Enumerating the number of degree d curves tangent to a planar conic

This question is based on a special case of the Coparaso Harris formula, as described in Counting curves on rational surfaces - R. Vakil.

Let $E$ be a non-singular planar conic.

Then every degree $d$ curve intersects it in $2d$ points (with mult.). Let us focus only on rational irreducible curves that have only simple tangencies with $E$ - these intersect $E$ at exactly $d$ points, each with multiplicity 2. If we fix $d-r$ of the intersection points-$\{p_i\}$ then it turns out that (using stable map terminology) the dimension of the moduli space of stable maps in $\overline{\mathcal{M}}_{0,d-r}(\mathbb{P}^2,dL)$ that map to curves satisfying our conditions (with the $k$-th marked point going to $p_k$) is $d+r-1$.

Fixing $d+r-1$ generic points on the plane for the curves to pass through, we are left with a finite set of points - $(d-r)!N^d(r)$ points to be exact.

$N^d(r)$ is the enumerative constant that counts the number of rational curves of degree $d$ which intersect $E$ only at simple tangencies, $d-r$ of them are fixed.

I'm looking for an asympotic upper bound for the sequence $N^d(r)$, $0\leq r\leq d$ as $d\to\infty$.

It turns out that there is a recursive formula which describes this sequence:

$N^1(0)=0$ and for all $d+r\geq 2$,

$N^d(r)=2(d-r+1)N^d(r-1)+\sum_{l=1}^{\min(r,d-r-2)}\{\sum\frac{1}{\sigma}\frac{(d+r-2)!}{(l+2)!}\prod_{i=1}^l (\frac{2r_iN^{d_i} (r_i)}{(d_i +r_i -1)!})\}$

Where,

1. $N^d(r)=0$ for $r<0$.

2. The inner sum runs over all sequences $d_1,\ldots,d_l$ and $r_1,\ldots,r_l$ where $d_i\geq r_i\geq 2,\,\forall i$ and $\sum _{i=1}^l r_i=r+l$, $\sum_{i=1}^l d_i=d-2$.

3. $\sigma$ equals to the order of the symmetry group of $<d_i| d_i=r_i>$

I'm pretty sure that $\log(N^d(r))<=2d\log d+O(d)$ or at most $\limsup\frac{\log N^d (r)}{2d\log d}\leq 1$, however, I don't know how to prove it.

Is this bound true, if not, what is the best bound that we can give? Any ideas?

• Giving some motivation or reducing to a simpler problem would help you get more attention. – John Jiang Mar 12 '16 at 6:16