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 21 Michael Herder 1 48.9335 22 Sturle Sunde 1 48.9324 23 Nayan Hajratwala 1 48.9067 24 Mark Molder 155 48.8708 25 Magnus Bergman 8 48.8090 26 Senji Yamashita 36.3333 48.7428 27 Scott Gilvey 3 48.6870 28 Dennis R. Gesker 1 48.6097 29 David Metcalfe 182.667 48.5355 30 Larry Soule 67 48.5097 31 David Mumper 1 48.4921 32 Thomas Ritschel 49.6667 48.4553 33 Lennart Vogel 153 48.4508 34 Martyn Elvy 1 48.3296 35 Grzegorz Granowski 103 48.3131 36 Raymond Schouten 95 48.2825 37 Jan Haller 2 48.1840 38 Bruce Dodson 92 48.1098 39 Lei Zhou 21 48.0647 40 Randall Scalise 77 48.0640

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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)).