 mersenneforum.org Number of octoproths per n
 Register FAQ Search Today's Posts Mark Forums Read 2006-01-13, 13:59 #1 Greenbank   Jul 2005 1100000102 Posts Number of octoproths per n All ranges checked from k=1 to 2^n (where 2^n-k 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 2006-01-20 at 16:04  2006-01-13, 15:36 #2 R. Gerbicz   "Robert Gerbicz" Oct 2005 Hungary 37·41 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*=(1-length(l)/p)/(1-1/p)^8);return(T) Then using it we can predict the total number of octoproths for a given n value by: Code: f(n)=floor(w(n)*2^n/(n*log(2))^8*1/16) Try it! 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 2006-01-13 at 15:39  2006-01-13, 16:00 #3 Greenbank   Jul 2005 2·193 Posts Nice work. f(49) = 55410 real count is 56905.  2006-01-14, 00:08   #4
fetofs

Aug 2005
Brazil

2×181 Posts Quote:
 Originally Posted by R. Gerbicz 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*=(1-length(l)/p)/(1-1/p)^8);return(T) Then using it we can predict the total number of octoproths for a given n value by: Code: f(n)=floor(w(n)*2^n/(n*log(2))^8*1/16) Try it! 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)
This may be stupid, but how do I enter the script on PARI (not the hard way, please)  2006-01-14, 01:10   #5
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

37×41 Posts Quote:
 Originally Posted by fetofs This may be stupid, but how do I enter the script on PARI (not the hard way, please)
I know only one way at the end of line type \  2006-01-14, 12:15 #6 robert44444uk   Jun 2003 Oxford, UK 111101110112 Posts 55 Greenbank Will take on n=55 Regards Robert Smith  2006-01-14, 17:44   #7
fetofs

Aug 2005
Brazil

2·181 Posts Quote:
 Originally Posted by R. Gerbicz I know only one way at the end of line type \
Typing IS the hard way!  2006-01-15, 23:08 #8 robert44444uk   Jun 2003 Oxford, UK 1,979 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  2006-01-16, 08:41 #9 Greenbank   Jul 2005 2·193 Posts 2^55 = 3.6E16 So if you've done to 3e15 then you've only done 8.3%  2006-01-16, 10:01   #10
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

37×41 Posts Quote:
 Originally Posted by Greenbank 2^55 = 3.6E16 So if you've done to 3e15 then you've only done 8.3%
No! 2^55>3.6E16
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 )  2006-01-16, 11:09 #11 Greenbank   Jul 2005 38610 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. Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post aketilander Operazione Doppi Mersennes 1 2012-11-09 21:16 ValerieVonck Octoproth Search 100 2007-02-16 23:43 ValerieVonck Octoproth Search 0 2007-02-14 07:24 robert44444uk Octoproth Search 268 2006-01-26 21:07 jasong Software 1 2005-05-10 20:08

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