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Old 2020-12-25, 16:04   #2
rogue's Avatar
Apr 2003
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Originally Posted by MattcAnderson View Post
A Fermat number is F(m) = 2^(2^m) + 1.
It has been shown that any factor of a Fermat number
has the form

k*2^n + 1.
with n greater than or equal to m+2
This information is at

if k is not odd then we have an equivalent representation

2*k*2^c + 1 = k*2^(c+1) + 1

I assume that mmff and other Fermat search programs
only search for odd k because the even cases will
be searched in increased exponent

Also, in the log of known Fermat factors,
All the k values are odd.

AFAIK, they eliminate even k. I know that gfndsieve does so I assume the others do as well.

Last fiddled with by rogue on 2020-12-25 at 16:04
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