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Old 2018-08-20, 19:17   #1
pepi37
 
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Default Poisson Distribution

Tell me how calculate Poisson Distribution for sequence 95*10^n-1

Know exponent (primes ) are

3,192,1133,4763,5812,48467,130620,466002


What will be next expected exponent that yield prime?
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Old 2018-08-20, 19:36   #2
GP2
 
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Quote:
Originally Posted by pepi37 View Post
Tell me how calculate Poisson Distribution for sequence 95*10^n-1

Know exponent (primes ) are

3,192,1133,4763,5812,48467,130620,466002


What will be next expected exponent that yield prime?
Maybe some of the links here will be interesting: OEIS sequence A257038

But assuming it's a Poisson process, can you really predict anything useful about where to look for the next one?
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Old 2018-08-20, 22:55   #3
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Originally Posted by GP2 View Post
Maybe some of the links here will be interesting: OEIS sequence A257038

But assuming it's a Poisson process, can you really predict anything useful about where to look for the next one?
No, I dont ask for somebody to pinpoint me place in sieve where next prime is, I am just curios to see result of Poisson proces.
I am working on this sequence for years, so I do know where prime is not :)
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Old 2018-08-21, 19:24   #4
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OK. The primes 2, 5, and 19 divide 95*10, so none of these will be a factor. This increases the chance your numbers will be prime by a factor of
\frac{2}{1} \cdot \frac{5}{4} \cdot \frac{19}{18} = \frac{95}{36} = \alpha/\log(10).

You could do more work with the cycle lengths of other primes to refine the value of \alpha, but I'll ignore those for now.

The 'chance' that 95*10^n-1 is prime can now be approximated by
\frac{\alpha/\log(10)}{\log(10^n)} = \frac{\alpha}{n}

Let's say you haven't tested past 466002. To get a symmetric 95% confidence interval you want a 2.5% chance on the left and a 2.5% chance on the right. How much Poisson mass do you need to make that happen? On the left, you want 0.025 = 1 - e^-x, so -log(1 - .025); on the right, -log(1 - .975). A little bit of calculus suggests that the answer is about
x \approx \exp\left(\frac{-\log(1 - .025)}{\alpha} + \log(466003)\right) \approx 467949 and
y \approx \exp\left(\frac{-\log(1 - .975)}{\alpha} + \log(466003)\right) \approx 855160 and so a prime 'should' appear between 467949 and 855160.

But these look pretty small and close together compared to my intuition, which suggests to me that the effect of the cycle lengths of the small primes is important and shouldn't be ignored.

Last fiddled with by CRGreathouse on 2018-08-21 at 19:25
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Old 2018-08-21, 20:08   #5
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Originally Posted by CRGreathouse View Post
OK. The primes 2, 5, and 19 divide 95*10, so none of these will be a factor. This increases the chance your numbers will be prime by a factor of
\frac{2}{1} \cdot \frac{5}{4} \cdot \frac{19}{18} = \frac{95}{36} = \alpha/\log(10).

You could do more work with the cycle lengths of other primes to refine the value of \alpha, but I'll ignore those for now.

The 'chance' that 95*10^n-1 is prime can now be approximated by
\frac{\alpha/\log(10)}{\log(10^n)} = \frac{\alpha}{n}

Let's say you haven't tested past 466002. To get a symmetric 95% confidence interval you want a 2.5% chance on the left and a 2.5% chance on the right. How much Poisson mass do you need to make that happen? On the left, you want 0.025 = 1 - e^-x, so -log(1 - .025); on the right, -log(1 - .975). A little bit of calculus suggests that the answer is about
x \approx \exp\left(\frac{-\log(1 - .025)}{\alpha} + \log(466003)\right) \approx 467949 and
y \approx \exp\left(\frac{-\log(1 - .975)}{\alpha} + \log(466003)\right) \approx 855160 and so a prime 'should' appear between 467949 and 855160.

But these look pretty small and close together compared to my intuition, which suggests to me that the effect of the cycle lengths of the small primes is important and shouldn't be ignored.

I also hope that will place where "prime" should appear, but now, I passed 1M5 without new prime.
Would this new data have impact on your calculation?
And thanks for answer!

Last fiddled with by pepi37 on 2018-08-21 at 20:10
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Old 2018-08-22, 12:17   #6
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Quote:
Originally Posted by pepi37 View Post
I also hope that will place where "prime" should appear, but now, I passed 1M5 without new prime.
Would this new data have impact on your calculation?
Sure -- it just takes the place of 466003 in the formulas. You get a 95% confidence interval of [1506263, 2752643] for 1500000 which is my best guess as to what you meant by 1M5 (I haven't seen that notation before).

But see the caveat above: I think more analysis is needed, the intervals should be wider and larger (start higher and end much higher) because of the effect of the other primes beside 2, 5, and 19.
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Old 2018-08-28, 02:10   #7
LaurV
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Quote:
Originally Posted by CRGreathouse View Post
1M5 (I haven't seen that notation before)
Ye're too much a math bookworm... ye may need to step out from the library from time to time, hehe... The notation is used a lot in engineering field, especially in electronics (values of components, resistors, etc, is usually written 6k8, because writing 6.8k can cause confusion on large schematics with small print where the dot will not really be visible - even here on the forum, is not, if you get closer to an age..), and it was used in the past repeatedly and frequently on the forum too.

OTOH, we like your post with the calculus, and we learned something interesting from it... (like usual, your posts are quite interesting). Few weeks ago we struggled to teach our little LaurV who's studying abroad, about confidence intervals (she has a half semester statistics, big pain for a future biologist, haha), using skype/paint as scratchpad, and it was a pain in the but, because we also forgot some stuff we didn't use in years, and it was a lot of "cries" on both sides... hehe... fun time...

Last fiddled with by LaurV on 2018-08-28 at 02:16
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