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 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 41 Tadashi Taura 11 46.4554 42 Thomas Ritschel 57 46.3560 43 Adam Sutton 31 46.3530 44 Roland Clarkson 1 46.3420 45 Peter Grobstich 25.3333 46.3353 46 Gordon Spence 1 46.2958 47 John Cosgrave 5 46.2621 48 Scott Yoshimura 1 46.1681 49 David Underbakke 34 46.1270 50 Michael Angel 8.6667 46.1172 51 Lennart Vogel 118 46.0816 52 Yves Gallot 5 46.0216 53 Dhumil Zaveri 14 46.0045 54 Karsten Bonath 41 45.9323 55 Vaughan Davies 96 45.8160

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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)).