I am trying to make 3bit counter with an arbitrary sequence but it gets stuck in a loop (Proteus8)
Let me suggest some thoughts to consider. First off, let's look over the states you need to output:
$$\begin{array}{c}
\text{State}\\\\
  {\begin{smallmatrix}\begin{array}{ccc}
  C &amp; B &amp; A\\\\
  0&amp;0&amp;1\\
  1&amp;1&amp;1\\
  0&amp;0&amp;0\\
  0&amp;1&amp;0\\
  0&amp;0&amp;0\\
  1&amp;0&amp;0\\
  0&amp;0&amp;0\\
  0&amp;0&amp;1\\
  0&amp;0&amp;0
  \end{array}\end{smallmatrix}}
\end{array}$$
The very first thing that crosses my mind is that the first state isn't 0. But when you power-up some JK FF's, the default power-up state (upon applying a \$\overline{RESET}\$ signal) should be all zeroes. So, I'd like to avoid any \$\overline{RESET}\$ difficulties. Which means I'd like to use \$\overline{Q_A}\$ instead of \$Q_A\$ to drive the observed output.
The new table is:
$$\begin{array}{c|c}
\text{Internal State} &amp; \text{Output State}\\\\
{\begin{smallmatrix}\begin{array}{cccc}
  Q_D &amp; Q_C &amp; Q_B &amp; Q_A\\\\
  0&amp;0&amp;0&amp;0\\
  0&amp;1&amp;1&amp;0\\
  0&amp;0&amp;0&amp;1\\
  0&amp;0&amp;1&amp;1\\
  0&amp;0&amp;0&amp;1\\
  0&amp;1&amp;0&amp;1\\
  0&amp;0&amp;0&amp;1\\
  0&amp;0&amp;0&amp;0\\
  0&amp;0&amp;0&amp;1
  \end{array}\end{smallmatrix}} &amp;
  {\begin{smallmatrix}\begin{array}{ccc}
  C &amp; B &amp; A=\overline{Q_A}\\\\
  0&amp;0&amp;1\\
  1&amp;1&amp;1\\
  0&amp;0&amp;0\\
  0&amp;1&amp;0\\
  0&amp;0&amp;0\\
  1&amp;0&amp;0\\
  0&amp;0&amp;0\\
  0&amp;0&amp;1\\
  0&amp;0&amp;0
  \end{array}\end{smallmatrix}}
\end{array}$$
Note that the reset state is perfect. Just what we want.
The next problem is all those zeroes. So let's assign the fact that if \$Q_D=1\$ then this fact alone will force the upper two bits of the output to zero. That way, we can just use \$Q_D=0\$ as an enable to allow the upper two FF states to pass to the output and when \$Q_D=1\$ those upper two FF states are ignored and zeroes are output, instead.
This also plays into the idea of the two &quot;one&quot; states you have. So the following table results, keeping in mind the states after \$\overline{RESET}\$:
$$\begin{array}{c|c|c}
\text{Internal State} &amp; \text{Conceptual State} &amp; \text{Output State}\\\\
{\begin{smallmatrix}\begin{array}{cccc}
  Q_D &amp; Q_C &amp; Q_B &amp; Q_A\\\\
  0&amp;y&amp;y&amp;0\\
  1&amp;1&amp;1&amp;0\\
  0&amp;x&amp;x&amp;1\\
  1&amp;0&amp;1&amp;1\\
  0&amp;x&amp;x&amp;1\\
  1&amp;1&amp;0&amp;1\\
  0&amp;x&amp;x&amp;1\\
  0&amp;y&amp;y&amp;0\\
  0&amp;x&amp;x&amp;1
  \end{array}\end{smallmatrix}} &amp;
{\begin{smallmatrix}\begin{array}{cccc}
  \overline{Q_D} &amp; Q_C &amp; Q_B &amp; \overline{Q_A}\\\\
  1&amp;y&amp;y&amp;1\\
  0&amp;1&amp;1&amp;1\\
  1&amp;x&amp;x&amp;0\\
  0&amp;0&amp;1&amp;0\\
  1&amp;x&amp;x&amp;0\\
  0&amp;1&amp;0&amp;0\\
  1&amp;x&amp;x&amp;0\\
  1&amp;y&amp;y&amp;1\\
  1&amp;x&amp;x&amp;0
  \end{array}\end{smallmatrix}} &amp;
  {\begin{smallmatrix}\begin{array}{cccc}
  C &amp; B &amp; A=\overline{Q_A}\\\\
  0&amp;0&amp;1\\
  1&amp;1&amp;1\\
  0&amp;0&amp;0\\
  0&amp;1&amp;0\\
  0&amp;0&amp;0\\
  1&amp;0&amp;0\\
  0&amp;0&amp;0\\
  0&amp;0&amp;1\\
  0&amp;0&amp;0
  \end{array}\end{smallmatrix}}
\end{array}$$
Here, I've assigned \$xx\$ where we have a state 0 case and \$yy\$ where there is a state 1 case. In the \$xx\$ situation, we have four 0 cases so we'll need all four permutations. But in the \$yy\$ case we only need two of four permutations.
