Thread: Number of octoproths per n View Single Post
 2006-01-13, 15:36 #2 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 5·11·29 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