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 501 Joe DeMaio 1 40.9345 502 Thomas Wolter 1 40.9117 503 Yves Minet 1 40.8807 504 Dale Andrews 1 40.8245 505 Michael Carpenter 1 40.8212 506 Brian Schroeder 1 40.8172 507 Kimmo Herranen 1 40.7879 508 Christian Arntzen 1 40.7638 509 Toshihiro Ohigashi 1 40.7493 510 Volker Milbrandt 1 40.7219 511 Anthony Stockalis 1 40.7181 512 Alexander Schüpstuhl 1 40.7024 513 Masakatu Morii 3 40.7006 514 John F. (Jack) Brennen 1 40.6962 515 Brage E. Elliott 1 40.6762 516 Harvey Patterson 1 40.6748 517 Jonathan Balza 1 40.6733 518 Kouhei Ohyanagi 1 40.6733 519 Jeff Stevens 1 40.6718 520 Martin Brouwers 1 40.6713

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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)).