Top person sorted by score

The Prover-Account Top 20
Persons by: number score normalized score
Programs by: number score normalized score
Projects by: number score normalized score

At this site we keep several lists of primes, most notably the list of the 5,000 largest known primes. Who found the most of these record primes? We keep separate counts for persons, projects and programs. To see these lists click on 'number' to the right.

Clearly one 100,000,000 digit prime is much harder to discover than quite a few 100,000 digit primes. Based on the usual estimates we score the top persons, provers and projects by adding ‎(log n)3 log log n‎ for each of their primes n. Click on 'score' to see these lists.

Finally, to make sense of the score values, we normalize them by dividing by the current score of the 5000th prime. See these by clicking on 'normalized score' in the table on the right.

rankpersonprimesscore
41 Detlef Lexut 17 50.9770
42 Michael Cameron 1 50.9234
43 Wes Hewitt 23 50.8920
44 Konstantin Agafonov 1 50.8197
45 Kai Presler 31 50.8129
46 Thomas Ritschel 60 50.7984
47 Predrag Kurtovic 35 50.6987
48 Jason Biggs 1 50.6636
49 Murray Sondergard 1 50.6521
50 Ken Glennie 2 50.6508
51 Heinrich Podsada 1 50.6352
52 Florian Piesker 110 50.6141
53 Seonghwan Kim 45 50.6074
54 Honza Cholt 26 50.6039
55 Marius Vultur 31 50.5867
56 Ivica Kelava 2 50.5849
57 Brian R Kaczala 2 50.5776
58 Michael Schulz 1 50.5434
59 Ed Goforth 8 50.5421
60 Kyle Gingrich 10 50.5215

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Notes:


Score for Primes

To find the score for a person, program or project's primes, we give each prime n the score (log n)3 log log n; and then find the sum of the scores of their primes. For persons (and for projects), if three go together to find the prime, each gets one-third of the score. Finally we take the log of the resulting sum to narrow the range of the resulting scores. (Throughout this page log is the natural logarithm.)

How did we settle on (log n)3 log log n? For most of the primes on the list the primality testing algorithms take roughly O(log(n)) steps where the steps each take a set number of multiplications. FFT multiplications take about

O( log n . log log n . log log log n )

operations. However, for practical purposes the O(log log log n) is a constant for this range number (it is the precision of numbers used during the FFT, 64 bits suffices for numbers under about 2,000,000 digits).

Next, by the prime number theorem, the number of integers we must test before finding a prime the size of n is O(log n) (only the constant is effected by prescreening using trial division).  So to get a rough estimate of the amount of time to find a prime the size of n, we just multiply these together and we get

O( (log n)3 log log n ).

Finally, for convenience when we add these scores, we take the log of the result.  This is because log n is roughly 2.3 times the number of digits in the prime n, so (log n)3 is quite large for many of the primes on the list. (The number of decimal digits in n is floor((log n)/(log 10)+1)).

Printed from the PrimePages <t5k.org> © Reginald McLean.