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 Borys Jaworski 20 49.8373 42 Derek Gordon 1 49.7454 43 Patrice Salah 1 49.7436 44 Honza Cholt 25 49.5324 45 Michael Goetz 4 49.5303 46 Grzegorz Granowski 70 49.3603 47 Ars Technica Team Prime Rib 1 49.2624 48 Ralf Terber 18 49.1469 49 Walter Darimont 2 49.1458 50 Randy Ready 13 49.1180 51 Andrew M Farrow 5 49.0549 52 Sean Humphries 7 49.0192 53 Denis Iakovlev 1 48.9974 54 Martin Vanc 1 48.9657 55 Senji Yamashita 28.3333 48.9570 56 Michael Herder 1 48.9335 57 Sturle Sunde 1 48.9324 58 Nayan Hajratwala 1 48.9067 59 Daniel Thonon 71 48.8993 60 Predrag Kurtovic 20 48.8789

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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)).