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 583 Jens Kruse Andersen 3.3334 40.5356 584 Stefan Gronow 1 40.5348 585 Nuutti Kuosa 1.3333 40.2057 586 Darren Smith 2 40.1303 587 Eric J. Sorensen 3 39.9527 588 Geoffrey Reynolds 1 39.9340 589 Bryan Koen 1 39.7459 590 George F. Woltman 0.3333 39.7147 590 Gennady N. Gusev 0.3333 39.7147 592 Mike Oakes 15.75 39.6150 593 Jeffrey Young 7 39.6095 594 Robert Burrowes 0.25 39.6074 594 Randy Dunaieff 0.25 39.6074 594 Chris Nash 0.25 39.6074 597 Wolfgang Straubinger 1 39.5652 598 Brian Rodenkirk 1 39.5652 599 Dipl.-Ing.(FH) Christian Kramer 1 39.5651 600 Robert Apps 1 39.5651 601 Leon Marchal 0.3333 39.5549 602 Ken Davis 32 39.4760

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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)).