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 1042 Dr. Michael Paridon 3 35.6868 1043 Charles H (Chuck) Lasher 4.8333 35.4451 1044 Laurent Facq 0.6667 35.4211 1045 Michael A Foreman I 0.5 35.3074 1046 Jeroen Doumen 1 34.8314 1046 Peter Beelen 1 34.8314 1048 Mathew Steine 1.5 34.8306 1049 Michael Angel 9 34.6934 1050 Dr. Dirk Augustin 8 34.6933 1051 Wilfrid Keller 11.8333 34.6507 1052 Dana Jacobsen 1 34.3765 1053 Stephan Schöler 2.5 34.2606 1054 Henri Lifchitz 0.5 33.8647 1055 François Morain 5.0833 33.4972 1056 Norman Luhn 26.6667 33.1167 1057 Gerd Lamprecht 6 33.0557 1058 Jens Kruse Andersen 0.3333 32.3531 1059 Michael Stocker 6 30.8285 1060 Alexander Gramolin 1 30.5348

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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)).