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Old 2015-03-11, 14:13   #34
dbaugh
 
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Douglas Staple programmed up a very good algorithm and he has some hella big computing power available. We should see pi(1e27) soon. I think pi(1e28) might best be a distributed project. That would be right up our alley.
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Old 2015-11-25, 10:00   #35
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Dear all,

Using Kim Walisch's primecount program and 34 days on my BigRig computer, I have now independently confirmed Staple's result of pi(10^26) = 1699246750872437141327603.

Best regards,

David Baugh
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Old 2015-11-25, 12:46   #36
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Thanks David!

Are few more details about the computation:

pi(10^26) = 1,699,246,750,872,437,141,327,603

The computation took 34 days on David's dual socket server (36 CPU
cores, Intel Xeon E5-2699 v3) which corresponds to 3.35 CPU cores
years. The peak memory usage was about 117 gigabytes.
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Old 2015-11-25, 13:40   #37
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Nice work, David! I'm glad to hear it.
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Old 2015-11-25, 16:39   #38
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Thanks David, and well done to Kim and Douglas.

Jan Büthe has been publishing some nice work on tighter bounds lately.
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Old 2015-11-29, 05:30   #39
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Wow , this number pi(10^26) is correct. Many months ago, I found in a eMail-comment with another methode this number.

1,699,246,750,872,437,141,327,603
found by Guillimin and Briarée in 2014 .. was long ago.

Norman


http://www.mersenneforum.org/showthr...918#post388918

Last fiddled with by Cybertronic on 2015-11-29 at 05:36
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Old 2015-11-30, 07:38   #40
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D. B. Staple first found the value using the supercomputers you mentioned. My post was to announce that it has now been independently confirmed. I think we both used the same essential method (combinatorial). I used Walisch's implementation and Staple used his own.
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Old 2015-12-02, 16:26   #41
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Quote:
Originally Posted by danaj View Post
Jan Büthe has been publishing some nice work on tighter bounds lately.
I hadn't seen arXiv:1511.02032 before. Do you know why (1.9) in Theorem 2 is weaker than Korollar 5.3 in his thesis? The former has 27.57 in place of the latter's 2*9.7 = 19.4.
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Old 2015-12-02, 20:44   #42
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Kor 5.3 is defined for 10^8 < x < 10^19, while 1.9 is 2 <= x <= 10^19.

Also 1.9's leading factor is $\frac{\sqrt{x}}{\log x}$ while Kor 5.3 would be $\sqrt{x} \over \log{\sqrt{x}}$ if I interpret the start of section 5.1 correctly.
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Old 2015-12-07, 03:52   #43
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Quote:
Originally Posted by danaj View Post
Kor 5.3 is defined for 10^8 < x < 10^19, while 1.9 is 2 <= x <= 10^19.

Also 1.9's leading factor is $\frac{\sqrt{x}}{\log x}$ while Kor 5.3 would be $\sqrt{x} \over \log{\sqrt{x}}$ if I interpret the start of section 5.1 correctly.
I accounted for the latter with my *2 above, but I didn't see the former. Thanks!
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Old 2016-12-25, 22:15   #44
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Quote:
Originally Posted by dbaugh View Post
We should see pi(1e27) soon.
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