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Old 2018-12-13, 03:08   #34
axn
 
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Quote:
Originally Posted by Prime95 View Post
maybe we haven't reached "asymptotically" yet.
For sure, this is certainly true. However, if you ignore the smallest primes (say p < 1e6), you might get a better agreement between theory and simulation.
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Old 2018-12-13, 04:20   #35
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Quote:
Originally Posted by ewmayer View Post
OTOH such a behavioral change in frequency of M-prime occurrences would seem to be tied to a corresponding change in factor-occurrence statistics, so perhaps some deep TF DB data-mining could be useful.
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Originally Posted by Prime95 View Post
I think the proposed analysis of TF factors is a great idea -- just needs a volunteer.
I like to data mine the database, but I need to figure out exactly what to look for, I guess it is this Wagstaff paper?
http://www.ams.org/journals/mcom/198...54-X/home.html
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Old 2018-12-13, 09:40   #36
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Originally Posted by JeppeSN View Post
Does anyone know Lenstra, Pomerance or Wagstaff personally? All three appear to be alive.
Professor Lenstra is now officially emeritus but in practice still highly active.
He will laugh if he sees your comment!
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Old 2018-12-13, 10:57   #37
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Quote:
Originally Posted by Prime95 View Post
In a run of 1000 simulations of testing all exponents to 85000000, 51+ Mersennes is quite normal - occurring 463 times. 13+ over 10 million is abnormal occurring only 5 times in 1000.
Throw a coin some times (we don't know if the coin is fair or not), and stop when we see ten consecutive heads say at the #throw 330-339. And then simulates, test only the throws at 330-339. From tests you would say that the coin is not fair.

It would be much fair to test other ranges also, not only one fixed, where we see a crowd of Mersenne primes.
So see if there is a T, where [T,8.5T] (for T<10e6) contains at least 13 Mersenne prime exponents or not.
Modifying your code and doing 10000 iterations: (spec=0 means only that in the previous run there was no crowd, otherwise it gives the last prime in the first crowd)
crowd: 2372, spec=0. 51 or more: 4848. 13 or more over 10M: 58 of 10000(27,2)(28,1)(29,1)(30,4)(31,4)(32,14)(33,12)(34,25)(35,42)(36,49)(37,76)(38,76)(39,140)(40,169)(41,216)(42,277)(43,371)(44,372)(45,484)(46,513)(47,538)(48,563)(49,588)(50,615)(51,617)(52,601)(53,518)(54,528)(55,422)(56,391)(57,366)(58,284)(59,254)(60,194)(61,165)(62,137)(63,99)(64,87)(65,61)(66,34)(67,28)(68,26)(69,11)(70,10)(71,6)(72,4)(73,2)(74,1)(76,1)(78,1)

So we have roughly 24 percentage to see at least one crowd, nothing that very especial.
Interestingly in the test the mode was at 51, ofcourse there is no full test up to 85e6, so "our" real count could be even higher.

To see a typical (crowded) run:
Simulated Mp: 2, 3, 5, 7, 11, 13, 17, 19, 31, 157, 251, 317, 397, 443, 457, 1193, 1789, 3623, 4201, 4603, 10061, 10477, 18919, 31573, 32363, 32887, 43451, 77029, 90821, 119173, 138427, 155377, 186551, 206051, 285071, 294313, 346139, 364213, 569251, 684889, 706337, 776453, 838091, 1312769, 1975511, 2507917, 4955761, 10955579, 15812651, 21375961, 39944243, 40921501, 52886513, 81814867. Total: 54
and here there is a crowd ending with p=706337, even ending with p=838091 there is a crowd with 14 primes, but nothing that very interesting in the (10m,85m) range: there are "only" 7 primes.

Last fiddled with by R. Gerbicz on 2018-12-13 at 10:58 Reason: typo
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Old 2018-12-13, 14:26   #38
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Quote:
Originally Posted by Prime95 View Post
Chris Caldwell's summary of Lenstra and Pommerance's conjecture is there are (e^gamma/log 2) * log log x Mersenne primes less than x.

Plugging in 85000000 for x, I get 46 expected Mersenne primes, whereas my simulator is averaging about 50.5. Something is amiss. Maybe my simulator is buggy or maybe we haven't reached "asymptotically" yet.
I think you mean, 85000000 for the exponent, which is (to a fare-thee-well) for a reasonably large Mersenne prime x, p = log(x)/log(2), so that log(x) = p*log(2). Hmm. I'd better check. I have an unfortunate tendency to put terms on the wrong side of the fraction bar.

