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 101 Jonathan Seeley 15 49.8958 102 Lei Zhou 10 49.8824 103 Takahiko Ogawa 19 49.8817 104 Anthony Templin 3 49.8611 105 Ricky L Hubbard 20 49.8582 106 Benoit Da Mota 5 49.8112 107 Latah Headrick 8 49.8093 108 Dennis Sydekum 5 49.8034 109 Frank Doornink 23 49.7952 110 David Åkesson 12 49.7938 111 Michael Goetz 5 49.7604 112 Louis Meuler 6 49.7492 113 Bob Calvin 13 49.7488 114 Jan Kožíšek 14 49.7480 115 Derek Gordon 1 49.7454 116 Patrice Salah 1 49.7436 117 Ken Ito 10 49.7376 118 John Hall 7 49.7311 119 Charles Jackson 20 49.7195 120 Margus Sõmer 10 49.7153

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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)).