View Single Post
Old 2013-08-14, 17:22   #9
Thomas11
 
Thomas11's Avatar
 
Feb 2003

22·32·53 Posts
Default

Quote:
Originally Posted by Citrix View Post
Since these numbers take too long to LLR- I turned the problem around... and took k numbers less than 2^18. Then I split the k into base 2^5760 sequences.

k1=k*2^1
k2=k*2^2
... and so on.

I tested their weight using the above covering set (for 2^5760-1) & sieving to 1200

Only a few k remained with less than 20 candidates up to 1 million (corresponding to nash weight of 1/4). I have tested some of these to 2 Million with no prime.
I agree with you that while looking for small weights one should try to keep the k as small as possible to benefit from faster LLR testing.

However, I must admit that I don't understand the explanation of your approach.

You say you're taking k<2^18, which is just k<262144 - pretty small. But there are no really low weight Ks in the interval, only a few have Nash weights < 100, the lowest weight being 29 for k=138847 - about 100 times larger than the 1/4 you mentioned.

Then you write:
k1=k*2^1
k2=k*2^2
...
which leaves k essentially unchanged. It's just a cycling through the n. k1, k2, ... all have the same weight (in practise: more or less, since the n range is shifted to higher n due to the multiplier 2^i).

Then you mention the covering set for 2^5760-1.
Given the divisors of 5760 (= 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 32, 36, 40, 45, 48, 60, 64, 72, 80, 90, 96, 120, 128, 144, 160, 180, 192, 240, 288, 320, 360, 384, 480, 576, 640, 720, 960, 1152, 1440, 1920, 2880, 5760), this means a huge covering set, leading to k much larger than 64 bits or even 128 or 256 bits.
Or did you mean a covering set of length 5760 (=2^7 * 3^2 * 5)? This, on the other hand, would be too small.

Most likely I missed some essential part of your idea.

Perhaps, if you would post a few of the Ks generated by your procedure, I could find out what I'm missing...
Thomas11 is offline   Reply With Quote