## Brun's constant in a different context

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• hi so I ve been trying for awhile to write an asymptotic formula for sum p_n
Message 1 of 2 , Apr 2, 2014
hi
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
x/ln(x)^2*(A*ln(ln(X))+(B*C_twin))+err.term, C_twin=0.66016...
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
• ... At large x, the probability of primality is asymptotic to 1/log(x) and hence the average gap between primes is log(x). The integral of 1/log(x)^2 diverges
Message 2 of 2 , Apr 3, 2014
"Punkish301" wrote:

> asymptotic formula for
> sum p_n<=x 1/(p_n - p(n-1))

At large x, the probability of primality is asymptotic to
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]:

[1000, 1.932409314]
[10000, 2.104903983]
[100000, 2.175330374]
[1000000, 2.252107255]
[10000000, 2.335029340]
[100000000, 2.402612252]
[1000000000, 2.468588116]
[10000000000, 2.530925912]

So it is likely that S(x) increases faster than x/log(x)^2.

David
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