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 1 Curtis Cooper 65 53.5301 2 Steven R. Boone 1 53.5273 3 Dr. Martin Nowak 1 52.9330 4 Josh Findley 1 52.6968 5 Michael Shafer 1 52.2829 6 Michael Cameron 1 50.9234 7 Konstantin Agafonov 1 50.8197 8 Derek Gordon 1 49.7454 9 Ars Technica Team Prime Rib 1 49.2624 10 Sturle Sunde 1 48.9324 11 Nayan Hajratwala 1 48.9067 12 Randy Sundquist 1 47.9205 13 Daniel Heuer 334 47.8100 14 Scott Gilvey 1 47.7989 15 Dylan Bennett 1 47.3783 16 Richard Hassler 5 47.3311 17 Stephen Scott 9.5 47.1161 18 Matthew J Thompson 1 47.0073 19 Peter Benson 417 46.8422 20 Paul Underwood 4.7 46.7909

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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)).