I have found the Gridpatterns page.
One can apply the next algoritm and obtaine the grid pattern "1-2-3".
- On square grid paper start in the middle.
- Draw a line 1-unit long.
- Turn a right angle clockwise.
- Draw a line 2-unit long.
- Turn a right angle clockwise.
- Draw a line 3-unit long.
- Repeat steps 1-6 four times.
Result is below on the left figure. As you can see we returned to the start point (red).
Edit. After the lesnik's answer. Your positions after step 6 will be four points forming a square (right figure).
Question. Is there a rule by which it can be determined that the pattern "a-b-c" will return (convergence) to the starting point? Here a, b, c are integer.

Looks like for any $a$, $b$ and $c$ following your algorithm you will return to the starting point.
After steps 1-6 you shift from original position by some vector $\vec{x_1}$ and turn by 90 degrees. Now you repeat steps 1-6, and shift from your current position by some vector $\vec{x_2}$. Which is exactly like $\vec{x_1}$, but turned by 90 degrees. And so on. Sum of vectors $\vec{x_1}$, $\vec{x_2}$, $\vec{x_3}$ and $\vec{x_4}$ is zero because all of them have the same length, and each one is turned by 90 degrees.
Your positions after step 6 will be four points forming a square.