Does base n have any Rotate-Left-Double numbers? Task

A Rotate-Left-Double number in base \$n\$ is a number \$m\$, when its base-\$n\$ digits are rotated left once, equals \$2m\$. The base-\$n\$ representation of \$m\$ cannot have leading zeros.

One example in base 7 is the number 480, or \$1254_7\$. When rotated left once, the value becomes \$2541_7 = 960\$.

Given the base \$n \ge 2\$, determine if there exists a Rotate-Left-Double number in base \$n\$.

You can use your language's convention to represent truthy/falsy, or use two distinct values for truthy and falsy respectively.

Scoring and winning criterion

Standard code-golf rules apply. Shortest code in bytes wins.

Test cases

n -&gt; answer (example if true)
-----------------------------
2 -&gt; false
3 -&gt; false
4 -&gt; false
5 -&gt; true (13 x 2 = 31)
6 -&gt; false
7 -&gt; true (1254 x 2 = 2541)
8 -&gt; true (25 x 2 = 52)
9 -&gt; true (125 x 2 = 251)
10 -&gt; false
11 -&gt; true [3,7]
12 -&gt; true [2,4,9,7]
13 -&gt; true [1,2,4,9,5,11,10,8,3,7]
14 -&gt; true [4,9]
15 -&gt; true [1,2,4,9,3,6,13,12,10,5,11,8]
16 -&gt; true [2,4,9]
17 -&gt; true [1,2,4,9]


Reference implementation in Python.
05AB1E, 6 5 bytes
ÍD&lt;&amp;Ā

-1 byte thanks to @xnor and @Noodle9.
Try it online or verify the first \$[2,100]\$ test cases.
Explanation:

Í       # Decrease the (implicit) input-integer by 2
        # Check that this input-2 is a power of 2 by:
 D      #  Duplicating it
  &lt;     #  Decrease the copy by 1 (so integer-3)
   &amp;    #  Take the bitwise-AND of input-2 and input-3
    Ā   #  Check that this is NOT 0
        # (after which the result is output implicitly)

But wait, I don't see any use of bases nor rotation!
When I saw the challenge in the Sandbox and was working on a solution, I noticed that the only falsey values in the first \$n=[2,500]\$ bases formed the sequence A056469: number of elements in the continued fraction for \$\sum_{k=0}^n (\frac{1}{2})^{2^k}\$, which could be simplified to \$a(n)=\left\lfloor2^{n-1}+2\right\rfloor\$. Here a copy of the first 25 numbers in that sequence as reference:
2, 3, 4, 6, 10, 18, 34, 66, 130, 258, 514, 1026, 2050, 4098, 8194, 16386, 32770, 65538, 131074, 262146, 524290, 1048578, 2097154, 4194306, 8388610

It can also be note that all the numbers in this sequence are of the form \$a(n)=2^n+2\$, so checking whether \$n-2\$ is a power of \$2\$ will verify whether it's in this sequence. Since we want to do the invert here, and having a falsey result if it's in this sequence (or truthy if it's NOT in this sequence), we'll do just that, resulting in the code above.
Mathematical proof that all falsey cases of the Left-Rotate-Double numbers are of the form \$2^n+2\$:
Quote from @saulspatz at the Math SE, who provided me with this Mathematical proof to back up my theory I based on the first \$n=[2,500]\$ test cases. So all credit for this proof goes to him/her.
If \$m\$ is a \$(d+1)\$-digit Rotate-Left-Double number in base \$n\$, then $$m=xn^d+y\tag1$$ where \$d\geq1,\ 0&lt;x&lt;n,\ 0\leq y&lt;n^d\$. (Includes the rule that the number can't start with \$0\$.) Rotating \$m\$ gives \$ny+x\$, so we have \$2xn^d+2y=ny+x\$ or $$(n-2)y=(2n^d-1)x\tag2$$ If \$n=2^k+2\$ then \$(2)\$ gives \$(n-2)|x\$ (which means \$x\$ is divisible by \$(n-2)\$), since \$2n^s-1\$ is odd. But then \$y\geq 2n^d-1\$ which contradicts \$y&lt;n^d\$.
To show that these are the only falsey numbers, let \$p\$ be an odd prime dividing \$n-2\$.  (Such a \$p\$ exists because \$n-2\$ is not a power of \$2\$.) In \$(2)\$ we can take \$x=\frac{n-2}p&lt;n\$ and we have to show that there exist an exponent \$d&gt;0\$ and \$0\leq y&lt;n^d\$ such that $$py = 2n^d-1$$ If we can find a \$d\$ such that \$p|(2n^d-1)\$, we are done, for we can take \$y = \frac{2n^d-1}p&lt;n^d\$.
By assumption, \$n-2\equiv0\pmod{p}\$ so \$n\equiv 2\pmod p\$. Therefore, $$2n^d\equiv1\iff 2\cdot2^d\equiv1 \iff 2^{d+1}\equiv 1\pmod p,$$ and by Fermat's little theorem, which states that \$a^{p-1}\equiv 1\pmod p\$, we can take \$d=p-2\$, because $$2^{p-2+1}\equiv 1 \iff 2^{p-1}\equiv 1 \pmod p$$
This completes the proof.