Oh, Computer, could you please ask Pari-GP?

? exp(Euler)*log(85000000*log(2))/log(2)

%1 = 45.973385236161997342391226401795683851

OK, looks good.

I vote for "haven't reached `asymptotically' yet." The exponents haven't even reached 108. My analytic number theory prof once said, "Analytic number theory begins at 1040."

What's my guess for the new exponent? I don't have one. I reckon, I'll know soon enough...
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Old 2018-12-13, 16:43   #39
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Quote:
Originally Posted by Prime95 View Post
For fun, (y'all are ruining my getaway and why are .c files illegal to upload), I wrote a simulation of wagstaff's conjecture.
Fixed!

Attached Files
File Type: c wagstaff.c (5.7 KB, 51 views)
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Old 2018-12-13, 21:44   #40
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Quote:
Originally Posted by Prime95 View Post
My conclusion (and I am NOT a mathematician): Wagstaff-Lenstra-Pomerance formula is still looking good (from the perspective of number of Mps found). We've been on a lucky streak that has reverted us to the expected norm.
Yes, based on the various simulation results it seems another case of "finding patterns in random data".

Quote:
Originally Posted by R. Gerbicz View Post
Throw a coin some times (we don't know if the coin is fair or not), and stop when we see ten consecutive heads say at the #throw 330-339. And then simulates, test only the throws at 330-339. From tests you would say that the coin is not fair.

It would be much fair to test other ranges also, not only one fixed, where we see a crowd of Mersenne primes.
So see if there is a T, where [T,8.5T] (for T<10e6) contains at least 13 Mersenne prime exponents or not.
Modifying your code and doing 10000 iterations: (spec=0 means only that in the previous run there was no crowd, otherwise it gives the last prime in the first crowd)
crowd: 2372, spec=0. 51 or more: 4848. 13 or more over 10M: 58 of 10000(27,2)(28,1)(29,1)(30,4)(31,4)(32,14)(33,12)(34,25)(35,42)(36,49)(37,76)(38,76)(39,140)(40,169)(41,216)(42,277)(43,371)(44,372)(45,484)(46,513)(47,538)(48,563)(49,588)(50,615)(51,617)(52,601)(53,518)(54,528)(55,422)(56,391)(57,366)(58,284)(59,254)(60,194)(61,165)(62,137)(63,99)(64,87)(65,61)(66,34)(67,28)(68,26)(69,11)(70,10)(71,6)(72,4)(73,2)(74,1)(76,1)(78,1)

So we have roughly 24 percentage to see at least one crowd, nothing that very especial.
Interestingly in the test the mode was at 51, ofcourse there is no full test up to 85e6, so "our" real count could be even higher.

To see a typical (crowded) run:
Simulated Mp: 2, 3, 5, 7, 11, 13, 17, 19, 31, 157, 251, 317, 397, 443, 457, 1193, 1789, 3623, 4201, 4603, 10061, 10477, 18919, 31573, 32363, 32887, 43451, 77029, 90821, 119173, 138427, 155377, 186551, 206051, 285071, 294313, 346139, 364213, 569251, 684889, 706337, 776453, 838091, 1312769, 1975511, 2507917, 4955761, 10955579, 15812651, 21375961, 39944243, 40921501, 52886513, 81814867. Total: 54
Plotting your 'crowded run' the same way I did the actual M-prime exponents, we now see the largest dozen exponents again forming a 'hockey stick', but this time an upward-sloping one:
Attached Files
File Type: pdf mers_clumpy_sim.pdf (6.5 KB, 93 views)

Last fiddled with by ewmayer on 2018-12-13 at 22:00
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Old 2018-12-13, 21:52   #41
chalsall
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Quote:
Originally Posted by ewmayer View Post
Plotting your 'crowded run' the same way I did the actual M-prime exponents:
With all due respect, haven't you already done enough?
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Old 2018-12-13, 22:04   #42
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With all due respect, haven't you already done enough?
Not nearly so - to date all my attempts to endanger national security have failed. There's still work to be done!
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Old 2018-12-13, 22:15   #43
chalsall
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Originally Posted by ewmayer View Post
Not nearly so - to date all my attempts to endanger national security have failed. There's still work to be done!
Cute. And funny...

Just asking... Have you been ever been body searched (at gunpoint) after demonstrating that a major telecoms provider might not wish to leave the default passwords on their deployed legacy Unix kit (which I paid for)?

I have (true story).
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Old 2018-12-14, 01:11   #44
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I've seen clumpy runs. Nothing special.
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