328418.
Sequence in context: A211889 A174276 A117849 * A290760 A088857 A099081
Adjacent sequences: A328414 A328415 A328416 * A328418 A328419 A328420


KEYWORD

nonn,hard,more


AUTHOR

Jianing Song, Oct 14 2019


STATUS

approved