Please note that in the \$xx\$ cases \$Q_A=1\$ and in the \$yy\$ cases \$Q_A=0\$.
Let's expand the table:
$$\begin{array}{c|c|c}
\text{Internal State} &amp; \text{Conceptual State} &amp; \text{Output State}\\\\
{\begin{smallmatrix}\begin{array}{cccc}
  Q_D &amp; Q_C &amp; Q_B &amp; Q_A\\\\
  0&amp;0&amp;0&amp;0\\
  1&amp;1&amp;1&amp;0\\
  0&amp;0 &amp;0&amp;1\\
  1&amp;0&amp;1&amp;1\\
  0&amp;0&amp;1&amp;1\\
  1&amp;1&amp;0&amp;1\\
  0&amp;1&amp;0&amp;1\\
  0&amp;1&amp;0&amp;0\\
  0&amp;1&amp;1&amp;1\\\\
  0&amp;0&amp;0&amp;0\\
  0&amp;0&amp;0&amp;1\\
  0&amp;0&amp;1&amp;0\\
  0&amp;1&amp;0&amp;0\\
  0&amp;1&amp;1&amp;1\\
  1&amp;0&amp;1&amp;0\\
  1&amp;1&amp;1&amp;0
  \end{array}\end{smallmatrix}} &amp;
{\begin{smallmatrix}\begin{array}{cccc}
  \overline{Q_D} &amp; Q_C &amp; Q_B &amp; \overline{Q_A}\\\\
  1&amp;0&amp;0&amp;1\\
  0&amp;1&amp;1&amp;1\\
  1&amp;0&amp;0&amp;0\\
  0&amp;0&amp;1&amp;0\\
  1&amp;0&amp;1&amp;0\\
  0&amp;1&amp;0&amp;0\\
  1&amp;1&amp;0&amp;0\\
  1&amp;1&amp;0&amp;1\\
  1&amp;1&amp;1&amp;0\\\\
  1&amp;0&amp;0&amp;1\\
  1&amp;0&amp;0&amp;0\\
  1&amp;0&amp;1&amp;1\\
  1&amp;1&amp;0&amp;1\\
  1&amp;1&amp;1&amp;0\\
  0&amp;0&amp;1&amp;1\\
  0&amp;1&amp;1&amp;1
  \end{array}\end{smallmatrix}} &amp;
  {\begin{smallmatrix}\begin{array}{ccc}
  C &amp; B &amp; A=\overline{Q_A}\\\\
  0&amp;0&amp;1\\
  1&amp;1&amp;1\\
  0&amp;0&amp;0\\
  0&amp;1&amp;0\\
  0&amp;0&amp;0\\
  1&amp;0&amp;0\\
  0&amp;0&amp;0\\
  0&amp;0&amp;1\\
  0&amp;0&amp;0\\\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x
  \end{array}\end{smallmatrix}}
\end{array}$$
From here, you know that \$C=Q_D \cdot Q_C\$ and \$B=Q_D \cdot Q_B\$ and \$A=\overline{Q_A}\$.
