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 Peter Kaiser 85.3333 51.0002
42 Detlef Lexut 17 50.9913
43 Marius Vultur 32 50.9627
44 Michael Cameron 1 50.9234
45 Wes Hewitt 23 50.8920
46 Jean-Luc Garambois 4 50.8503
47 Konstantin Agafonov 1 50.8197
48 Thomas Ritschel 55 50.7756
49 Zack Friedrichsen 32 50.6902
50 Predrag Kurtovic 33 50.6893
51 Jason Biggs 2 50.6879
52 Zimai Wu 1 50.6754
53 Murray Sondergard 1 50.6521
54 Ken Glennie 2 50.6508
55 Heinrich Podsada 1 50.6352
56 Honza Cholt 25 50.5984
57 Seonghwan Kim 43 50.5965
58 Ivica Kelava 2 50.5849
59 Brian R Kaczala 2 50.5776
60 Kyle Gingrich 11 50.5579

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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.