Top person sorted by score

At this site we keep several lists of primes, most notably the list of the 5,000 largest known primes. Who found the most of these record primes? We keep separate counts for persons, projects and programs. To see these lists click on 'number' to the right.

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 221 Jonathan Gehrke 9 48.8820 222 KaiMing Poon 3 48.8760 223 Kyle J Johnson 3 48.8629 224 Wolfram Glatte 7 48.8524 225 Jiri Jaros 1 48.8448 226 Yifei Gao 7 48.8368 227 Brook Harste 2 48.8333 228 Km Kim 1 48.8277 229 Dr. sc. nat. Bernhard Pabsch 1 48.8146 230 Prescott Thompson 7 48.8045 231 Martin Neujahr 1 48.8035 232 Pavel Rosický 1 48.7976 233 Jacques Molné 3 48.7940 234 Alex Huetter 2 48.7837 235 Senji Yamashita 8.3333 48.7790 236 Christian Probst 1 48.7771 237 Magnus Bergman 1 48.7753 238 Zoltan Kemenes 2 48.7583 239 Richard R Wheeler 1 48.7473 240 Michael Garrison 1 48.7423

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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)).