Now, all you need is a transition table to provide the logic required to move from state to state. It's actually not that hard. For example, see here for a possible process to get there from here.
Feel free to ask questions about the above steps or about how to proceed beyond this point towards an answer that implements the desired output.

Appended Per Your Request
JK FF vs T FF
All JK FF can be turned into a T FF by simply wiring the J and K inputs together. This is a choice you can make on your own. The only difference might be some advantages in the combinatorial logic. If you want to go to the work of finding out, feel free. But the following table lists the only four possible cases you might care to think about:
$$\begin{array}{c|c}
\text{Transition} &amp; \text{JK FF} &amp; \text{T FF}\\\\
{\begin{smallmatrix}\begin{array}{c}
\text{start }\to\text{ end}\\\\
  0 \quad \to \quad 0\\
  1 \quad \to \quad 1\\
  0 \quad \to \quad 1\\
  1 \quad \to \quad 0
  \end{array}\end{smallmatrix}} &amp;
{\begin{smallmatrix}\begin{array}{cc}
  J &amp; K \\\\
  0&amp;x\\
  x&amp;0\\
  1&amp;x\\
  x&amp;1
  \end{array}\end{smallmatrix}} &amp;
  {\begin{smallmatrix}\begin{array}{c}
  T\\\\
  0\\
  0\\
  1\\
  1
  \end{array}\end{smallmatrix}}
\end{array}$$
Please take a moment and carefully examine the above table until you fully understand it. You should be able to readily see why it is a correct statement of the situation.
It's a lot easier to just wire both J and K together and focus on only one control bit. But that doesn't mean you cannot lay out the fuller table using both J and K, separately. If you do, you get some &quot;don't cares&quot; in your final k-maps which may work to your favor with some of the logic. My modest experience is that it's not worth the extra bother for educational purposes. However, if I were doing this professionally (and two decades ago before there was super-fancy software that is free to use and can work this out in moments for you) then I probably would go to the effort. But then, I might not be using full JK FF, either, in that case. So... who knows?
Bottom line? Just pick your poison and go forward from there.
K-Maps
The internal and conceptual states mentioned above are entirely equivalent. It really doesn't matter which you choose to work with. But for simplicity, I'd just stick with the internal states listed above to avoid confusion when talking about it.
Let's lay out the transitions:
$$\begin{array}{c|c|c}
\text{Internal State} &amp; \text{Excitations} &amp; \text{Output State}\\\hline\\
{\begin{smallmatrix}\begin{array}{cccc}
  Q_D &amp; Q_C &amp; Q_B &amp; Q_A\\
  \vphantom{\left.\overbrace{\begin{array}{cc}J &amp; K\end{array} } \right.}\\
  0&amp;0&amp;0&amp;0\\
  1&amp;1&amp;1&amp;0\\
  0&amp;0&amp;0&amp;1\\
  1&amp;0&amp;1&amp;1\\
  0&amp;0&amp;1&amp;1\\
  1&amp;1&amp;0&amp;1\\
  0&amp;1&amp;0&amp;1\\
  0&amp;1&amp;0&amp;0\\
  0&amp;1&amp;1&amp;1\\\\
  0&amp;0&amp;0&amp;0\\
  0&amp;0&amp;0&amp;1\\
  0&amp;0&amp;1&amp;0\\
  0&amp;1&amp;0&amp;0\\
  0&amp;1&amp;1&amp;1\\
  1&amp;0&amp;1&amp;0\\
  1&amp;1&amp;1&amp;0
  \end{array}\end{smallmatrix}} &amp;
{\begin{smallmatrix}\begin{array}{cccc}
  Q_D &amp; Q_C &amp; Q_B &amp; Q_A\\
  \left.\overbrace{\begin{array}{cc}J &amp; K\\
    1&amp;x\\
    x&amp;1\\
    1&amp;x\\
    x&amp;1\\
