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 Wolfgang Schwieger 79 51.8464 22 Marc Wiseler 12 51.8138 23 Diego Bertolotti 1 51.6397 24 Rudi Tapper 6 51.6228 25 Stefan Larsson 142 51.3783 26 Brian D. Niegocki 41 51.3370 27 Randall Scalise 148 51.2334 28 Peter Kaiser 84.3333 51.0046 29 Hiroyuki Okazaki 32 50.9621 30 Michael Cameron 1 50.9234 31 Alen Kecic 14 50.9065 32 Thomas Ritschel 88 50.8707 33 Konstantin Agafonov 1 50.8197 34 Peter Benson 110 50.7155 35 Erik Veit 35 50.6305 36 Michael Schulz 1 50.5434 37 Karsten Klopffleisch 1 50.5009 38 Roman Vogt 3 50.4948 39 Ed Goforth 9 50.4785 40 Barry Schnur 4 50.4538

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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)).