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 181 Bernardo Boncompagni 11 43.3255 182 Ludwig Berndt 1 43.2973 183 Dmitry Domanov 5 43.2929 184 Harvey Dubner 35.8333 43.2734 185 Arnold Padro 1 43.2618 186 Klaus Bodenstein 1 43.2281 187 Chris Siegert 5 43.2122 188 David Stevens 10 43.2116 189 Holger Meissner 1 43.1951 190 Jeffrey Sax 7 43.1635 191 Shane Findley 3 43.1326 192 Peter Tibbott 7 43.1294 193 Robert M Weis 6 43.1279 194 John Metcalf 4 43.1097 195 Christ van Willegen 4.6667 43.0918 196 Thomas Leps 4 43.0854 197 Karen Dowd 0.5 43.0792 198 Syoichiro Yamada 0.3333 43.0767 199 Bruce Blair 9 43.0712 200 Jochen Beck 4 43.0618

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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)).