A Restricted Domain Lucas Probable Prime Test paper

[paulunderwood]

Originally Posted by Nick

This is not my area (as you know!) but I would say broadly speaking that you have 2 paths forward: either a mathematical proof that your method performs better or, alternatively, using formal statistical methods to show that the testing you have done is sufficient to be significant.

I have gone for a mixture of what you wrote. The outlandish claims are back! The paper has undergone a major rewrite.

Enjoy the latest incarnation found in post #1.

[paulunderwood]

Originally Posted by paulunderwood

Enjoy the latest incarnation found in post #1.

A new copy has been uploaded with more statistics. I am waiting for the final data to come in. ETA: a few weeks

[paulunderwood]

The row of data for 9 digits has been added. The data for 10 digits will take another month of so.

I have tried to make the English simpler. So it is worth downloading the latest copy from post #1. Please enjoy the read -- it is less that 3 pages long -- and let me know about any improvements that could be made.

[paulunderwood]

I am still gathering data, but post #1 has been updated with the latest paper. The idea of segmenting P is introduced with the idea of unlikely geometric progression of passes of the test. I also offer £100 for a composite that passes for any "r".

EDIT: I have removed the wishy-washy paragraph about segmentation.

[paulunderwood]

Originally Posted by paulunderwood

EDIT: I have removed the wishy-washy paragraph about segmentation.

I can now clarify. Take the example n=2499327041 with 30258 P <= (n-1)/2 values that give rise to counterexamples. The multiplicative order of 2 is 560 meaning a single 2^r solution would give rise to 2231542 solutions in total, as r goes up to (n-1)/2. Maybe this is not the correct reasoning

[paulunderwood]

I have collected all data for 10e10 for a "linear choice" for the parameter P in x^2-P*x+2, which are attached to post #1.

There might be one more revision to the paper with better wording and a more granular calculation of "expectation" from the data.

Feel free to download the data and draw your own conclusion -- maybe post that here.

[paulunderwood]

Here is a more granular set of statistics for various digit lengths:

? allocatemem(100000000)                          
  ***   Warning: new stack size = 100000000 (95.367 Mbytes).     
? V=readvec("data_10e10.txt");                          
? for(d=5,10,c=0;ac=0.0;for(v=1,#V,[n,P]=V[v];if(10^(d-1)<n&&n<10^d,z=znorder(Mod(2,n));ac=ac+z/n/2;c++));print([d,c,ac]))      
[5, 26, 0.0085802083661410656672684116520217542767]              
[6, 98, 0.0049638320851750657858193213094107895834]
[7, 314, 0.0058614515164310000251931880685920512458]
[8, 1608, 0.0070816941908042819767139890946524717003]
[9, 15072, 0.030554972381921342409809684878746893179]
[10, 101630, 0.021620787552043991286872665384670317646]

Extrapolating the expectations, a pseudoprime to the basic test -- without the extra factor of 0.16 -- might be around 120 digits. So a non-minimal "r" pseudoprime for the test RDPRP would be about 700 digits

[paulunderwood]

I used ac=ac+z/n/2 when it should have been ac=ac+2*z/n and for RDPRP ac=ac+0.16*2*z/n.

Along with accumulated figures:

[5, 26, 0.03432083, 0.03432083]                                  
[6, 98, 0.01985533, 0.05417616]
[7, 314, 0.02344581, 0.07762197]
[8, 1608, 0.02832678, 0.1059487]
[9, 15072, 0.1222199, 0.2281686]
[10, 101630, 0.08648315, 0.3146518]

For RDPRP:

[5, 26, 0.005491333, 0.005491333]                                
[6, 98, 0.003176853, 0.008668186]
[7, 314, 0.003751329, 0.01241951]
[8, 1608, 0.004532284, 0.01695180]
[9, 15072, 0.01955518, 0.03650698]
[10, 101630, 0.01383730, 0.05034429]

So pseudoprimes can be expected at about 14 and 19 digits (resp.). The prize money is within reach!