# Top person 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.

rankpersonprimesscore
21 Frank Matillek 38 52.0332
22 Wolfgang Schwieger 82 51.8819
23 Marc Wiseler 12 51.8138
24 Diego Bertolotti 1 51.6397
25 Rudi Tapper 5 51.6218
27 Brian D. Niegocki 34 51.3290
28 Randall Scalise 148 51.2527
29 Hiroyuki Okazaki 41 51.0935
30 Peter Kaiser 84.3333 51.0046
31 Alen Kecic 15 50.9255
32 Michael Cameron 1 50.9234
33 Thomas Ritschel 87 50.8688
34 Konstantin Agafonov 1 50.8197
35 Erik Veit 41 50.7636
36 Peter Benson 101 50.6949
37 Michael Schulz 1 50.5434
38 Ed Goforth 11 50.5262
39 Karsten Klopffleisch 1 50.5009
40 Vaughan Davies 59 50.4954

#### 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)).