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 124 Matthias Brod 9 44.1150 125 Marco Crosa 2 44.1133 126 Dale Laluk 20 44.0816 127 Peter Riesen 3 44.0584 128 Andreas Kobara 13 44.0529 129 Tomas Bulka 10.5 44.0504 130 Bob Bruen 12 44.0355 131 Nathan A. Mathew 8 44.0346 132 Michael Gillion 8 44.0279 133 Peter van Hoof 1 44.0109 134 Gerald C. Snyder 6 43.9915 135 Tim Rickard 14 43.9867 136 Joel Armengaud 1 43.9762 137 Steve Anderegg 7 43.9244 138 Richard Kapek 4 43.9153 139 Patrick W. McKibbon 12.6667 43.9042 140 Harsh Aggarwal 1 43.9024 141 Daniel Petat 3 43.8927 142 Steven Tjung 10 43.8809 143 Olivier Haeberle 10 43.8667

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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)).