Top person 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 41 Max Dettweiler 13 51.2065 42 Alen Kecic 23 51.1520 43 Marius Vultur 40 51.1339 44 Randall Scalise 82 51.1219 45 Erik Veit 56 51.1117 46 Wes Hewitt 33 51.0781 47 Detlef Lexut 20 51.0687 48 Brandon Lee 12 51.0074 49 Peter Kaiser 84.3333 50.9960 50 Jann Kickler 10 50.9923 51 Gregory Coscia 9 50.9654 52 Bruce Marler 12 50.9321 53 Michael Cameron 1 50.9234 54 Predrag Kurtovic 28 50.8561 55 Jean-Luc Garambois 4 50.8503 56 Dmitry Domanov 18 50.8237 57 Konstantin Agafonov 1 50.8197 58 Zack Friedrichsen 35 50.7632 59 Leo LaiHao Wei 1 50.7496 60 Georges Vinotte 1 50.7474

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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)).