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Old 2014-12-06, 03:43   #12
CRGreathouse
 
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Quote:
Originally Posted by ldesnogu View Post
Perhaps someone can add them to OEIS?
The OEIS already has them (except 2^82 which was just posted).
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Old 2014-12-06, 15:39   #13
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Quote:
Originally Posted by CRGreathouse View Post
The OEIS already has them (except 2^82 which was just posted).
Oops, I had missed the table was in the links section...
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Old 2014-12-07, 00:15   #14
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Quote:
Originally Posted by ldesnogu View Post
Oops, I had missed the table was in the links section...
Yes, /list (always) just gives the first few lines, the bulk of the sequence is in the b-file.
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Old 2014-12-10, 14:33   #15
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Why is the pi(2^m) "only" at 2^81 when pi(10^n) is at 10^26 ~ 2^86.37 ?
Several people have commented on the small stature of 281 as compared to 1026.

The reason I hadn't calculated larger powers of two was priority. If I wanted to simultaneously release π(1026), π(281), π(282), π(283), π(284), π(285), and π(286), then it would have delayed the announcement of π(1026) by several weeks. I estimated the probability of someone else computing π(1026) as perhaps 1% per week, and I viewed this as an unacceptably high risk.

Here are some additional powers of two:
π(283) = 171136408646923240987028
π(284) = 338124238545210097236684
π(285) = 668150111666935905701562

For esoteric scheduling reasons, π(284) has been computed twice with varying hardware and parameters, but π(283) and π(285) have only been computed once each. At this point, it is extremely improbable that the redundant calculations will come back inconsistent.

I will also say that I am in the process of calculating π(286), and will post the result here when I have it.
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Old 2014-12-10, 14:59   #16
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Also, ldesnogu, you were right on the money: I was happy to have π(281) on hand and the others in the works, in case releasing π(1026) started some kind of arms race with one or multiple of Franke, Kleinjung, Büthe and Jost, or perhaps David Platt, or some other dark-horse candidate besides myself. FYI, since releasing the record, the expected but unknown candidate has identified themselves as Kim Walisch. Kim has a very impressive π(x) implementation, and sent me quite a friendly email congratulating me.

Do any of you know Franke, Kleinjung, Büthe, or Jost, or David Platt? I think it is quite interesting that we're now in a situation where two separate lineages of algorithms are very close in capability. Also, if any of them (any of you) are "close" to being able to calculate π(1027), then I suggest that you just tell me that. For my own mental health, I would prefer not to enter into an adversarial relationship with the kind of people that factor RSA-768. Instead, it would be very cool to be able to compute π(1027) in two completely different ways.
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Old 2014-12-10, 18:37   #17
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Quote:
Originally Posted by D. B. Staple View Post
Do any of you know Franke, Kleinjung
I've met with Jens and Thorsten on quite a few occasions. Not sure how useful that is, other than providing minor name-dropping rights

PM me if you think it may be useful.

Paul
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Old 2014-12-10, 20:01   #18
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Not mentioned here, but probably you know that there is a fast algorithm to compute the parity of primepi(n), the above results match with these, see http://www.primenumbers.net/Henri/us/ParPius.htm.
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Old 2014-12-10, 20:27   #19
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Quote:
Originally Posted by D. B. Staple View Post
FYI, since releasing the record, the expected but unknown candidate has identified themselves as Kim Walisch. Kim has a very impressive π(x) implementation, and sent me quite a friendly email congratulating me.
Google showed this up: kimwalish/primecount. This is a Deléglise-Rivat implementation. I wonder how your implementation (based on Tomás Oliveira e Silva article IIUC) compares.

Quote:
Do any of you know Franke, Kleinjung, Büthe, or Jost, or David Platt? I think it is quite interesting that we're now in a situation where two separate lineages of algorithms are very close in capability. Also, if any of them (any of you) are "close" to being able to calculate π(1027), then I suggest that you just tell me that. For my own mental health, I would prefer not to enter into an adversarial relationship with the kind of people that factor RSA-768. Instead, it would be very cool to be able to compute π(1027) in two completely different ways.
I think this kind of competition is good because there are basically two kinds of algorithms (I'm in fact surprised the Meissel et al variant is stillcompetitive, I wonder where the crossover is). This could lead both camps to try and improve their algorithms.
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Old 2014-12-11, 13:36   #20
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Quote:
Not mentioned here, but probably you know that there is a fast algorithm to compute the parity of primepi(n), the above results match with these, see http://www.primenumbers.net/Henri/us/ParPius.htm.
Honestly, I remember learning that it is relatively easy to check the parity of π(x), however, I somehow didn't think of this recently. I first worked on π(x) calculations in 2007, then gave up on it until I picked it up again in 2012 and slowly improved the algorithm and implementation until I was able to compute π(1026). So this was forgotten somewhere between 2007 and now. In any case, obviously a very useful fact. Thank you for reminding me of this and for checking the parity of my calculations.

While we're on the subject of triple- and quadruple-checks, it has to be said that the values I posted are in close agreement with the logarithmic integral: approximately half of the digits are identical, as is typical for these large values of x.
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Old 2014-12-18, 01:14   #21
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I have calculated one more power of two:
π(286) = 1320486952377516565496055

The double checks for π(283) and π(285) have now finished and agree with my previously reported values. The double-check for π(286) is underway; I'll post back here when it's complete.
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Old 2014-12-19, 13:14   #22
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Could you give us an idea of the timings of your runs for 10^n, 2^n, 5^n ?

How high do you intend to go?

Have you considered: 3^n, 7^n, k^n ?
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