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 Peter Benson 231 50.8958 22 Tom Greer 69 50.8272 23 Konstantin Agafonov 1 50.8197 24 Hiroyuki Okazaki 33 50.6794 25 Michael Schulz 1 50.5434 26 Wolfgang Schwieger 35 50.5417 27 Stefan Larsson 59 50.5274 28 Karsten Klopffleisch 1 50.5009 29 Roman Vogt 3 50.4948 30 Ronny Willig 155 50.4913 31 Serhiy Gushchak 1 50.4356 32 Ed Goforth 14 50.3947 33 Yair Givoni 1 50.3617 34 Sai Yik Tang 15 50.3508 35 Cesare Marini 1 50.3029 36 Peter Kaiser 47.3333 50.2596 37 Thomas Ritschel 93 50.2473 38 David Metcalfe 159 50.0082 39 Masashi Kumagai 2 49.9806 40 Dmitry Domanov 46 49.9299

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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)).