 Alternating dyadic colouring excludes symmetric weighted pair sums

Define a finite colouring of the positive integers by $c(n)=\lfloor {\log }_{2}n\rfloor (\mathrm{mod}2)$. No increasing infinite sequence has all the numbers $x_i+2x_j$, $i\ne j$, in one colour class.

Write $\delta (y)=2^{\lfloor {\log }_{2}y\rfloor +1}-y$. If $\delta$ is unbounded on the sequence, fix a term $x$ and choose a later term $y>2x$ with $\delta (y)>2x$. With $L=2^{\lfloor {\log }_{2}y\rfloor }$, the numbers $y+2x$ and $2y+x$ lie respectively in the adjacent dyadic intervals $[L,2L)$ and $[2L,4L)$, so their colours differ. If $\delta$ is bounded by $D$ on the sequence, choose $x>2D$ and then $y>2x$. Now $y+2x=2L-\delta (y)+2x$ lies in $[2L,4L)$, whereas $2y+x=4L-2\delta (y)+x$ lies in $[4L,8L)$. Their colours again differ. The finite colouring therefore excludes the symmetric weighted-pair configuration.