Returning to Example 11.3, consider the function $f_{\mathfrak{d}}$ attached to the ray of slope $1$ for the cases $\ell=1,2$ and $3$. In each case, the surface $X$ is $\mathbb{P}^{2}$, with coordinate axes $D_{1},D_{2}$ and $D_{\mathrm{out}}$. Then $\widetilde{X}$ is obtained by blowing up $\ell$ points on each of $D_{1}$ and $D_{2}$.

Considering first the case of $\ell=1$, we note that for $\beta=dH$, the class of a degree $d$ curve in $\mathbb{P}^{2}$, the only relevant choice of ${\bf P}$ is ${\bf P}_{1}=d$, ${\bf P}_{2}=d$, and thus we have

This represents the class of a curve of degree $d$ passing through the two blown-up points $d$ times each. It is easy to see that the only choice for such a curve is a $d$-fold cover of a line passing through the two points. Furthermore, this cover must be totally ramified over $D_{\mathrm{out}}$ to guarantee the required order of tangency with $D_{\mathrm{out}}$. This requires a virtual count, and the relevant localization calculations are carried out in [27], giving a value of $N_{{\bf P}}=(-1)^{d+1}/d^{2}$. Thus we get

Exponentiating one finds $f_{\mathfrak{d}}=1+t^{2}x^{-1}y^{-1}$, agreeing with Example 11.3. So here we are just counting the one line through two points in $\mathbb{P}^{2}$ along with certain multiple covers of this line.

Going to $\ell=2$, and $\beta=dH$, one finds four choices for the partition in the case $d=1$, ${\bf P}=(1+0,1+0),(1+0,0+1),(0+1,1+0)$, and $(0+1,0+1)$. Each corresponds to a choice of one point on each of $D_{1}$, $D_{2}$, and one has one line through each of these pairs of points. Thus $N_{\bf P}=1$ for each choice of such ${\bf P}$. As in the case $\ell=1$, each of these lines also contributes to higher degree via multiple covers, with, say, ${\bf P}=(d+0,d+0)$ contributing $N_{\bf P}=(-1)^{d+1}/d^{2}$. For $d=2$, one sees there are no curves for ${\bf P}=(2+0,1+1)$, say, as this would require a conic with a node on $D_{1}$ and tangent to $D_{\mathrm{out}}$; such does not exist. But with ${\bf P}=(1+1,1+1)$, we look at conics passing through all four points and tangent to $D_{\mathrm{out}}$. It is very easy to see there are two such conics.

One can then check that the only other curves contributing are multiple covers of one of the four lines or two conics. The multiple cover contribution for conics is actually different than for lines, because the order of tangency with $D_{\mathrm{out}}$ is different. It turns out the correct contribution for a $d$-fold cover of a conic is $1/d^{2}$. Hence we find

and exponentiating we get

In the case that $\ell=3$, one expects $3\times 3=9$ lines, as there is one line passing through each pair of choices of one point on $D_{1}$ and one point on $D_{2}$. For conics, one has double covers of these lines, for a contribution of $-9/4$, and $2\times 3\times 3=18$ conics. Here one needs to choose two points on $D_{1}$ and two points on $D_{2}$, and then there are two conics passing through these four points tangent to $D_{\mathrm{out}}$.

For cubics, there is the contribution of triple covers of lines, for a total of $9/9$, and a number of contributions from plane cubics. It turns out that for ${\bf P}=(1+1+1,1+1+1)$, $N_{\bf P}=18$. Note this gives a count of nodal plane cubics passing through $6$ fixed points and for which $D_{\mathrm{out}}$ is a tri-tangent. On the other hand, for ${\bf P}=(1+2+0,1+1+1)$, $N_{\bf P}=3$. Note that there are a total of $12$ partitions of this shape. This latter count represents nodal cubics with the node at one of the chosen points, passing also through four other chosen points, with $D_{\mathrm{out}}$ being tritangent. One concludes that

A direct comparision with the value given in Example 11.3 gives agreement.