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#23 | |
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Feb 2005
22×32×7 Posts |
Quote:
Note that 357913942 = 1 + (2^30 - 1)/3 seems to be related to ord(3,Mq) which equals (Mq - 1)/3 for q=31. |
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#24 |
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Feb 2004
France
22×229 Posts |
Hello,
I've updated the 4 files (.tex, .dvi, .ps, .pdf) listed in post#1 with the draft of proof from maxal and with the summary of the computations done by maxal and myself. See chapters 11 to 14. T. |
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#25 |
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Feb 2004
France
22×229 Posts |
Better.
Let say: DN(q) = number of distinct numbers of the form 3^n+1/3^n modulo Mq Mq=2^q-1 Code:
For q=3,5,7,17,19 we have: order(3,Mq) /DN(q) = 2 For q=13, 31 we have: (order(3,Mq)+2)/DN(q) = 2 T. |
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#26 | |
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Feb 2004
France
22·229 Posts |
Quote:
T. |
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#27 | |
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Feb 2005
22·32·7 Posts |
Quote:
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