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 41 Ken Glennie 2 50.6508 42 Honza Cholt 25 50.5779 43 Michael Schulz 1 50.5434 44 Ed Goforth 7 50.5142 45 Peter Benson 41 50.5082 46 Karsten Klopffleisch 1 50.5009 47 Roman Vogt 3 50.4948 48 Barry Schnur 4 50.4782 49 Florian Piesker 113 50.4737 50 Predrag Kurtovic 38 50.4543 51 Max Dettweiler 59 50.4386 52 Serhiy Gushchak 1 50.4356 53 Peter Harvey 3 50.4233 54 Borys Jaworski 15 50.4207 55 Michael Millerick 12 50.4119 56 Marius Vultur 19 50.4054 57 Seonghwan Kim 43 50.3675 58 Yair Givoni 1 50.3617 59 James Krauss 9 50.3521 60 Sai Yik Tang 10 50.3491

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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)).