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Old 2006-05-12, 08:17   #23
maxal
 
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Quote:
Originally Posted by T.Rex
Have you run it with q=31 ?
I've written a C++ program for that. And it has reported 357913942 distinct numbers of the form 3^n + 3^(-n) mod Mq.
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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Old 2006-05-13, 14:03   #24
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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.
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Old 2006-05-13, 19:15   #25
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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
q=13 and q=31 are special cases.
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Old 2006-05-14, 12:19   #26
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Quote:
Originally Posted by maxal
The following pari/gp program computes \eta(3,M_q) the number of distinct numbers 3^n+1/3^n modulo Mq for q around 20 in a matter of seconds.
Code:
{ t(q) = local(M);
M=2^q-1;
length(Set(vector(2^(q-1),n,lift(Mod(3,M)^n+Mod(3,M)^(-n)))))
}
Hum. ZetaX from AOPS forum has proven that 2 \eta(3,M_q) = ord(3,M_q)+2, although this PARI/gp program misses 1 element for q=3,5,7,17,19. So maybe PARI/gp has a bug. (I've checked by hand that eta(3,M_5)=16 while PARI/gp says 15).
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Old 2006-05-14, 14:50   #27
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Quote:
Originally Posted by T.Rex
So maybe PARI/gp has a bug. (I've checked by hand that eta(3,M_5)=16 while PARI/gp says 15).
That's because we count elements for n=1..2^(q-1) while the missing one has n=0.
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