Just write the word!

Let $a_n$ be the number of ways to get to the $n$ -th "layer". We know $a_2=8$ and we want to find $a_6$ .

We make the following observations about $a_{n+1}$ :

1. Out of $a_n$ ways, 4 of them end at corners. From each of these corners, there are 5 neighbouring letters.

2. The rest of the $a_n-4$ ways have 3 neighbouring letters.

Hence $a_{n+1}=3(a_n-4)+4*5$ $\Rightarrow a_{n+1}=3a_n+8$

Given $a_2=8$ , we get $a_6=\boxed{968}$

Has anyone not noticed that "SOLDAG" backwards is "GLADOS"?

Anyways, we consides the top quarter of the figure first, i.e. how many ways to get to the G's to the top. Starting from the S, you can move $5$ times upward either in a north-east, north or north-west direction so $3$ choices and $5$ moves, giving us $3^5 = 243$ ways to get to the top layer of G's.

We multiply this number by $4$ to obtain the number of ways to get to all the G's on the sides. However, this double-counts the number of ways to get to the corners. Since there is exactly $1$ way to get to any corner G, and there are $4$ such corners, we subtract the result by $4$ .

Our answer is then $(4)(243)-4 = \boxed{968}$ .

I made the same observations as Shaun Leong, but took a different path afterwards (3 ways everywhere and 2 additional ways at the 4 corners (each layer).

I ended up with the following closed (non-recursive) formula:

$8 × 3^4 + 4 × 2 × 3^3 + 4 × 2 × 3^2 + 4 × 2 × 3 + 4 × 2 = \boxed {968}$

If we also find, that 4 × 2 = 8 and therefore, we are dealing with a geometric series, we can get a general formula for n letters:

$8 × {{3^{n-1}-1} \over {3 - 1}} = 4 × (3^{n-1} -1) = 4 × 3^ {n-1} - 4$

And this is the general form of the formula, what Manuel Kahayon used, as for n = 6 ("SODALG" has 6 letters), we get

$4 × 3^5 - 4 = \boxed {968}$