Class of languages recognizable by single-tape 3-state TMs
The beast is extremely powerful, for example we can build a TM $M$ that  accepts every string of the form

$L_Y = \{ r\; 0^n \; 1^m\;A \mid m \leq n \}$

and rejects every string of the form

$L_N = \{ r\; 0^n \; 1^m\;A \mid m &gt; n \}$

The idea is to transform the first $r$ into $R$ and then let the head reach the middle of the string then do a "stateless"  zig-zag; if it reaches $A$ before $R$ then it accepts.

Informal description:


on $r$ write $R$ and move right (prepare for rejecting)
on $0$ write $c$ and move right (moving toward the center)
on $1$ write $&gt;$ and move left (mark a $1$, prepare for next left-to-right crossing, and go left)
on $c$ write $&lt;$ and move right (mark a $0$, prepare for the next right-to-left crossing, and go right)
on $&lt;$ write $&gt;$ and go left (left-to-right crossing, prepare for next  right-to-left)
on $&gt;$ write $&lt;$ and go right (right-to-left crossing, prepare for next  left-to-right)
on $A$ accept, on $R$ reject
on blank $b$ reject


Example:

  :r 0 0 0 1 1 A
   R:0 0 0 1 1 A
   R c:0 0 1 1 A
   R c c:0 1 1 A
   R c c c:1 1 A
   R c c:c &gt; 1 A
   R c c &lt;:&gt; 1 A
   R c c &lt; &lt;:1 A
   R c c &lt;:&lt; &gt; A
   R c c:&lt; &gt; &gt; A
   R c:c &gt; &gt; &gt; A
   R c &lt;:&gt; &gt; &gt; A
   ...
   R c &lt; &lt; &lt; &lt;:A ACCEPT


This zig-zag technique is used in the smallest 2-states Universal Turing machine (it has 18 symbols) constructed by Rogozhin (Rogozhin 1996. TCS, 168(2):215–240)).

Some attention should be paid in order to prove that $M$ halts on all inputs (just note that it rejects on blank input and all non-halting symbols "converge" to  $&lt;$ or $&gt;$ which cannot lead to an infinite loop).

As commented by DavidG, the language $L(M)$ recognized by $M$ is a superset of $L_Y$ (i.e. $L_Y \subset L(M)$) but it doesn't contain any string from $L_N$ (i.e. $L(M) \cap L_N = \emptyset$) and - as commented by MikhailRudoy - this is enough to prove that $L(M)$ is not regular.

Indeed if $L(M)$ is regular then also $L(M) \cap \{ r0^*1^*A \} = L_Y = \{ r0^n 1^mA \mid m \leq n \}$ would be regular (which is not by a simple application of the pumping lemma); leading to a contradiction.

So $L(M)$ is not regular.

... But like all superheroes $C_3$ has an Achille's heel: it cannot even recognize the regular:

$L = \{ 1^{2n} \}$

Informally it can use only the leftmost $b 1 ...$ ($b$ is the blank symbol) or the rightmist $1b...$ as a hook and "expand" in the other direction; but the final Accept or Reject cannot be on the blank symbol of the opposite side, so it can be done only on the inner part of the $1$s and starting from a long enough input it can be "faked" stretching it by one.

Finally - after reading the paper - I can confirm that the one-state TM described in it matches your $C_3$ class ... (so nothing new here :-) ...and the langauge used by the author to prove that it contains non-regular languages is very similar to mine.