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If we change the setup and require that each line have infinitely many points of every color, we can extend Anthony Quas's idea to show that you can add any two patterns. That is, if there is a $C_d(n_1,k_1)$ and a $C_d(n_2, k_2)$, there is a $C_d(n_1 + n_2, k_1 + k_2)$. Overlay the two patterns, and at each lattice point, pick randomly which pattern to display. With probability 1, every line will contain infinitely many points of each color it contained in the original patterns. In particular, there is a $C_d(1,1)$, so given a $C_d(n,k)$, we can find a $C_d(n+1,k+1)$.

If we only want periodic patterns, you can replace the random mask we chose with another periodic pattern. Pick a prime which doesn't divide the periods of the two patterns, and construct a $C_d(2,2)$, then overlay the three patterns and choose which of the two original ones to display based on the color of the third. (Note that this doesn't require the periods of the original patterns to be relatively prime.)

If we do have relatively prime periods for our $C_d(n_1,k_1)$ and $C_d(n_2, k_2)$, then we can produce a $C_d(n_1 n_2, k_1 k_2)$, just by overlaying the two and taking the new colors to be ordered pairs of colors from the two patterns.

Like I said in the comments, I don't think your simplex construction actually works for $d>2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $x$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$.

Since we're free to add any number to both $n$ and $k$, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. By adding these to the rainbow constructions which give us $C_2(k^2,k)$, we can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $k \leq n < C k^2$ for some $C < 1$. There are no $C_2(n,k)$ for $n > k^2$, so we have the asymptotic answer.

If we change the setup and require that each line have infinitely many points of every color, we can extend Anthony Quas's idea to show that you can add any two patterns. That is, if there is a $C_d(n_1,k_1)$ and a $C_d(n_2, k_2)$, there is a $C_d(n_1 + n_2, k_1 + k_2)$. Overlay the two patterns, and at each lattice point, pick randomly which pattern to display. With probability 1, every line will contain infinitely many points of each color it contained in the original patterns. In particular, there is a $C_d(1,1)$, so given a $C_d(n,k)$, we can find a $C_d(n+1,k+1)$.

If we only want periodic patterns, you can replace the random mask we chose with another periodic pattern. Pick a prime which doesn't divide the periods of the two patterns, and construct a $C_d(2,2)$ with that period, then overlay the three patterns and choose which of the two original ones to display based on the color of the third. (Note that this doesn't require the periods of the original patterns to be relatively prime.)

If we do have relatively prime periods for our $C_d(n_1,k_1)$ and $C_d(n_2, k_2)$, then we can produce a $C_d(n_1 n_2, k_1 k_2)$, just by overlaying the two and taking the new colors to be ordered pairs of colors from the two patterns.

Like I said in the comments, I don't think your simplex construction actually works for $d>2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $x$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$.

Since we're free to add $1$ to both $n$ and $k$, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. By adding these to the rainbow constructions which give us $C_2(k^2,k)$, we can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $k \leq n < C k^2$ for some $C < 1$. There are no $C_2(n,k)$ for $n > k^2$, so we have the asymptotic answer.

If we change the setup and require that each line have infinitely many points of every color, we can extend Anthony Quas's idea to show that you can add any two patterns. That is, if there is a $C_d(n_1,k_1)$ and a $C_d(n_2, k_2)$, there is a $C_d(n_1 + n_2, k_1 + k_2)$. Overlay the two patterns, and at each lattice point, pick randomly which pattern to display. With probability 1, every line will contain infinitely many points of each color it contained in the original patterns.

If we only want periodic patterns, you can replace the random mask we chose with another periodic pattern. Pick a prime which doesn't divide the periods of the two patterns, and construct a $C_d(2,2)$, then overlay the three patterns and choose which of the two original ones to display based on the color of the third. (Note that this doesn't require the periods of the original patterns to be relatively prime.)

If we do have relatively prime periods for our $C_d(n_1,k_1)$ and $C_d(n_2, k_2)$, then we can produce a $C_d(n_1 n_2, k_1 k_2)$, just by overlaying the two and taking the new colors to be ordered pairs of colors from the two patterns.

Like I said in the comments, I don't think your simplex construction actually works for $d>2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $k$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$.

