View Single Post
Old 2008-12-17, 08:16   #1
gd_barnes's Avatar
May 2007
Kansas; USA

52·11·37 Posts
Default Generalizing algebraic factors on Riesel bases

This will be an important bit of information for any of you that work on new Riesel bases in the future:

After extensive analysis of the patterns of algebraic factors on Riesel bases < 100, I have concluded the following:

For all bases b where b == (4 mod 5), algebraic factors on even-n combine with a numeric factor on odd-n in the following scenario:

k=m^2 and m==(2 or 3 mod 5)

Factors to:

for even-n, let n=2q:
odd-n: factor of 5

In "English terms" , this states that for any base that leaves a remainder of 4 after dividing by 5 (i.e. 4, 9, 14, 19, etc.), any k that is a perfect square of an integer that leaves a remainder of 2 or 3 after dividing by 5, i.e. k=2^2, 3^2, 7^2, 8^2, etc. can be eliminated from testing on these bases because it is proven composite for all n.

Now that I've updated the pages to include many Riesel bases up to 125, you can see that k=4 and 9 are eliminated on several bases as a result of the above scenario.

You won't see these algebraic factors listed on the web pages for ALL bases that leave a remainder of 4 after dividing by 5. I only list the algebraic factors that are applicable for the base in question and generalized to the greatest extent. For instance on bases 4 and 9, ALL k's that are perfect squares can be removed so I show that instead. Bases that are b==(14 mod 15) have a conjecture of k=4 so algebraic factors that make a full covering set are not applicable.

Having this generalization available would have made Riesel base 24 much easier to begin with. With two different "kinds" of algebraic factors on that base as shown on the pages, having this first "kind" above readily available would have made the second kind (k=6*m^2) much easier to come up with.

A less common but equally important generalization will follow in the next post.


Last fiddled with by gd_barnes on 2008-12-17 at 08:31
gd_barnes is online now   Reply With Quote