20141206, 03:43  #12 
Aug 2006
1011101100001_{2} Posts 

20141206, 15:39  #13 
Jan 2008
France
3·181 Posts 

20141207, 00:15  #14 
Aug 2006
1761_{16} Posts 

20141210, 14:33  #15  
Nov 2007
Halifax, Nova Scotia
56_{10} Posts 
Quote:
The reason I hadn't calculated larger powers of two was priority. If I wanted to simultaneously release π(10^{26}), π(2^{81}), π(2^{82}), π(2^{83}), π(2^{84}), π(2^{85}), and π(2^{86}), then it would have delayed the announcement of π(10^{26}) by several weeks. I estimated the probability of someone else computing π(10^{26}) as perhaps 1% per week, and I viewed this as an unacceptably high risk. Here are some additional powers of two: π(2^{83}) = 171136408646923240987028 π(2^{84}) = 338124238545210097236684 π(2^{85}) = 668150111666935905701562 For esoteric scheduling reasons, π(2^{84}) has been computed twice with varying hardware and parameters, but π(2^{83}) and π(2^{85}) 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(2^{86}), and will post the result here when I have it. 

20141210, 14:59  #16 
Nov 2007
Halifax, Nova Scotia
2^{3}×7 Posts 
Also, ldesnogu, you were right on the money: I was happy to have π(2^{81}) on hand and the others in the works, in case releasing π(10^{26}) started some kind of arms race with one or multiple of Franke, Kleinjung, Büthe and Jost, or perhaps David Platt, or some other darkhorse 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(10^{27}), 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 RSA768. Instead, it would be very cool to be able to compute π(10^{27}) in two completely different ways. 
20141210, 18:37  #17 
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
2^{4}×5×7×19 Posts 

20141210, 20:01  #18 
"Robert Gerbicz"
Oct 2005
Hungary
2×727 Posts 
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.

20141210, 20:27  #19  
Jan 2008
France
3×181 Posts 
Quote:
Quote:


20141211, 13:36  #20  
Nov 2007
Halifax, Nova Scotia
70_{8} Posts 
Quote:
While we're on the subject of triple and quadruplechecks, 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. 

20141218, 01:14  #21 
Nov 2007
Halifax, Nova Scotia
56_{10} Posts 
I have calculated one more power of two:
π(2^{86}) = 1320486952377516565496055 The double checks for π(2^{83}) and π(2^{85}) have now finished and agree with my previously reported values. The doublecheck for π(2^{86}) is underway; I'll post back here when it's complete. 
20141219, 13:14  #22 
May 2004
New York City
2^{3}·23^{2} Posts 
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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