While looking at various ways of computing $\binom{2n}{n}$ for various values of $n$, I tried the following method: First, define the sequence $T_0=\{n+1,...,2n\}$, then for $k=1,...,n$, given $T_{k-1}=\{a
^{(k-1)}_1,...,a^{(k-1)}_n\}$, define:
$$
t_{0}:=k , \\\,\\
t_{i}:=\frac{t_{i-1}}{\gcd\left(t_{i-1},\,a_i^{(k-1)}\right)}, \\ \,\\
a_{i}^{(k)}:=\frac{a_i^{(k-1)}}{\gcd\left(t_{i-1},\,a_i^{(k-1)}\right)}
$$ for $i=1,...,n$. Eventually, we get $\binom{2n}{n}=a_1^{(n)}\cdots a_n^{(n)}$.
By indicating for each $k=1,...,n$ which indices are affected by it (i.e. those $i$ for which $a_i^{(k-1)}\neq a_i^{(k)}$), we get a patterns like the following:
($n=10$)
The graph of the case $n=10$ is gotten as follows.
In this case we have:
$$
T_0=\{11,12,13,14,15,16,17,18,19,20\} \\
T_1=\{11,12,13,14,15,16,17,18,19,20\} \\
T_2=\{11,6,13,14,15,16,17,18,19,20\} \\
T_3=\{11,2,13,14,15,16,17,18,19,20\} \\
T_4=\{11,1,13,7,15,16,17,18,19,20\} \\
\cdots
$$
So the first column won't have any points, because the first factor $11$ never changes. The second and third column would have a point at height $2$, because the second number changes $12\to6\to2$ while we cancel $2$ and $3$ respectively. The fourth column would have 2 points at height $2$ and $4$, because while cancelling the factor $4$ the second entry $2\to1$ and the fourth entry $14\to7$ etc..
($n=100$)
($n=1000$)
($n=10000$)
($n=40000$)
Interestingly, by removing the entries for prime $k$'s from the graph we get the following image:
($n=40000$)
Furthermore by coloring $k$'s of the form $p$, $2p$ and $3p$ in blue, green and cyan respectively, where $p$ is a prime, we get the following plot:
($n=40000$)
Here's the same image without the red $k$'s:
($n=40000$)
And here's with additionally $4p$, $5p$ and $6p$ as magenta, yellow and black, respectively:
($n=40000$)
Are there any known numerical properties of that can explain the above pattern?