20070620, 01:38  #23 
"Jason Goatcher"
Mar 2005
6663_{8} Posts 
Please think of any of the following that are between parentheses as "subscripted."
F(0),F(1),F(2),F(3), and F(4) are all prime F(n) for n=5 to 32 have a known factor f(33) is the lowest one which is an unknown Status unknown for F(n) with nvalues of 3335, 4041,4447,4950 
20070620, 07:44  #25  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Quote:
Thank you ATH for updating my data. Well these types of records are like grave stones. At least one is added every day and they with time disappear into oblivion ! Lets go in for more lasting things to disagree with! BTW: I appreciate your trouble and in no way am disagreeing with you Mally 

20070620, 13:53  #26 
Tribal Bullet
Oct 2004
DE3_{16} Posts 
The trouble is not representing Fn, it is performing arithmetic modulo Fn. The things you're doing arithmetic on are random bitstrings that have the same number of bits as Fn (it's just like LL tests, Prime95 doesn't bother trying to compress the numbers it works with because it wouldn't do any good).

20070620, 17:21  #27 
∂^{2}ω=0
Sep 2002
República de California
5·2,351 Posts 
...except in the base2 PRP case I mentioned above  there, the bitstrings that occur during the repeated squareandmodding are not random. But it's alas not useful for rigorously establishing primality, it only tells you which of the numbers of the general form 2^{n}+1 *could* be prime.

20070620, 18:16  #28  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,877 Posts 
Quote:
number of bits) x mod 2^n + 1 > Put x = A*2^n + B then x mod 2^n+1 = A*2^n + B + A  A mod 2^n+1 = A*(2^n+1) + BA mod 2^n + 1 = BA mod 2^n + 1 

20070620, 18:55  #29  
Feb 2007
2^{4}·3^{3} Posts 
thanks for pleading in favour of my idea (indirectly...)
Quote:
A less trivial question : what are the nonzero bits of 3^k or 5^k mod Fn? are they really random ? if so for 3 and 5, maybe not for some other admissible "seed" (as TRex would certainly call it ;) ? 

20070620, 18:59  #30 
∂^{2}ω=0
Sep 2002
República de California
5×2,351 Posts 
I'm pretty certain Jason P was referring to the *squaring* of a lengthF_n random bitstring and/or the accompanying implicit DWTbased mod, not the trivial mod you refer to.
Last fiddled with by ewmayer on 20070620 at 19:00 
20070620, 19:54  #31  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,877 Posts 
Quote:
"The trouble is not representing Fn, it is performing arithmetic modulo Fn" This is pretty clear. He said nothing about squaring a random bitstring. You are referring to the difficulty of doing a multiplication, A*B. Once one has the product doing the reduction is easy. The difficulty is getting the product. 

20070620, 20:02  #32  
∂^{2}ω=0
Sep 2002
República de California
11755_{10} Posts 
Quote:
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20070620, 20:32  #33  
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Quote:
Alex Last fiddled with by akruppa on 20070620 at 20:34 

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