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Old 2018-12-13, 12:40   #89
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Quote:
Originally Posted by Madpoo View Post
and made sure the only diff was that last digit or two (see... I'm not even giving away if the last digit is a 9 by saying maybe the last 2 changed).
I think it safe to say that a power \(2^p\) of two, cannot end in a digit zero, because the prime 5 cannot divide a power of the prime 2. Therefore, when you subtract one from a power of two, only the last decimal digit changes.

Besides, they already gave us the exponent modulo 4, so we can infer that your correction consisted in replacing a final digit 2 with a 1.

/JeppeSN
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Old 2018-12-13, 15:33   #90
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Quote:
Originally Posted by Dr Sardonicus View Post
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...
Yeah, if you try with 10M, that is:

(exp(Euler)/log(2))*log(log(2.0^10e6))

you get 40.47, which is too much (there are only 38 Mersennes there in reality). The, as you say, with 85M:

(exp(Euler)/log(2))*log(log(2.0^85e6))

it says 45.97 which is too little, since we know there are 51 or more.

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Old 2018-12-13, 16:01   #91
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Originally Posted by JeppeSN View Post
Besides, they already gave us the exponent modulo 4, so we can infer that your correction consisted in replacing a final digit 2 with a 1.
/JeppeSN
Umm aren't 9, 13 and 17 also 1 modulo 4, making the last digit either 1, 3, 7 or 9?
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Old 2018-12-13, 16:36   #92
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Quote:
Originally Posted by R. Gerbicz View Post
So we have roughly 24 percentage to see at least one crowd, nothing that very especial.
You are probably right.

But every time we convince ourselves of that, someone goes and discovers yet another Mersenne prime, and then the speculation starts all over again.

Maybe if we look hard enough for coincidences, we find them everywhere, like "six nines in pi".
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Old 2018-12-13, 16:50   #93
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Originally Posted by Madpoo View Post
I don't think it was long before people just trivially looked at the exponents they were already looking at, go the last digit, and I do remember seeing someone guess the correct one in that thread. Well, maybe they would have got lucky anyway.
Yeah... And further, someone edited the entry at Wikipedia about MPs listing the 50th MP, but they had guessed incorrectly! This was fortunately noticed quickly and reverted, but it could have diluted the authority of the official announcement.

Again, let me please implore people to let George et al announce officially. If done correctly it can result in a huge increase in throughput for GIMPS.

Admittedly the increase is a surge which then tapers off as the news reports do, but some newly acquired participants do stick around. Further, because of the new assignment rules, this should help decrease the gap between the DC and LL wave-fronts.
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Old 2018-12-13, 17:26   #94
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Quote:
Originally Posted by chalsall View Post
Yeah... And further, someone edited the entry at Wikipedia about MPs listing the 50th MP, but they had guessed incorrectly! This was fortunately noticed quickly and reverted, but it could have diluted the authority of the official announcement.

Again, let me please implore people to let George et al announce officially. If done correctly it can result in a huge increase in throughput for GIMPS.

Admittedly the increase is a surge which then tapers off as the news reports do, but some newly acquired participants do stick around. Further, because of the new assignment rules, this should help decrease the gap between the DC and LL wave-fronts.
Plus, the new version of Prime95 stores interim partial residues along the way. Although we're not currently doing anything with them, if someone starts but doesn't finish a test, we may still be able to use those interim residues to help gauge whether a completed test seems accurate or not. If they diverge at any point, I'd say those were candidates for the early double-checking project.
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Old 2018-12-13, 18:17   #95
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Originally Posted by Dr Sardonicus View Post
I vote for "haven't reached `asymptotically' yet." .
Me too. Hard to believe that dealing with numbers of 20+ million digits we haven't reached asymptotic yet. But, "log log" is a powerful "reducer".

The difference between asymptotic behavior and current behavior is the "a" in the "log ap" calculation (where a is 2 or 6 and p is the exponent). If p is large, a can be ignored. Right now, log ap = log a + log p, and log p is about 18 and log 2 is 0.7 and log 6 is 1.8. Thus we should be finding new Mersenne primes about 4% or 10% more often than the asymptotic behavior. This well explains my simulation getting 50.5 hits vs. the asymptotic expected 46.

The section of Wagstaff's paper where he compares the slope of known Mersenne primes in 1983 to his asymptotic formula for expected number of Mersenne primes is pretty much rubbish as back then we were far far away from asymptotic behavior. I think Chris's web page could also use some modifications.

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.

Now, back to drinks by the poolside.
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Old 2018-12-13, 18:55   #96
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Quote:
Originally Posted by GP2 View Post
You are probably right.

But every time we convince ourselves of that, someone goes and discovers yet another Mersenne prime, and then the speculation starts all over again.

Maybe if we look hard enough for coincidences, we find them everywhere, like "six nines in pi".
A couple of fun things about the number pi; one related to Feynman, the other to whether pi is a "normal number" (both of these being mentioned in the above link):

From Surely You're Joking, Mr. Feynman! ("Lucky numbers"):
Quote:
One day I was feeling my oats. It was lunch time in the technical area, and I don't know how I got the idea, but I announced, "I can work out in sixty seconds the answer to any problem that anybody can state in ten seconds, to 10 percent!" People started giving me problems they thought were difficult, such as integrating a function like 1/(1 + x4 ), which hardly changed over the range they gave me. The hardest one somebody gave me was

the binomial coefficient of x10 in (1 + x)20; I got that just in time.

They were all giving me problems and I was feeling great, when Paul Olum walked by in the hall. Paul had worked with me for a while at Princeton before coming out to Los Alamos, and he was always cleverer than I was.

...

So Paul is walking past the lunch place and these guys are all excited. "Hey, Paul!" they call out. "Feynman's terrific! We give him a problem that can be stated in ten seconds, and in a minute he gets the answer to 10 percent. Why don't you give him one?" Without hardly stopping, he says, "The tangent of 10 to the 100th." I was sunk: you have to divide by pi to 100 decimal places! It was hopeless.
There's a passage in a book called The Lore of Large Numbers by Philip J. Davis, in a chapter about the number pi, which mentions a question dreamed up by the great mathematician L. E. J. Brouwer. His question was, whether in the decimal expansion of pi there were ever a thousand consecutive zeros. AFAIK the answer is not known. But if pi is a "normal number" to the base ten, then such a block of digits would occur with frequency abouyt 1 in 101000.

Last fiddled with by Dr Sardonicus on 2018-12-13 at 18:56
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Old 2018-12-13, 23:33   #97
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Originally Posted by mvusse View Post
Umm aren't 9, 13 and 17 also 1 modulo 4, making the last digit either 1, 3, 7 or 9?
If you look at this table:
\[\begin{matrix}
p: & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots \\
2^p: & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 & \dots
\end{matrix}\]
you will notice that the last digit goes like \(2,4,8,6,2,4,8,6,2,4,\ldots\). Every fourth final digit is a \(2\). In fact, if you disregard \(M_2=2^2-1=4-1=3\), every Mersenne prime will be one less than a number ending in either \(2\) or \(8\), because only the odd exponents \(p\) matter. /JeppeSN

Last fiddled with by JeppeSN on 2018-12-13 at 23:35
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Old 2018-12-13, 23:36   #98
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very nice explanation!
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Old 2018-12-14, 15:51   #99
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No surprise by this point, but my Prime95 run of the discovery confirms as well - it's a new prime!
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