Combined with the above, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. Using the rainbow construction, we can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $k \leq n < C k^{2}$ for some $C < 1$.

If we change the setup and require that each line have infinitely many points of every color, we can extend Anthony Quas's idea to show that you can add any two patterns. That is, if there is a $C_d(n_1,k_1)$ and a $C_d(n_2, k_2)$, there is a $C_d(n_1 + n_2, k_1 + k_2)$. Overlay the two patterns, and at each lattice point, pick randomly which pattern to display. With probability 1, every line will contain infinitely many points of each color it contained in the original patterns. In particular, there is a $C_d(1,1)$, so given a $C_d(n,k)$, we can find a $C_d(n+1,k+1)$.

If we only want periodic patterns, you can replace the random mask we chose with another periodic pattern. Pick a prime which doesn't divide the periods of the two patterns, and construct a $C_d(2,2)$, then overlay the three patterns and choose which of the two original ones to display based on the color of the third. (Note that this doesn't require the periods of the original patterns to be relatively prime.)

If we do have relatively prime periods for our $C_d(n_1,k_1)$ and $C_d(n_2, k_2)$, then we can produce a $C_d(n_1 n_2, k_1 k_2)$, just by overlaying the two and taking the new colors to be ordered pairs of colors from the two patterns.

Like I said in the comments, I don't think your simplex construction actually works for $d>2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $x$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$.

Since we're free to add any number to both $n$ and $k$, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. By adding these to the rainbow constructions which give us $C_2(k^2,k)$, we can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $k \leq n < C k^2$ for some $C < 1$. There are no $C_2(n,k)$ for $n > k^2$, so we have the asymptotic answer.

If we change the setup and require that each line have infinitely many points of every color, we can extend Anthony Quas's idea to show that you can add any two patterns. That is, if there is a $C_d(n_1,k_1)$ and a $C_d(n_2, k_2)$, there is a $C_d(n_1 + n_2, k_1 + k_2)$. Overlay the two patterns, and at each lattice point, pick randomly which pattern to display. With probability 1, every line will contain infinitely many points of each color it contained in the original patterns.

If we only want periodic patterns, you can replace the random mask we chose with another periodic pattern. Pick a prime which doesn't divide the periods of the two patterns, and construct a $C_d(2,2)$, then overlay the three patterns and choose which of the two original ones to display based on the color of the third. (Note that this doesn't require the periods of the original patterns to be relatively prime.)

If we do have relatively prime periods for our $C_d(n_1,k_1)$ and $C_d(n_2, k_2)$, then we can produce a $C_d(n_1 n_2, k_1 k_2)$, just by overlaying the two and taking the new colors to be ordered pairs of colors from the two patterns.

Like I said in the comments, I don't think your simplex construction actually works for $d>2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $k$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$.

Combined with the above, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. We can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $n \leq C_\varepsilon k^{2-\varepsilon}$ for each $\varepsilon$.

If we change the setup and require that each line have infinitely many points of every color, we can extend Anthony Quas's idea to show that you can add any two patterns. That is, if there is a $C_d(n_1,k_1)$ and a $C_d(n_2, k_2)$, there is a $C_d(n_1 + n_2, k_1 + k_2)$. Overlay the two patterns, and at each lattice point, pick randomly which pattern to display. With probability 1, every line will contain infinitely many points of each color it contained in the original patterns.

If we only want periodic patterns, you can replace the random mask we chose with another periodic pattern. Pick a prime which doesn't divide the periods of the two patterns, and construct a $C_d(2,2)$, then overlay the three patterns and choose which of the two original ones to display based on the color of the third. (Note that this doesn't require the periods of the original patterns to be relatively prime.)

If we do have relatively prime periods for our $C_d(n_1,k_1)$ and $C_d(n_2, k_2)$, then we can produce a $C_d(n_1 n_2, k_1 k_2)$, just by overlaying the two and taking the new colors to be ordered pairs of colors from the two patterns.

Like I said in the comments, I don't think your simplex construction actually works for $d>2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $k$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$.

Combined with the above, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. Using the rainbow construction, we can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $k \leq n < C k^{2}$ for some $C < 1$.