Prime path

[Gottfried Helms]

Am 09.10.2006 13:20 schrieb Eric Angelini:
> Hello Math-Fun and SeqFan,
> I have the feeling this is (very) old hat... Anyway:
> http://www.cetteadressecomportecinquantesignes.com/PrimePath.htm
> Best,
> É.
> 
> 
> 
The highest numbers at the right hand of a node n  is something

   n
 sum (k) - (n-2)  =  n(n+1)/2 -  n + 2 = (n^2 - n + 4)/2   = N1
  k=1

The length of the interval is d = n-1

The smallest number is N1 - d = (n^2 - 3n +6)/2 = N0

But if we assume the conjecture, that between m^2 and (m-1)^2 is
always a prime, then in each interval d here is a prime and thus
there are infinitely many prime-pathes. Actually the
number of primes between N1 and N0 is about

  N1/ln(N1)  - N0/ln(N0)

or so, so this is then the number of prime-pathes branching
to the right in each interval.
I don't understand, what you mean with "unique"?


Gottfried