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 21 Stefan Larsson 191 52.0906 22 Ben Maloney 1 52.0371 23 Frank Matillek 10 52.0287 24 Wolfgang Schwieger 90 51.9419 25 Marc Wiseler 9 51.8176 26 Diego Bertolotti 1 51.6397 27 Rudi Tapper 4 51.6208 28 Brian D. Niegocki 26 51.4097 29 Randall Scalise 155 51.3747 30 Hiroyuki Okazaki 50 51.2153 31 Antonio Lucendo 20 51.1797 32 Peter Kaiser 81.3333 51.0002 33 Vaughan Davies 54 50.9892 34 Erik Veit 53 50.9788 35 Alen Kecic 18 50.9744 36 Michael Cameron 1 50.9234 37 Thomas Ritschel 69 50.8313 38 Konstantin Agafonov 1 50.8197 39 Valter Cavecchia 36 50.7079 40 Wes Hewitt 17 50.7077

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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)).