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 81 Michael Goetz 5 49.7604 82 Patrice Salah 1 49.7436 83 Ken Ito 10 49.7376 84 Frank Doornink 24 49.7375 85 Detlef Lexut 14 49.7359 86 Charles Jackson 21 49.7285 87 Bob Calvin 12 49.6855 88 Ian Johns 18 49.6614 89 Will Steinbach 10 49.6322 90 Hans Joachim Böhm 15 49.5956 91 Greg Miller 9 49.5805 92 LeRoy Blanchard 11 49.5724 93 Jan Kožíšek 13 49.5720 94 Per Provencher 7 49.5637 95 Jochen Beck 11 49.5585 96 Ralf Terber 20 49.4960 97 Philipp Bliedung 3 49.4832 98 Christian Wallbaum 11 49.4770 99 Igor Karpenko 1 49.4585 100 Jordan Romaidis 6 49.4237

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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)).