20060113, 13:59  #1 
Jul 2005
2·193 Posts 
Number of octoproths per n
All ranges checked from k=1 to 2^n (where 2^nk would be 0).
All primes verified with PARI 2.1.7. n<=26 op=0 n=27 op=1 n=28 op=2 n=29 op=1 n=30 op=1 n=31 op=2 n=32 op=6 n=33 op=2 n=34 op=13 n=35 op=26 n=36 op=11 n=37 op=92 n=38 op=28 n=39 op=83 n=40 op=331 n=41 op=110 n=42 op=453 n=43 op=632 n=44 op=1297 n=45 op=2129 n=46 op=5017 n=47 op=5278 n=48 op=3979 n=49 op=56905 n=50 op=18547 n=51 op=16870 n=52 op=219117 n=53 op=60620 n=54 op=230143 n=55 op=786971 n=56 op=285415 Full list of Octoproths in completed ranges (up to n=54): Raw text: http://octoproth.greenbank.org/downl...o_complete.txt 12024KB (12MB!) ZIP: http://octoproth.greenbank.org/downl...o_complete.zip 4922KB (4.9MB) TODO  Download link for user discovered Octoproths (with attributions) Last fiddled with by Greenbank on 20060120 at 16:04 
20060113, 15:36  #2 
"Robert Gerbicz"
Oct 2005
Hungary
10101100010_{2} Posts 
Very good work!
It is very interesting that we can predict the total number of octoproths for a given n!!! I've worked out it today: by modifying some very hard conjectures, first I define for every n value the "weight" of n: in (PARI): Code:
w(n)=T=128.0;forprime(p=3,10^4,l=listcreate(8);g=Mod(2,p)^n;h=1/g;a=[g,g,h,h,2*g,2*g,h/2,h/2];\ a=lift(a);for(i=1,8,listput(l,a[i],i));l=listsort(l,1);T*=(1length(l)/p)/(11/p)^8);return(T) Code:
f(n)=floor(w(n)*2^n/(n*log(2))^8*1/16) For n=51 it gives that f(n)=16537 It is a very good approximation because Greenbank has calculated that the true number is 16870 ps you'll need also w() to use f() Note that in w() the w(n) is also a prediction because it is using primes up to 10^4 ( to become faster the computation) Last fiddled with by R. Gerbicz on 20060113 at 15:39 
20060113, 16:00  #3 
Jul 2005
110000010_{2} Posts 
Nice work.
f(49) = 55410 real count is 56905. 
20060114, 00:08  #4  
Aug 2005
Brazil
2·181 Posts 
Quote:


20060114, 01:10  #5  
"Robert Gerbicz"
Oct 2005
Hungary
2×13×53 Posts 
Quote:


20060114, 12:15  #6 
Jun 2003
Oxford, UK
3×5^{4} Posts 
55
Greenbank
Will take on n=55 Regards Robert Smith 
20060114, 17:44  #7  
Aug 2005
Brazil
2·181 Posts 
Quote:


20060115, 23:08  #8 
Jun 2003
Oxford, UK
3·5^{4} Posts 
55 almost there
Greenbank
I have done 55 up to 3e15, so a little more to do overnight. The file is enormous, how do I get it to you? Maybe you can send me a private message with your email address and I will send. Regards Robert Smith 
20060116, 08:41  #9 
Jul 2005
2×193 Posts 
2^55 = 3.6E16
So if you've done to 3e15 then you've only done 8.3% 
20060116, 10:01  #10  
"Robert Gerbicz"
Oct 2005
Hungary
1378_{10} Posts 
Quote:
You have to use a larger number than 1+2^n, if you want to search the full range for n. I think Robert444444uk correctly search this interval for n=55, but he mistyped here ( just seeing previous submissions from him ), 3e15 isn't a large range for him. And you can also use my formula to give a prediction for the total number. Using my formula it'll be about: f(55)=772430 ( if we are using rounding instead of floor ) So the size of the file will be larger than the size of all previous files altogether ( up to n=54 ) 

20060116, 11:09  #11 
Jul 2005
110000010_{2} Posts 
Yeah, I meant to say 2^55 ~ 3.6E16. I was just trying to point out that 3E15 was an order of magnitude out.
2^55 = 36028797018963968 3.6E16 = 36000000000000000 robert44444uk, I'll send you an email about getting the n=55 stuff. 
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