Top person sorted by score

The Prover-Account Top 20
Persons by: number score normalized score
Programs by: number score normalized score
Projects by: number score normalized 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.

rankpersonprimesscore
441 Sabrina Tarson 2 47.5524
442 Ondrej Hajek 3 47.5460
443 Nèstor Abad 2 47.5409
444 Keishi Toda 2 47.5342
445 Timothy Sorbera 1 47.5331
446 Jonathan Floyd 2 47.5271
447 Sean Hsu 3 47.5201
448 Ryo Takahashi 2 47.5170
449 Andrei Piaive 2 47.5112
450 Nicholas Vrontakis 3 47.5106
451 Huiwon Kim 1 47.5087
452 Michael Oppliger 1 47.5029
453 Örvar Andrésson 2 47.5002
454 Alexander Falk 3 47.4991
455 Marc Monnin 2 47.4946
456 Raphaël Gervais Lavoie 1 47.4921
457 Roman Trunov 7 47.4916
458 Joseph Hsieh 3 47.4894
459 Yuuichi Hoonoki 2 47.4799
460 Eric Liskay 1 47.4781

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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)).

Printed from the PrimePages <t5k.org> © Reginald McLean.