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 Tim Terry 14 49.9596 102 Willie Black 9 49.9573 103 Ronny Willig 24 49.9362 104 Tim McArdle 1 49.9091 105 Grzegorz Granowski 14 49.9011 106 Peyton Hayslette 1 49.8982 107 Hans Joachim Böhm 20 49.8975 108 Jonathan Seeley 15 49.8958 109 Kellen Shenton 14 49.8788 110 Lei Zhou 9 49.8692 111 Jan Kožíšek 16 49.8663 112 Anthony Templin 3 49.8611 113 Benoit Da Mota 5 49.8112 114 Latah Headrick 8 49.8093 115 Dennis Sydekum 5 49.8034 116 Frank Doornink 23 49.7952 117 David Åkesson 12 49.7938 118 Michael Goetz 5 49.7604 119 Louis Meuler 6 49.7492 120 Bob Calvin 13 49.7488

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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)).