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 108 Robert L. Hillegas, Jr. 15 44.4037 109 Max Dettweiler 15 44.3867 110 Ed Goforth 5 44.3824 111 Alan J Hidy 15 44.3430 112 Jay Berg 7 44.3354 113 Souichi Murata 18.3333 44.3033 114 Takahiro Nohara 3.3333 44.2918 115 Boris Iskra 5 44.2592 116 Satoshi Noda 1.3333 44.2465 117 Willem Siemelink 3.3333 44.2329 118 Dr. James Scott Brown 19 44.2114 119 Jean Penné 3.2 44.1978 120 Franz Hagel 2 44.1825 121 Guido Stolz 2 44.1770 122 Yary Hluchan 2 44.1735 123 Maksym Voznyy 14 44.1597 124 Matthias Brod 9 44.1150 125 Marco Crosa 2 44.1133 126 Dale Laluk 20 44.0816 127 Peter Riesen 3 44.0584

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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)).