    1&amp;x\\
    x&amp;1\\
    0&amp;x\\
    0&amp;x\\
    0&amp;x\\\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x
  \end{array} } \right. &amp;
  \left.\overbrace{\begin{array}{cc}J &amp; K\\
    1&amp;x\\
    x&amp;1\\
    0&amp;x\\
    0&amp;x\\
    1&amp;x\\
    x&amp;0\\
    x&amp;0\\
    x&amp;0\\
    x&amp;1\\\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x
  \end{array} } \right. &amp;
  \left.\overbrace{\begin{array}{cc}J &amp; K\\
    1&amp;x\\
    x&amp;1\\
    1&amp;x\\
    x&amp;0\\
    x&amp;1\\
    0&amp;x\\
    0&amp;x\\
    1&amp;x\\
    x&amp;1\\\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x
  \end{array} } \right. &amp;
  \left.\overbrace{\begin{array}{cc}J &amp; K\\
    0&amp;x\\
    1&amp;x\\
    x&amp;0\\
    x&amp;0\\
    x&amp;0\\
    x&amp;0\\
    x&amp;1\\
    1&amp;x\\
    x&amp;1\\\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x\\
    x&amp;x
  \end{array} } \right. \\
  \end{array}\end{smallmatrix}} &amp;
  {\begin{smallmatrix}\begin{array}{ccc}
  C &amp; B &amp; A\\
  \vphantom{\left.\overbrace{\begin{array}{cc}J &amp; K\end{array} } \right.}\\
  0&amp;0&amp;1\\
  1&amp;1&amp;1\\
  0&amp;0&amp;0\\
  0&amp;1&amp;0\\
  0&amp;0&amp;0\\
  1&amp;0&amp;0\\
  0&amp;0&amp;0\\
  0&amp;0&amp;1\\
  0&amp;0&amp;0\\\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x\\
  x&amp;x&amp;x
  \end{array}\end{smallmatrix}}
\end{array}$$
At this point, you can map the above into the following k-maps. Anywhere you see \$x\$, it means an explicit &quot;don't care&quot; comes from the above table. Anywhere you see nothing at all, it means an implicit &quot;don't care&quot; because it's from the impossible states at the bottom of the table. The rest are, of course, explicit values:
$$\begin{array}{rl}
\begin{smallmatrix}\begin{array}{r|cccc}
Q_D\text{ }J&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;1&amp;1&amp;1&amp;\\
\overline{Q_D}\:Q_C&amp;0&amp;0&amp;0&amp;\\
Q_D\: Q_C&amp;&amp;x&amp;&amp;x\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;x&amp;
\end{array}\end{smallmatrix}
&amp;
\begin{smallmatrix}\begin{array}{r|cccc}
Q_D\text{ }K&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;x&amp;x&amp;x&amp;\\
\overline{Q_D}\:Q_C&amp;x&amp;x&amp;x&amp;\\
Q_D\: Q_C&amp;&amp;1&amp;&amp;1\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;1&amp;
\end{array}\end{smallmatrix}\\\\
\begin{smallmatrix}\begin{array}{r|cccc}
Q_C\text{ }J&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;1&amp;0&amp;1&amp;\\
\overline{Q_D}\:Q_C&amp;x&amp;x&amp;x&amp;\\
Q_D\: Q_C&amp;&amp;x&amp;&amp;x\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;0&amp;
\end{array}\end{smallmatrix}
&amp;
\begin{smallmatrix}\begin{array}{r|cccc}
Q_C\text{ }K&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;x&amp;x&amp;x&amp;\\
\overline{Q_D}\:Q_C&amp;0&amp;0&amp;1&amp;\\
Q_D\: Q_C&amp;&amp;0&amp;&amp;1\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;x&amp;
\end{array}\end{smallmatrix}\\\\
\begin{smallmatrix}\begin{array}{r|cccc}
Q_B\text{ }J&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;1&amp;1&amp;x&amp;\\
\overline{Q_D}\:Q_C&amp;1&amp;0&amp;x&amp;\\
Q_D\: Q_C&amp;&amp;0&amp;&amp;x\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;x&amp;
\end{array}\end{smallmatrix}
&amp;
\begin{smallmatrix}\begin{array}{r|cccc}
