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.
