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 401 Stuart Mullage 6 47.7313 402 Nicholas Liu 3 47.7308 403 Jonathan Bush 1 47.7235 404 Miroslav Kupka 2 47.7076 405 Marian Brockerhoff 1 47.6942 406 Uwe Pilz 2 47.6882 407 Louis Helm 3 47.6880 408 Jiri Eisler 3 47.6849 409 Gang Zhou 6 47.6823 410 Tapio Rajala 1 47.6789 411 Dr. Karsten Steffens 2 47.6634 412 Ruediger K. Eckhard 3 47.6621 413 Lukas Sandhop 2 47.6555 414 James Jayaputera 1 47.6535 415 Erwin Doescher 4 47.6520 416 Roger Allen 2 47.6494 417 Mark Jones 3 47.6493 418 Charlie McDonald 2 47.6491 419 Martin Garnier 1 47.6485 420 Mincong Liang 2 47.6432

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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)).