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 161 Zoltan Kemenes 3 49.4905 162 Philipp Bliedung 3 49.4832 163 Lorin Arnold 8 49.4733 164 Lukas Plätz 2 49.4619 165 Ruben Steinberg 12 49.4598 166 Igor Karpenko 1 49.4585 167 Brandon Wharton 12 49.4583 168 Gary Barnes 19 49.4424 169 Daniel Brockwell 19.6667 49.4370 170 Itsuki Kadowaki 4 49.4361 171 Alex Meister 7 49.3929 172 Robert Gelhar 10 49.3863 173 Göran Schmidt 18.3334 49.3766 174 Logan Hood 5 49.3642 175 Matt Jurach 7 49.3580 176 Johannes David 5 49.3554 177 Reginald McLean 6 49.3475 178 Yuki Mizusawa 11 49.3258 179 Hans Sveen 6 49.3168 180 Anthony Ayiomamitis 5 49.2978

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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)).