20211107, 02:18  #12  
Sep 2002
Database er0rr
2×19×113 Posts 
Quote:
Enjoy the latest incarnation found in post #1. 

20211120, 14:01  #14 
Sep 2002
Database er0rr
10C6_{16} Posts 
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. 
20220107, 22:32  #15 
Sep 2002
Database er0rr
2·19·113 Posts 
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 wishywashy paragraph about segmentation. Last fiddled with by paulunderwood on 20220108 at 12:38 
20220115, 16:57  #16 
Sep 2002
Database er0rr
4294_{10} Posts 
I can now clarify. Take the example n=2499327041 with 30258 P <= (n1)/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 (n1)/2. Maybe this is not the correct reasoning
Last fiddled with by paulunderwood on 20220115 at 17:58 
20220219, 13:20  #17 
Sep 2002
Database er0rr
10C6_{16} Posts 
I have collected all data for 10e10 for a "linear choice" for the parameter P in x^2P*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. Last fiddled with by paulunderwood on 20220219 at 15:52 
20220220, 07:35  #18 
Sep 2002
Database er0rr
1000011000110_{2} Posts 
Here is a more granular set of statistics for various digit lengths:
Code:
? 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^(d1)<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] Last fiddled with by paulunderwood on 20220220 at 08:04 
20220220, 15:29  #19 
Sep 2002
Database er0rr
2·19·113 Posts 
Gross over estimation in previous post
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: Code:
[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] Code:
[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] 
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