Thread: Octoproths
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Old 2006-01-26, 21:07   #269
R. Gerbicz
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"Robert Gerbicz"
Oct 2005

72·31 Posts
Default number of octoproths for n

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?


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:
See the famous Hardy-Littlewood Prime k-tuple Conjecture.

Actually an octoproth isn't a k-tuplet but it is very similar to that.
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