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 )
More information at:
http://www.ltkz.demon.co.uk/ktuplets.htm#mathbg
See the famous HardyLittlewood Prime ktuple Conjecture.
Actually an octoproth isn't a ktuplet but it is very similar to that.