trj
Posts:
91
From:
Korea
Registered:
2/23/12


Re: Request for help with short argument
Posted:
Nov 15, 2012 8:04 AM


> Thanks for looking at it, trj. > > I need to think about your point  initially not sure > I get your point properly, so I might be confused. > > I am visualising each n as in the centre of a stretch > of p_nmany numbers for each given pn. > > So n2, n1, n, n+1, n+2 for multiples of 5 > then n3, n2, n1, n, n+1, n+2, n+3 for multiples of > 7 > and so on. > > Do the probabilities _need_ to be independent of each > other in the sense you mean? Surely if we look at a > range of numbers n, and test each one like this, it > remains true that 3/5 of the total n in the range > will be in the neighbourhood of a multiple of 5 that > is not n1 or n+1, while 3/5 . 5/7 of those n will be > in the neighbourhood both of a multiple of 5 not at > n1 or n+1 and at the same time in the neighbourhood > of a a multiple of 7 not at n1 or n+1 either? > > I agree it is probably at this point in the argument > that I am overlooking something.
I would not say "overlooking". Your reasoning is sound, though it appears you base it on assumptions that cannot be taken for granted. Look, when you have a factor of 1/2, it is because half of the numbers in an interval have the property that both their neighbours are odd, i.e. they are themselves even. When you have a factor of 1/3, then it because one third of the numbers in an interval, and in this case also one third of any consecutive sequence of even numbers, have the property that both their neighbours are not multiples of 3, i.e. the numbers themselves are multiples of 3. Altogether you deduce that the "center" n of any pair n1,n+1 of twin primes has to be divisible by 6. If you continue this line of thought for the following primes 5,7,11,... then you get congruency conditions modulo 30,210,2310,... etc. for n. But as you also noted, this product of consecutive primes grows very fast. So what happens is that you are looking at a modest stretch of numbers starting from the square of some prime and ending at the square of the next following prime, and the modulus of the congruency condition is huge in comparison.If you think about it, then it means that you cannot be sure that there is one of the numbers in your limited interval which satisfies the congruency condition, unless you know that the primes are extremely uniformly distributed locally.

