 2006-01-26, 21:07   #269
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

1,429 Posts number of octoproths for n

Quote:
 Originally Posted by robert44444uk I am still mystified why some values of n give many more octoproths and dodecaproths than others. I am also mystified as to why Robert and others are able to forecast the number of octos. If it was variable k I would understand, (modular arithmetic), but I am a bit stumped here. Can someone furnish a non code explanation? Regards Robert Smith
OK here it an explanation: note that this is only a conjecture, because we can't make much weaker statment that f(n)>0 because it would mean that we know that there are infinitely many twin primes, however there is no proof for that

By prime number theorem a random k integer is prime by about 1/log(k) probability, because there are about k/log(k) prime numbers up to k. If we are searching for octoproths then we are searching 8 prime numbers where k<2^n, so if these numbers are independent then there are about 2^n/(log(2^n)^4*log(2^(2*n))^4)=2^n/((n*log(2))^8*16) octoproths ( because 4 forms are about 2^n in almost all cases the other 4 forms are much larger: about 2^(2*n) )

But these numbers are not independent! We have to remultiple by some factors, count the remainders modulo p..., this factor for each n is the weight of n ( for octoproth ) 