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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 $i$.

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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 $i\mapsto\infty$.

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I think if you try this a bit, you'll find that to get the points where the degree is lowest, you can just apply the same permutation to the points for every value of $i$. The procedure of making a Cremona transformation at points $p_7$, $p_8$, and $p_9$, and then reordering the points so those come first, will achieve the sequence of transformations that you want.

Once you do that, to get the degree and multiplicity of $C_i$ is just a matter of applying some explicit $10 \times 10$ matrix a bunch of times. This matrix will have a $3 \times 3$ Jordan block associated to the eigenvalue $1$ (probably, I didn't check), and so you get the growth of degrees that you want. If you use 10 points instead, it will have an eigenvalue greater than $1$.

To illustrate, here are the first few $C_i$. I write "$d;m_1,m_2,\dots,m_9$" for a curve of degree $d$ with multiplicities $m_i$. (Sorry, don't know how to make a table.)

$C_0$: 1;1,1,0,0,0,0,0,0,0

$C_1$: 2;1,1,1,1,1,0,0,0,0

$C_2$: 4;2,2,2,1,1,1,1,1,0

$C_3$: 6;3,3,2,2,2,2,1,1,1

$C_4$: 9;4,4,4,3,3,2,2,2,2

$C_5$: 12;5,5,5,4,4,4,3,3,2

$C_6$: 16;7,7,6,5,5,5,4,4,4

Each row I get by making a Cremona transformation at the last three points, computing the new degree, and then moving those three points to the beginning. This has the convenient property that I always use the points of least multiplicity, like you want. Going from one row to the next is a linear map (the one saying what the new degree/mults are, composed with the one for permuting). The degrees are growing quadratically, and figuring out an explicit formula is simple linear algebra.

You can also do this using facts about the Weyl group of a certain root system. The Asterisque volume by Dolgachev and Ortland is a good source. Another good reference is "Automorphisms of Blow-ups of $\mathbf P^2$" by McMullen, where you will see a similar procedure. (The application of the "Coxeter element" there corresponds to doing the same thing, but with a different permutation rule, so it won't always hit the three points of lowest multiplicity. Nevertheless the degrees of the $C_i$ grow unboundedly.)

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