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 541 Kyle Polansky 2 47.2323 542 Giovanni Valentino 3 47.2291 543 Roman Voskoboynikov 2 47.2225 544 Dylan Delgado 2 47.2147 545 Rick Channing 2 47.2005 546 Gang Zhou 3 47.1932 547 Wei Wei 1 47.1911 548 Dale Laluk 3 47.1773 549 Tyler Bredl 2 47.1740 550 LATGE 2 47.1738 551 Rolf Henrik Nilsson 2 47.1734 552 Rusty Clark 2 47.1701 553 Jean-Loup Saladin 1 47.1699 554 Chaichontat Sriworarat 1 47.1680 555 Nico Puada 1 47.1662 556 Gary Craig 2 47.1651 557 William Donovan 2 47.1534 558 Scott Cox 2 47.1485 559 Rashid Naimi 1 47.1258 560 Ardo van Rangelrooij 2 47.1081

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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)).