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Old 2018-12-29, 19:08   #89
science_man_88
 
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Quote:
Originally Posted by JeppeSN View Post
I do not know what "empirically" means in the preceding posts, but for any odd \(p\) we have \[2^p\equiv(-1)^p = -1 \pmod 3\] and also for any \(p\ge 3\) we have \[2^p=2^3\cdot 2^{p-3}\equiv 0\cdot 2^{p-3}=0 \pmod 8\] so if we combine, whenever \(p\) is both odd and at least three, then \(2^p\equiv 8 \pmod {24}\). Therefore \(2^p-1\equiv 7 \pmod {24}\) and \(2^p+1\equiv 9 \pmod {24}\).

/JeppeSN
we can show more:

\[ 1^w7^x17^y23^z\bmod(24)\equiv 1^w7^{\min(y,z)+x}17^{y-\min(y,z)}23^{z-\min(y,z)}\] and \[\max(y,z)-\min(y,z)\equiv 0 \bmod 2\] and \[x+\min(y,z)\equiv 1 \bmod 2\]

probably more but I'll leave it for now.

Last fiddled with by science_man_88 on 2018-12-29 at 19:10
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Old 2018-12-30, 13:52   #90
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Quote:
Originally Posted by science_man_88 View Post
we can show more:

...

probably more but I'll leave it for now.
mod 120 this is equivalent to:

\[1^a7^b17^c23^d31^e41^f47^g49^h71^i73^j79^k89^l97^m103^n113^o119^p\]

and

\[\max(c+f+l+o,d+g+i+p)-\min(c+f+l+o,d+g+i+p)\equiv 0\bmod 2\]

and

\[b+e+k+n +\min(c+f+l+o,d+g+i+p)\equiv 1 \bmod 2 \]

As to what else can be said I'm not sure.

Last fiddled with by science_man_88 on 2018-12-30 at 13:54
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Old 2018-12-30, 15:16   #91
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Quote:
Originally Posted by science_man_88 View Post
and \[\max(y,z)-\min(y,z)\equiv 0 \bmod 2\] and \[x+\min(y,z)\equiv 1 \bmod 2\]
I am not sure I understand. What are \(x\), \(y\) and \(z\) here? If one of \(y\) and \(z\) is odd and the other one even, the first one is false. And the second one is false if I pick the parity of \(x\) (in)correctly. I may be missing some context? /JeppeSN
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Old 2018-12-30, 15:20   #92
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Quote:
Originally Posted by JeppeSN View Post
I am not sure I understand. What are \(x\), \(y\) and \(z\) here? If one of \(y\) and \(z\) is odd and the other one even, the first one is false. And the second one is false if I pick the parity of \(x\) (in)correctly. I may be missing some context? /JeppeSN
They are the number of times a prime factor of a certain mod 24 ( or in the latest version mod 120)
appear in fully factored mersenne numbers.

Last fiddled with by science_man_88 on 2018-12-30 at 15:42
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Old 2018-12-30, 18:32   #93
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Quote:
Originally Posted by science_man_88 View Post
\[\max(y,z)-\min(y,z)\equiv 0 \bmod 2\]
I don't know what's going on but this is just

\[y \equiv z \pmod 2.\]
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Old 2018-12-30, 18:56   #94
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I don't know what's going on but this is just

\[y \equiv z \pmod 2.\]
yeah I know but we need it in min form for use with x later.

I was just giving natural powers to the possible remainders of prime factors, and saying that since any factorization even using conposites needs the power on 7 to be odd in the end. since 17 times 23 is 7 mod 24 we can pair them up and get rid of the minimal one used. we then have 17^2 and 23^2 being 1 mod 24 so the difference in powers needs be even.

Last fiddled with by science_man_88 on 2018-12-30 at 18:56
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Old 2019-01-07, 07:28   #95
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Originally Posted by ET_ View Post
How (im)probable is to find such a cluster of Mersenne primes grouped together instead of laying on the linear regression line?
By the Mersenne heuristic, the probability of getting at least 13 primes in an arbitrary interval from x to 10x is about 0.79%:

Code:
atleast(x,lambda)=1-sum(k=0,ceil(x)-1,lambda^k/k!)/exp(lambda);
addhelp(atleast, "atleast(x,lambda): Given a Poisson event with parameter lambda, finds the probability of getting at least x events.");
atleast(13, exp(Euler)*log(10)/log(2))
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Old 2019-01-07, 07:51   #96
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Quote:
Originally Posted by CRGreathouse View Post
By the Mersenne heuristic, the probability of getting at least 13 primes in an arbitrary interval from x to 10x is about 0.79%:

Code:
atleast(x,lambda)=1-sum(k=0,ceil(x)-1,lambda^k/k!)/exp(lambda);
addhelp(atleast, "atleast(x,lambda): Given a Poisson event with parameter lambda, finds the probability of getting at least x events.");
atleast(13, exp(Euler)*log(10)/log(2))
If we get one more MP before 100M, the probability of a 14-interval would be 0.32%

And the same formula (if I'm using it correctly) seems to indicate a 38% probability of finding at least one more between 83M and 100M

atleast(1, exp(Euler)*log(100/83.)/log(2)) == 0.38
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Old 2019-01-07, 15:02   #97
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Originally Posted by preda View Post
If we get one more MP before 100M, the probability of a 14-interval would be 0.32%
Yes.

Of course, having found as many as we have already, the conditional probability is much higher. Someone wanna surf report_exponent and tell me how much?
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