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 Cesare Marini 1 50.3029 82 Dmitry Domanov 23 50.2855 83 Igor Keller 8 50.2812 84 HyeongShin Kang 18 50.2791 85 Shawn Schafer 11 50.2742 86 Scott Lee 7 50.1935 87 Bill Cavnaugh 20 50.1112 88 Ian Johns 26 50.0818 89 Jonathan Sipes 20 50.0276 90 Andrew M Farrow 4 49.9997 91 Michael Curtis 28 49.9985 92 James Winskill 3 49.9926 93 Masashi Kumagai 1 49.9772 94 Tim Terry 14 49.9596 95 Ronny Willig 26 49.9595 96 Jonathan Seeley 17 49.9191 97 Tim McArdle 1 49.9091 98 Peyton Hayslette 1 49.8982 99 Hans Joachim Böhm 20 49.8975 100 Lei Zhou 10 49.8824

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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)).