Number coloring : same color sequence The natural numbers are colored with two colors. Prove that there exists a sequence of natural numbers $ k_1 < k_2 < ... < k_n < ...$ with the property that the sequence $2k_1 < k_1 + k_2 < 2k_2 < k_2 + k_3 < 2k_3 < ...$ is of the same color.
Please suggest.
 A: If the two colors are red and blue, we begin by making two preliminary attempts at picking sequences like this:


*

*Pick an arbitrary value $r_1$ such that $2r_1$ is red, and keep trying to pick numbers to continue it: for as long as you can pick a number $r_{i+1}$ bigger than the previous $r_i$ such that $r_i + r_{i+1}$ and $2r_{i+1}$ are both red, do so.

*Pick an arbitrary value $b_1$ such that $2b_1$ is blue, and keep trying to pick numbers to continue it: for as long as you can pick a number $b_{i+1}$ bigger than the previous $b_i$ such that $b_i + b_{i+1}$ and $2b_{i+1}$ are both blue, do so.


It's possible that we might be lucky, and one (or both!) of these sequences can be continued forever. If not, we eventually end up with sequences $r_1 < r_2 < \dots < r_n$ and $b_1 < b_2 < \dots < b_m$ that cannot be extended further. This means:


*

*For all $x > r_n$, at least one of $\{x + r_n, 2x\}$ must be blue.

*For all $x > b_m$, at least one of $\{x + b_m, 2x\}$ must be red.


We assume, without loss of generality, that $b_m > r_n$, because either $b_m > r_n$ or $b_m < r_n$ must hold. (We can't have $b_m = r_n$, because $2b_m$ is blue and $2r_n$ is red.)
Choose some number $z > r_n + b_m$ that is even and colored red. (We'll later see how to deal with the case that all sufficiently large even numbers are blue.) Letting $x = z - r_n$, we know the following:


*

*One of $\{x + r_n, 2x\}$ must be blue. But $x + r_n = z$ is red, so $2x$ is blue.

*One of $\{x + b_m, 2x\}$ must be red. But $2x$ is blue, so $x + b_m$ is red.


Since $x + b_m = (z - r_n) + b_m = z + (b_m - r_n)$, we've shown that whenever $z > r_n + b_m$ is red, so is $z + (b_m - r_n)$. But by repeating this argument, we conclude that every term of the infinite arithmetic progression $$z, z + (b_m - r_n), z + 2(b_m - r_n), z + 3(b_m - r_n), \dots$$ is red.
Now we are ready to choose the sequence we're looking for. Let $k_i = \frac z2 + i(b_m - r_n)$. Then we have $k_1 < k_2 < k_3 + \cdots$, and all numbers of the form $2k_i$ or $k_i + k_j$ are red, because they're elements of this infinite arithmetic progression.
(Earlier, we set the "all sufficiently large even numbers are blue" case aside for later. We can now come back to it: if all even numbers $\ge N$ are blue, let $k_i = \frac{N}{2} + 2i$. Then all numbers of the form $2k_i$ or $k_i + k_j$ are even and greater than $N$, so they're all blue.)
