Brun's constant in a different context
so I've been trying for awhile to write an asymptotic formula for
sum p_n<=x 1/(p_n - p(n-1))
(the sum of reciprocals of prime gaps)
where p_n means nth prime and n>1.
my attempt is this
where A=approx 0.37787 and B is Brun's constant.
The value of A is ln(2*C_twin)+1/10, and the error term is very small even for large x. Can someone verify this please? I can send you a spreadsheet if you like, I've tested this up to 500,000..
thanks from Guy
- "Punkish301" wrote:
> asymptotic formula forAt large x, the probability of primality is asymptotic to
> sum p_n<=x 1/(p_n - p(n-1))
1/log(x) and hence the average gap between primes is log(x).
The integral of 1/log(x)^2 diverges as x/log(x)^2.
However, your sum favours small gaps. So let us consider the
role of the twin primes. They have a conjectural density of
order 1/log(x)^2. Hence their contribution to the sum is, by
itself, of order x/log(x^2). So one might expect that the
full sum diverges faster than x/log(x)^2.
> I've tested this up to 500,000.We may easily go further.In a few minutes,
I obtained results up to x = 10^10.
For prime p > 2, let g(p) = p - q be the gap between p and
the previous prime, q. Then consider this sum over prime p:
S(x) = sum(2 < p < x, 1/g(p))
Here is a table of [x, S(x)*log(x)^2/x]:
So it is likely that S(x) increases faster than x/log(x)^2.