I am working through the paper by Atkin and Bernstein, Prime sieves using binary quadratic forms, see https://www.ams.org/journals/mcom/2004-73-246/S0025-5718-03-01501-1/S0025-5718-03-01501-1.pdf
Atkin and Bernstein use three binary quadratic forms in order to set up a fast sieve for primes in an interval. Depending on the residue class modulo 60 they choose the quadratic form appropriately. In particular, for the four classes $\{11,23,47,59\bmod 60\}$ they use the quadratic form $3x^2-y^2$. Atkin and Bernstein give the constants of the asymptotical number of steps needed, but with very brief explanation. As one varies over $x$ and $y$ in certain ranges, and certain residue classes, this comes down to counting lattice points in a specified region.
Right now I am trying to figure out the constants that Atkin and Bernstein give on the number of these lattice points ("iterations of step 3") for their binary quadratic forms (given in section 3 and shortly explained in section 4). I managed to get the results for the two elliptic forms (ie. $4x^2+y^2$ and $3x^2+y^2$) but I failed to reproduce their result for the hyperbolic form $3x^2-y^2$.
To compute the area of the hyperbola I am using Zagier "Zetafunktionen und quadratische Körper" chapter 8 (a quite similar result is in Davenport "Multiplicative Number Theory" chapter 6) which gives $A=\frac{N}{\sqrt{D}}\, \log(\epsilon)$ (This area of the hyperbola) where $N$ is the upper bound for the primes I am interested in, $D=b^2-4ac$ is the discriminant of the form $f(x,y)=ax^2+bxy+cy^2$ and $\epsilon= \frac{t+u\sqrt{D}}{2}\ $ where the pair $(t,u)$ is a minimal solution to Pell's equation $t^2-Du^2=4$.
If applied to the form $3x^2-y^2$ I get that $D=12$ and $(t,u)=(4,1)$ thus $\epsilon=\frac{4+\sqrt{12}}{2}=2+\sqrt{3}$ which gives $A=\frac{\log(2+\sqrt{3})}{\sqrt{12}}\, \, N$. For each of the four residue classes $\delta \in \{11,23,47,59\}\bmod 60$ there are 24 possible choices for (x mod 10, y mod 30) out of a total 300 choices. This gives the approximation for the number of lattice points $ \frac{24}{300}A=0.0304N$ and since Atkin and Bernstein consider an interval length of $60B$ this yields $1.8248B$ which is exactly twice the bound given in the paper namely $\sqrt{1.92}\log(\sqrt{0.5}+\sqrt{1.5})B=0.9124B$.
So it seems that somewhere in my computation I am missing a factor even though I am mostly plugging values into formulas.
By now I have found the constant I was missing in my computations. It turns out that one only needs to consider values of $x$ and $y$ such that $x>y>0$ which follows from the proof of Theorem 6.3 in the paper. This gives the factor $1/2$ that is not there in the general formula for the hyperbola.