Let us consider nine general points $p\_1,...,p\_9\in\mathbb{P}^2$ and the line $L = \left\langle p\_1,p\_2\right\rangle$. Take the standard Cremona $f\_1$ centred in $p\_3,p\_4,p\_5$, then $C\_1 = f\_1(L)$ is a conic through $p\_1,p\_2,p\_3,p\_4,p\_5$ (where I keep denoting by $p\_i$ the images of the contracted lines). Now, take a Cremona $f\_2$ centred in $p\_6,p\_7,p\_8$, then $C\_2 = f\_2(C\_1)$ is a curve of degree four passing through $p\_1,...,p\_8$. Now we iterate this process by taking at the step $i$ the Cremona centred in three points $p\_i,p\_j,p\_k$ such that the sum of the multiplicities of $C\_{i-1}$ in $p\_i,p\_j,p\_k$ is the lowest possible so that the degree of $C\_i = f\_i(C\_{i-1})$ is the biggest possible. 


I would like to know if either there exists a close formula for the degree of the curve $C\_i$ obtained at the step $i$ or if there is a simple way to prove that $deg(C\_i)>deg(C\_{i-1})$ for any $i$. 



Standard plane Cremona transformation
We begin with the line $C\_2 = \left\langle p\_1,p\_2\right\rangle$, and consider the standard Cremona $f\_2:\mathbb{P}^2\dashrightarrow\mathbb{P}^2$ centred in $p\_7,p\_8,p\_9$. The curve $C\_3 = f\_2(C\_2)$ is a conic through $p\_1,...,p\_5$. We proceed recursively by taking at the step $i$ the standard Cremona $f\_{i-1}:\mathbb{P}^2\dashrightarrow\mathbb{P}^2$ centred at the three points among $p\_1,...,p\_9$ of lowest multiplicity for the curve $C\_i = f\_{i-1}(C\_{i-1})$. We denote by $d$ the degree and by $m\_i$ the degree and the multiplicity in $p\_i$ of these curves. The following table displays the step from $i = 2$ to $i = 16$ of the iteration.  

$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline 
\textit{i} & \textit{d} & \textit{m}\_1 & \textit{m}\_2 & \textit{m}\_3 & \textit{m}\_4 & \textit{m}\_5 & \textit{m}\_6 & \textit{m}\_7 & \textit{m}\_8 & \textit{m}\_9 \\ 
\hline 
2 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 
\hline 
3 & 2 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 
\hline 
4 & 4 & 2 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 0 \\ 
\hline 
5 & 6 & 3 & 3 & 2 & 2 & 2 & 2 & 1 & 1 & 1 \\ 
\hline 
6 & 9 & 4 & 4 & 4 & 3 & 3 & 2 & 2 & 2 & 2 \\ 
\hline 
7 & 12 & 5 & 5 & 5 & 4 & 4 & 4 & 3 & 3 & 2 \\ 
\hline 
8 & 16 & 7 & 7 & 6 & 5 & 5 & 5 & 4 & 4 & 4 \\ 
\hline 
9 & 20 & 8 & 8 & 8 & 7 & 7 & 6 & 5 & 5 & 5 \\ 
\hline 
10 & 25 & 10 & 10 & 10 & 8 & 8 & 8 & 7 & 7 & 6 \\ 
\hline 
11 & 30 & 12 & 12 & 11 & 10 & 10 & 10 & 8 & 8 & 8 \\ 
\hline 
12 & 36 & 14 & 14 & 14 & 12 & 12 & 11 & 10 & 10 & 10 \\ 
\hline 
13 & 42 & 16 & 16 & 16 & 14 & 14 & 14 & 12 & 12 & 11 \\ 
\hline 
14 & 49 & 19 & 19 & 18 & 16 & 16 & 16 & 14 & 14 & 14 \\ 
\hline 
15 & 56 & 21 & 21 & 21 & 19 & 19 & 18 & 16 & 16 & 16 \\
\hline 
16 & 64 & 24 & 24 & 24 & 21 & 21 & 21 & 19 & 19 & 18 \\ 
\hline 
\end{array} 
$$
Now, we want to prove that:
$$deg(C\_i) \sim \frac{i^2}{4}$$
and $mult\_{p\_j}C\_i \sim \frac{(i+2)^2-2(i+2)-1}{12}$ for $j = 1,2,3$, $mult\_{p\_j}C\_i \sim \frac{(i+1)^2-2(i+1)-1}{12}$ for $j = ,4,5,6$, $mult\_{p\_j}C\_i \sim \frac{i^2-2i-1}{12}$ for $j = 7,8,9$, where $\sim$ means that the values differs at most by a rational number $-1 <\epsilon <1$. This is verified for all the steps in the table. Let us assume it is true at the step $i$. Then 
$$\deg(C\_{i+1}) = 2\deg(C\_i)-m\_7(C\_i)-m\_8(C\_i)-m\_9(C\_i) \sim 2\frac{i^2}{4}-3\frac{i^2-2i-1}{12} = \frac{(i+1)^2}{4}.$$
From the table we see that
$$m\_j(C\_{i+1}) = m\_{j-3}(C\_{i})\sim \frac{(i+2)^2-2(i+2)-1}{12}$$
for $j = 4,5,6$. Furthermore 
$$m\_j(C\_{i+1}) = m\_{j-3}(C\_{i})\sim \frac{(i+1)^2-2(i+1)-1}{12}$$
for $j = 7,8,9$. Finally
$$m\_j(C\_{i+1}) \sim \frac{i^2}{4}-2\frac{i^2-2i-1}{12} = \frac{i^2+4i+2}{12} = \frac{(i+3)^2-2(i+3)-1}{12}.$$
Then $\deg(C\_i)\sim \frac{i^2}{4}\rightarrow\infty$ for $i\mapsto\infty$.


