Question

Suppose $p_n$ is $n$-th prime, $g_n=p_{n+1}-p_n$ is the corresponding prime gap. What is the highest number $C$, such that $p_N>C$ can be proven for $N=\min\{n\mid g_n\geq 1.609\cdot 10^{18}\}$.

Motivation: I've read about Goldbach's weak conjecture. The number $C$ above is obvious lower bound for the first odd number, which does not admit a representation as a sum of three primes, which follows from check of Goldbach's conjecture up to $1.609\cdot 10^{18}$, which is done already by computers. I just want to know, how big is it.

Answer 1

The title and body ask different questions, so I'll address both.

A reasonable estimate for the first prime gap of length $L$ is $e^{\sqrt L}$, so no prime gap this large would be expected below $\exp(1.268\cdot10^9)$, a 550,886,759-digit number.

As for a lower bound, Dusart 2010 [1] shows that for $x\ge396,738$, there is a prime between x and $x\left(1+\frac{1}{25\log^2x}\right)$, so 113353896002617492536754 (about $1.13\cdot10^{23}$) is a lower bound.

Under the Riemann hypothesis ([2]), the bound can be improved to 15373988432858515871940264945439 (about $1.5\cdot10^{31}$).

[1] http://arxiv.org/abs/1002.0442

[2] Lowell Schoenfeld, 'Sharper Bounds for the Chebyshev Functions $\theta(x)$ and $\psi(x)$. II'. Mathematics of Computation, Vol 30, No 134 (Apr 1976), pp. 337-360.

Answer 2

From the Wikipedia page on Bertrand's postulate, Dusart (1998) showed that for all $x > 3275$, there exists a prime between $x$ and $x \left( 1 + \frac{1}{2 \ln^2 x} \right)$. You are looking for the largest $N$ such that $\frac{N}{2\ln^2 N} < 1.609 \cdot 10^{18}$.

A quick computation yields about $8.193\cdot 10^{21}$.

Answer 3

Well, Wikipedia's page on Bertrand's postulate which Scott referred to in his answer does not cite the strongest estimates on this problem. The paper of Olivier Ramaré and Yannick Saouter MR 2004a:11095 "Short effective intervals containing primes. J. Number Theory 98 (2003), no. 1, 10–33 yield stronger result. They prove that the interval $[x(1-1/\Delta), x]$ contains a prime for $x \geq 10 726 905 041$ and $\Delta=28 313 999$. In fact they have a table that gives even stronger result in the relevant range for this problem. If we look at Table 1 from their paper we get that for $x \geq e^{60}$ we can choose $\Delta=209 267 308$. We thus need to determine $x$ so that $$ \frac x {209 267 308} <1.609 \cdot 10^{18}$$ Computation shows that $x<3.367\cdot 10^{26}$. Since $\log(3.367\cdot 10^{26})=61.1>60$ this is permissable. This gives a better estimate than Charles and Scott's answers above.

It should be remarked that already Ramaré-Saouter used these estimates for the Ternary Goldbach problem. However due to a shorter range where the Goldbach problem had been checked at that time they did get the shorter range $1.13256\cdot 10^{22}$ (Corrollary 1) than using the recent computer checked bounds for the Goldbach problem.