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 81 Senji Yamashita 39.6667 45.1356 82 Drew Bishop 4 45.1063 83 Lasse Mejling Andersen 1 45.0970 84 Darren Wallace 9 45.0273 85 Chris Chatfield 31 44.9879 86 David W Linton 3 44.9635 87 Will Fisher 2 44.9550 88 Manfred Toplic 5 44.9485 89 Kevin Odermatt 11 44.8587 90 Andy Penrose 8 44.8233 91 Paul van den Berg 7 44.8004 92 Thomas Masser 24 44.7912 93 Jiong Sun 31 44.7898 94 Greg Childers 1 44.7534 95 Tetsuya Taniguchi 23 44.7146 96 Giovanni Di Maria 5 44.7126 97 Kimihisa Ayuha 9 44.7078 98 Bruno Courty 26 44.6980 99 Phil Carmody 25.7833 44.6803 100 Ray Ballinger 10 44.6630

move up list

move down list

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)).