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 577 Werner Klein 1 40.6246 578 Paul Ziebell 1 40.6243 579 Steven F. Bohl 1 40.6242 580 Ian Abraham 1 40.6226 581 Ariel BaratLeloup 1 40.6221 582 Adam Cisler 1 40.6216 583 Takanori Morinaga 1 40.6204 584 Jens Kruse Andersen 3.3334 40.5356 585 Stefan Gronow 1 40.5348 586 Nuutti Kuosa 1.3333 40.2057 587 Darren Smith 2 40.1303 588 Eric J. Sorensen 3 39.9527 589 Geoffrey Reynolds 1 39.9340 590 Bryan Koen 1 39.7459 591 George F. Woltman 0.3333 39.7147 591 Gennady N. Gusev 0.3333 39.7147 593 Mike Oakes 15.75 39.6150 594 Jeffrey Young 7 39.6095 595 Robert Burrowes 0.25 39.6074 595 Randy Dunaieff 0.25 39.6074 595 Chris Nash 0.25 39.6074

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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)).