20080114, 21:56  #1 
Sep 2002
Database er0rr
3×11×107 Posts 
Prediction for the next prime
This is a simple statistical prediction for the next 3*2^n1 prime based on historical evidence. It assumes no primes have been missed. The statistics do not take into account the "prime number theorem" Please criticize.
Code:
"n" Increase over the previous exponent   1 0.0000000 2 2.0000000 3 1.5000000 4 1.3333333 6 1.5000000 7 1.1666667 11 1.5714286 18 1.6363636 34 1.8888889 38 1.1176471 43 1.1315789 55 1.2790698 64 1.1636364 76 1.1875000 94 1.2368421 103 1.0957447 143 1.3883495 206 1.4405594 216 1.0485437 306 1.4166667 324 1.0588235 391 1.2067901 458 1.1713555 470 1.0262009 827 1.7595745 1274 1.5405079 3276 2.5714286 4204 1.2832723 5134 1.2212179 7559 1.4723413 12676 1.6769414 14898 1.1752919 18123 1.2164720 18819 1.0384042 25690 1.3651097 26459 1.0299338 41628 1.5733021 51387 1.2344336 71783 1.3969097 81330 1.1329981 85687 1.0535719 88171 1.0289892 97063 1.1008495 123630 1.2737088 155930 1.2612634 164987 1.0580838 234760 1.4229000 414840 1.7670813 584995 1.4101702 702038 1.2000752 727699 1.0365522 992700 1.3641629 1201046 1.2098781 1232255 1.0259848 2312734 1.8768307 3136255 1.3560812  av 1.3400057 std.dev 0.2971180 std.error 0.04 95% 1.26 3948985.29 +95% 1.42 4456213.62 
20080115, 00:51  #2 
Jun 2003
11352_{8} Posts 
And what is the probability that no primes would be found between now and 4456213.62?
EDIT: What happens if you take Geometric Mean (or use log of the ratios?) Last fiddled with by axn on 20080115 at 00:52 
20080115, 05:29  #3  
Sep 2002
Database er0rr
3×11×107 Posts 
Quote:
The geometric mean is 1.31. My original assumption is based on a standard error for a normal distribution, which it is not the distribution of this data. Here is a stem and leaf diagram betrig: Code:
1.0 9,4,5,2,3,2,5,2,5,3,2 1.1 6,1,3,6,8,7,7,3,0,0 1.2 7,3,0,8,2,1,3,7,6,0 1.3 3,8,6,9,6,5 1.4 4,1,7,2,1 1.5 0,0,7,4,7 1.6 3,7 1.7 5,6 1.8 8,7 1.9 2.0 0 2.1 2.2 2.3 2.4 2.5 7 

20080525, 07:37  #4  
May 2007
Kansas; USA
2827_{16} Posts 
Quote:
In checking that list vs. your list, I see a typo. Your prime at n=81330 should be n=80330, which would also affect 2 of your ratios. Other then that, your list should be absolutely correct up to n=260K. NPLB plans to doublecheck all k=31001 up to n=260K by late 2009. So far, we have completed all k=325 and 401417. Gary 

20080525, 12:05  #5 
Sep 2002
Database er0rr
3·11·107 Posts 
A well spotted typo, Gary. The typo only appears in the spreadsheet and makes little difference to the crude calculation.
Last fiddled with by paulunderwood on 20080525 at 12:06 
20080525, 17:31  #6 
May 2007
Kansas; USA
19×541 Posts 
Agreed since you're mostly dealing with the geometric mean. It likely would have a small 'normalizing' effect on the standard error from normal distribution but inconsequential.

20080616, 20:41  #7 
Sep 2002
Database er0rr
3×11×107 Posts 
Here is the latest "crude method" for predicting the next prime based on the last prime. Any modifications such a possion distribution or criticisms are welcome
Code:
1 0.0000000 2 2.0000000 3 1.5000000 4 1.3333333 6 1.5000000 7 1.1666667 11 1.5714286 18 1.6363636 34 1.8888889 38 1.1176471 43 1.1315789 55 1.2790698 64 1.1636364 76 1.1875000 94 1.2368421 103 1.0957447 143 1.3883495 206 1.4405594 216 1.0485437 306 1.4166667 324 1.0588235 391 1.2067901 458 1.1713555 470 1.0262009 827 1.7595745 1274 1.5405079 3276 2.5714286 4204 1.2832723 5134 1.2212179 7559 1.4723413 12676 1.6769414 14898 1.1752919 18123 1.2164720 18819 1.0384042 25690 1.3651097 26459 1.0299338 41628 1.5733021 51387 1.2344336 71783 1.3969097 80330 1.1190672 85687 1.0666874 88171 1.0289892 97063 1.1008495 123630 1.2737088 155930 1.2612634 164987 1.0580838 234760 1.4229000 414840 1.7670813 584995 1.4101702 702038 1.2000752 727699 1.0365522 992700 1.3641629 1201046 1.2098781 1232255 1.0259848 2312734 1.8768307 3136255 1.3560812 4235414 1.3504686 av 1.3401779 std.dev 0.2943653 std.error 0.04 95% 1.26 5342998.07 95% 1.42 6009418.75 mean ānā 5676208.41 
20080620, 10:31  #8  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
the rank order of each prime fits a straight line well. We conjecture from this that the expected number of primes between n1 and n2 is c*ln(n2/n1). If we take n2/n1 to be your average ratio, we choose c such that the expected number of primes is one. We can use this to construct a poll where the ranges represent the 25% percentiles. The four ranges offered in the poll are: n<a a<=n<b b<=n<c c<=n The "fair" choice of ranges has a 75% chance of no primes before a 50% chance of no primes before b 25% chance of no primes before c The construction of this fair poll (or one with more options) gives the clearest possible answer to "where is the next prime" IMO. David Last fiddled with by davieddy on 20080620 at 11:06 

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