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 231 Morihiko Tamai 5 42.8314 232 Pietari Snow 5 42.8273 233 Ben Rhodes 5 42.8209 234 Pierre Cami 4 42.8032 235 Pascal Vandaele 3 42.7977 236 Wes Cure 3 42.7977 237 Keith P. Adney 2 42.7905 238 Thierry Gaillard 1.5 42.7791 239 Ian J Brown 6 42.7380 240 Catherine Chaton 4 42.6980 241 Colin Nish 3 42.6799 242 Rob Binnekamp 4 42.6262 243 Helmut Michel 1 42.6260 244 Douglas Stones 4 42.6182 245 Peter O'braian 6 42.6142 246 Yuriy A. Choliy 2 42.6127 247 Jeff Webster 5 42.5946 248 Michael Spall 3 42.5935 249 Mike Ingram 4 42.5766 250 Oleg Garbarenko 3 42.5725

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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)).