Q_B\text{ }K&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;x&amp;x&amp;1&amp;\\
\overline{Q_D}\:Q_C&amp;x&amp;x&amp;1&amp;\\
Q_D\: Q_C&amp;&amp;x&amp;&amp;1\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;0&amp;
\end{array}\end{smallmatrix}\\\\
\begin{smallmatrix}\begin{array}{r|cccc}
Q_A\text{ }J&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;0&amp;x&amp;x&amp;\\
\overline{Q_D}\:Q_C&amp;1&amp;x&amp;x&amp;\\
Q_D\: Q_C&amp;&amp;x&amp;&amp;1\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;x&amp;
\end{array}\end{smallmatrix}
&amp;
\begin{smallmatrix}\begin{array}{r|cccc}
Q_A\text{ }K&amp;\overline{Q_B}\:\overline{Q_A}&amp;\overline{Q_B}\: Q_A&amp;Q_B \:Q_A&amp;Q_B \:\overline{Q_A}\\
\hline
\overline{Q_D}\:\overline{Q_C}&amp;x&amp;0&amp;0&amp;\\
\overline{Q_D}\:Q_C&amp;x&amp;1&amp;1&amp;\\
Q_D\: Q_C&amp;&amp;0&amp;&amp;x\\
Q_D\:\overline{Q_C}&amp;&amp;&amp;0&amp;
\end{array}\end{smallmatrix}
\end{array}$$
You should be able to work out the optimal logic for each of the J and K inputs of each of the four JK FF, now.
And please satisfy yourself about how I developed the above tables (and check my work while you are at it.)
If you do work it out well, I think you'll find needing not more than (4) AND gates and (3) OR gates, in addition to the (4) JK FF. If using 74xx series parts, perhaps (2) 7473, (1) 7408, and (1) 7432.
(You can't imagine how much of this I had to do in order to design and then wire-wrap and build the execution unit of my own CPU back circa 1974. I could only wish we had EEPROMS back then. But no luck.)
A Final Result -- Added After OP's Homework Completed
The next step was to skim the K-Maps and try to minimize the required logic for each. Others may choose differently, but without spending more than a few minutes (and then a few more minutes to capture this picture), I came up with the following choices (red circles the &quot;1&quot;s and blue circles the &quot;0&quot;s):

This leads to the following results:
$$\begin{array}{c|c}
&amp; \text{Expression}\\\hline
{\begin{smallmatrix}\begin{array}{r}
  Q_D \left\{\begin{array}{c}
        \vphantom{\overline{Q_C}} J\\
        \vphantom{1} K
      \end{array}
      \right.\\
  Q_C \left\{\begin{array}{c}
        \vphantom{\overline{Q_D}\cdot\left(Q_B +\overline{Q_A}\right)} J\\
        \vphantom{Q_B} K\end{array}
      \right.\\
  Q_B \left\{\begin{array}{c}
        \vphantom{\overline{Q_A}+\overline{Q_C}} J\\
        \vphantom{Q_C+\overline{Q_D}} K\end{array}
      \right.\\
  Q_A \left\{\begin{array}{c}
        \vphantom{Q_C} J\\
        \vphantom{Q_C\cdot\overline{Q_D}} K\end{array}
      \right.
  \end{array}\end{smallmatrix}} &amp;
{\begin{smallmatrix}\begin{array}{c}
  \left.\begin{array}{c}\overline{Q_C}\\1\end{array}\right.\\
  \left.\begin{array}{c}\left(\overline{Q_A}\, +\,Q_B\right)\,\cdot\,\overline{Q_D}\\Q_B\end{array}\right.\\
  \left.\begin{array}{c}\overline{Q_A}\,+\,\overline{Q_C}\\Q_C\,+\,\overline{Q_D}\end{array}\right.\\
  \left.\begin{array}{c}Q_C\\Q_C\,\cdot\,\overline{Q_D}\end{array}\right.
  \end{array}\end{smallmatrix}}
\end{array}$$
And then this implementation in Neemann's Digital program:

This validates the operation just simulating the basic parts on the left and using 7400 parts on the right. They both produce exactly your desired state transitions, perfectly.