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 841 Bart van Rooijen 2 46.3381 842 Eric Vitiello 1 46.3348 843 Claudius Dräger 1 46.3302 844 Jeffrey Wright 1 46.3231 845 Howard Abbey 1 46.3230 846 Michael Ruber 1 46.3226 847 Chika Yamamoto 1 46.3151 848 Andreas Reich 1 46.3119 849 Michael Thanry 1 46.3007 850 Gordon Spence 1 46.2958 851 Jess Donovan 1 46.2824 852 Bradley Skinner 1 46.2615 853 Jonathan Bubloski 1 46.2555 854 John Cosgrave 2 46.2303 855 Brent Thorson 1 46.2297 856 Henning Busacker 1 46.1842 857 Stefan Straleger 1 46.1840 858 Lee Stephens 1 46.1825 859 James Steichen 1 46.1813 860 Matthew K Williams 1 46.1723

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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)).