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 601 Timothy D. Dickey 1 47.1051 602 Jay Parangalan 2 47.1048 603 Galih Santosa 1 47.0887 604 Gary B. Gostin 2 47.0882 605 Joachim Haberer 1 47.0665 606 Stephen Scott 1 47.0636 607 Mischa Rodermond 1 47.0507 608 Jakub Łuszczek 1 47.0494 609 Dmitrij Sesok 1.5 47.0480 610 Homero Pieritz 1 47.0466 611 Timothy Yarham 1 47.0400 612 Ken Davis 85.6667 47.0352 613 Grzegorz Sikorski 1 47.0350 614 Emil Hristoskov 1 47.0349 615 Roger Corron 1 47.0344 616 Donald Ward 1 47.0337 617 Jared Dickson 1 47.0333 618 Yunfei Cheng 1 47.0324 619 Wilhelm Drehling, c't 1 47.0322 620 Michael Wang 1 47.0319

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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)).