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 Rod Skinner 4 49.2391 162 Vincent Diepeveen 5 49.2227 163 A.J.Brech 13 49.2218 164 Frank Meador 5 49.2169 165 Liam McGonegal 7 49.2133 166 William Byerly 8 49.2100 167 Alvaro Morera 8 49.1899 168 Eric Nietering 11 49.1864 169 Alex Hogan 4 49.1798 170 Paul Underwood 49.8333 49.1463 171 Walter Darimont 2 49.1458 172 David Hua 4 49.1448 173 Kellen Shenton 11 49.1435 174 Brandon Wharton 10 49.1411 175 Darren Li 5 49.1393 176 Keith Reinhardt 13 49.1384 177 Randy Ready 8 49.1258 178 Daniel Thonon 14 49.1188 179 Eudy Silva 6 49.1029 180 Dao Heng Liu 10 49.0798

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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)).