Combination of smartphones' pattern password

I believe the answer can be found in OEIS. You have to add the paths of length $4$ through $9$ on a $3\times3$ grid, so $80+104+128+112+112+40=576$

I have validated the $80$, $4$ number paths. If we number the grid $$\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9 \end{array}$$

The paths starting $12$ are $1236, 1254, 1258, 1256$ and there were $8$ choices of corner/direction, so $32$ paths start at a corner. Starting at $2$, there are $2145,2147,2369,2365,2541,2547,2587,2589,2563,2569$ for $10$ and there are $4$ edge cells, so $40$ start at an edge. Starting at $5$, there are $8$ paths-four choices of first direction and two choices of which way to turn

Added per user3123's comment that cycles are allowed: unfortunately in OEIS there are a huge number of series titled "Number of n-step walks on square lattice" and "Number of walks on square lattice", and there is no specific definition to tell one from another. For $4$ steps, it adds $32$ more paths-four squares to go around, four places to start in each square, and two directions to cycle. So the $4$ step count goes up to $112$. For longer paths, the increase will be larger. But there still will not be too many.

I don't have the answer as "how to mathematically demonstrate the number of combinations". Still, if that helps, I brute-forced it, and here are the results.

• $1$ dot: $9$
• $2$ dots: $56$
• $3$ dots: $320$
• $4$ dots: $1624$
• $5$ dots: $7152$
• $6$ dots: $26016$
• $7$ dots: $72912$
• $8$ dots: $140704$
• $9$ dots: $140704$

Total for $4$ to $9$ digits $:389,112$ combinations