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 21 Daniel Heuer 207 47.7583 22 Dylan Bennett 1 47.3783 23 Richard Hassler 3 47.3294 24 Lei Zhou 126 47.1642 25 Stephen Scott 9.5 47.1579 26 James P. Burt 224 47.1472 27 Steven Harvey 217 47.1074 28 Ian M Gunn 204 47.0102 29 Matthew J Thompson 1 47.0073 30 Chris Cardall 1 46.8877 31 David Metcalfe 226 46.8823 32 Paul Underwood 3.7 46.7901 33 Gary Barnes 247.5 46.7690 34 Brian Lody 308 46.7314 35 Predrag Minovic 115.2 46.7106 36 Joseph Bohanon 63 46.5988 37 Michael Curtis 40 46.5973 38 Borys Jaworski 54 46.5608 39 Szymon Banka 5 46.5298 40 Ian Keogh 1 46.5194

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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)).