Top persons 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 21 Diego Bertolotti 1 51.6397 22 Rudi Tapper 7 51.6235 23 Brian D. Niegocki 38 51.3121 24 Randall Scalise 158 51.1483 25 Stefan Larsson 96 51.0402 26 Peter Kaiser 81.3333 50.9806 27 Michael Cameron 1 50.9234 28 Hiroyuki Okazaki 31 50.9180 29 Alen Kecic 13 50.8639 30 Thomas Ritschel 91 50.8615 31 Konstantin Agafonov 1 50.8197 32 Pavel Atnashev 5 50.7982 33 Peter Benson 140 50.7742 34 Michael Schulz 1 50.5434 35 Karsten Klopffleisch 1 50.5009 36 Roman Vogt 3 50.4948 37 Barry Schnur 4 50.4538 38 Serhiy Gushchak 1 50.4356 39 Borys Jaworski 18 50.4293 40 Peter Harvey 3 50.4233

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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)).