TITLE: Arithmetic Progressions in slowly oscillating sequences
QUESTION [26 upvotes]: An infinite sequence ($a_0$, $a_1$, ...) is such that the absolute value of the difference between any 2 consecutive terms is equal to $1$. Is there a length-8 subsequence such that the terms are equally spaced on the original sequence and the terms form an arithmetic sequence from left to right?
Clarifications:
1. The Common difference can be negative or 0
Example:
the sequence 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 7, 6, 7, 8, 7, 8, 9, 8, 9, 10
works because the 3rd term is 3, 6th term is 4, 9th term is 5, ..., 24th term is 10. 
and 3rd, 6th, ..., 24th terms are equally spaced. They also form an arithmetic sequence.
I am thinking about Szekeres theorem but idk if that would work
EDIT: I was able to show it for $n=$. I am actually more interested in indefinitely long ones. But hey, it could be that one can construct something that an indefinitely long one will never happen.

REPLY [6 votes]: There exist a sequence that has no length $54$ equally spaced arithmetic subsequence. This is a modified counterexample from this link.
Let $$u(x)=\frac{3}{2\pi}\sum_{n=1}^\infty 18^n\sin(\frac{2\pi x}{3(36)^n})$$ and put $a_n$ to be an even number in the interval $[u(x)-1,u(x)+1)$ when $n$ is even and an odd number when $n$ is odd.
Then we have \begin{align}|a_k-a_{k+1}|\leq|u(k)-u(k+1)|+2&\leq\frac{3}{2\pi}\sum_{n=1}^\infty 18^n\left|\sin(\frac{2\pi k}{3(36)^n})-\sin(\frac{2\pi(k+1)}{3(36)^n})\right|+2\\&<\frac{3}{2\pi}\sum_{n=1}^\infty18^n\frac{2\pi}{3}\frac1{36^n}+2=3\end{align}
Since $a_k-a_{k+1}$ is odd, we see that $|a_k-a_{k+1}|=1$
Now let $k_1, k_2,\ldots,k_{54}$ be an arithmetic sequence with common difference $d>0$. Define $m$ to be $36^{m-1}\le d<36^{m}$ and $h\le18$ as the smallest integer that $36^m/2\le hd\le 36^m$.  Then $\frac{2\pi k_{19}}{3(36)^m},\frac{2\pi k_{20}}{3(36)^m},\ldots,\frac{2\pi k_{36}}{3(36)^m}$ is an arithmetic sequence with common difference at least $\pi/54$ but less than $2\pi/3$. So one of $\frac{2\pi k_{19}}{3(36)^m},\frac{2\pi k_{20}}{3(36)^m},\ldots,\frac{2\pi k_{36}}{3(36)^m}$ must be in the interval $[\pi/6,5\pi/6]$ or $[7\pi/6,11\pi/6]$ in $\pmod{2\pi}$. Let $\frac{2\pi k_i}{3(36)^m}$ be one of it.
We now show that $a_{k_i-h},a_{k_i},a_{k_i+h}$ is not an arithmetic sequence.
Let $K=2\pi k_i/3$, $D=2\pi hd/3$. First we have $$\sin\frac{D}{2(36)^m},\left|\sin\frac{K}{36^m}\right|\geq\sin\frac\pi6$$
Now
\begin{align}
&|a_{k_i-h}-2a_{k_i}+a_{k_i+h}|\\&\ge |u(k_i-h)-2u(k_i)+u(k_i+h)|-4\\&\ge\frac{3(18)^m}{2\pi}\left|\sin\frac{K-D}{36^m}-2\sin \frac{K}{36^m}+\sin\frac{K+D}{36^m}\right|-\sum_{n=1}^{m-1}\frac{3(18)^n}{2\pi}\left|\sin\frac{K-D}{36^n}-2\sin\frac{K}{36^n}+\sin\frac{K+D}{36^n}\right|-\sum_{n=m+1}^\infty\frac{3(18)^n}{2\pi} \left|\sin\frac{K-D}{36^n}-2\sin\frac{K}{36^n}+\sin\frac{K+D}{36^n}\right|-4\\&\geq \frac{3(18)^m}{2\pi}4\sin^2\frac{D}{2(36)^m}\left|\sin\frac{K}{36^m}\right|-\sum_{n=1}^{m-1}\frac{3(18)^n}{2\pi}4-\sum_{n=m+1}^\infty \frac{3(18)^n}{2\pi} 4\sin^2\frac{D}{2(36)^n}\left|\sin\frac{K}{36^n}\right|-4\\&\geq\frac{3(18)^m}{2\pi}4\sin^2\frac\pi6\sin\frac\pi6-\frac{3}{2\pi}\frac{4(18)^m}{17}-\sum_{n=m+1}^\infty \frac{3(18)^n}{2\pi}4\left(\frac{36^m}{2(36)^n}\right)^2-4\\&=\frac{3(18)^m}{2\pi}\left(\frac12-\frac4{17}-\frac1{71}\right)-4
\end{align}
So if $m\ge2$, $$|a_{k_i-h}-2a_{k_i}+a_{k_i+h}|\geq\frac{3(18)^2}{2\pi}\left(\frac12-\frac4{17}-\frac1{71}\right)-4>0$$
If $m=1$, we can actually have $$|a_{k_i-h}-2a_{k_i}+a_{k_i+h}|\geq\frac{3(18)^1}{2\pi}\left(\frac12-\frac1{71}\right)-4>0$$ as the term $$\sum_{n=1}^{m-1}\frac{3(18)^n}{2\pi}\left|\sin\frac{K-D}{36^n}-2\sin\frac{K}{36^n}+\sin\frac{K+D}{36^n}\right|$$ doesn't occur.
Therefore we always have $a_{k_i-h}-2a_{k_i}+a_{k_i+h}\ne0$ which proves the assertion.

Remark
The argument can be improved a bit more to prove for $36$ using the same equation. As shown in the argument, it is easy to prove for arithmetic sequences whose common difference is large(compare the cases $m=1$ and $m=2$). This post will be updated soon with some more tweaks attached to it.