Top person sorted by 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.

rank person primes score 41 Willie Black 12 51.2226 42 Alen Kecic 21 51.1137 43 Erik Veit 54 51.0467 44 Peter Kaiser 84.3333 50.9960 45 Jann Kickler 10 50.9923 46 Detlef Lexut 17 50.9913 47 Marius Vultur 32 50.9627 48 Michael Cameron 1 50.9234 49 Bruce Marler 11 50.9123 50 Wes Hewitt 24 50.9116 51 Predrag Kurtovic 34 50.8653 52 Dmitry Domanov 22 50.8543 53 Jean-Luc Garambois 4 50.8503 54 Konstantin Agafonov 1 50.8197 55 Thomas Ritschel 45 50.7170 56 Zack Friedrichsen 32 50.6902 57 Jason Biggs 2 50.6879 58 Kellen Shenton 18 50.6779 59 Murray Sondergard 2 50.6771 60 Zimai Wu 1 50.6754

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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)).