Top persons sorted by score
| The Prover-Account Top 20 | |||
|---|---|---|---|
| Persons by: | number | score | normalized score |
| Programs by: | number | score | normalized score |
| Projects by: | number | score | normalized 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 21 Wolfgang Schwieger 40 51.4039 22 Brian D. Niegocki 28 51.2601 23 Randall Scalise 157 50.9911 24 Hiroyuki Okazaki 75 50.9471 25 Michael Cameron 1 50.9234 26 Thomas Ritschel 95.5 50.8681 27 Alen Kecic 16 50.8289 28 Konstantin Agafonov 1 50.8197 29 Peter Benson 167 50.8176 30 Peter Kaiser 68.8333 50.8134 31 Stefan Larsson 82 50.8095 32 Pavel Atnashev 5 50.7982 33 Michael Schulz 1 50.5434 34 Karsten Klopffleisch 1 50.5009 35 Roman Vogt 3 50.4948 36 Ronny Willig 127 50.4404 37 Serhiy Gushchak 1 50.4356 38 Peter Harvey 4 50.4255 39 Ed Goforth 8 50.4072 40 Yair Givoni 1 50.3617
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)).