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| A328417
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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.
(history;
published version)
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#8 by N. J. A. Sloane at Fri Oct 18 17:07:01 EDT 2019
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#7 by Jianing Song at Fri Oct 18 10:30:39 EDT 2019
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#6 by Jianing Song at Fri Oct 18 10:10:51 EDT 2019
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#5 by Jianing Song at Fri Oct 18 10:09:50 EDT 2019
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| KEYWORD
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nonn,hard,more,changed
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#4 by Jianing Song at Fri Oct 18 10:09:29 EDT 2019
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| COMMENTS
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Conjecture: this sequence is infinite. That is to say, A328412 is unbounded.
It seems that a(n) == 2 (mod 4) for n > 1.
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| LINKS
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Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>
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| EXAMPLE
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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.
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#3 by Jianing Song at Tue Oct 15 01:43:02 EDT 2019
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| NAME
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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.
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#2 by Jianing Song at Mon Oct 14 23:39:28 EDT 2019
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| NAME
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allocatedNumbers 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 Jianingall Songk' < k.
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| DATA
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1, 2, 6, 30, 78, 210, 690, 1050, 4830
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| OFFSET
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1,2
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| PROG
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(PARI) my(t=0); for(k=1, 5000, if(A328412(k)>t, print1(k, ", "); t=A328412(k))) \\ See A328412 for its program
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| KEYWORD
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allocated
nonn,more
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| AUTHOR
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Jianing Song, Oct 14 2019
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| STATUS
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approved
editing
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#1 by Jianing Song at Mon Oct 14 23:02:05 EDT 2019
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| NAME
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allocated for Jianing Song
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| KEYWORD
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allocated
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| STATUS
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approved
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