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 Tim McArdle 1 49.9091 42 Michael Curtis 47 49.9069 43 Peyton Hayslette 1 49.8982 44 Borys Jaworski 19 49.8344 45 Derek Gordon 1 49.7454 46 Patrice Salah 1 49.7436 47 Honza Cholt 33 49.7157 48 Michael Goetz 4 49.5303 49 Ken Ito 13 49.4907 50 Grzegorz Granowski 69 49.3556 51 Ars Technica Team Prime Rib 1 49.2624 52 Ralf Terber 18 49.1632 53 Walter Darimont 2 49.1458 54 Randy Ready 13 49.1180 55 Daniel Thonon 81 49.1120 56 Predrag Kurtovic 24 49.0525 57 Andrew M Farrow 4 49.0483 58 Sean Humphries 7 49.0192 59 Vaughan Davies 60 49.0092 60 Denis Iakovlev 1 48.9974

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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)).