How many hexagonal paths?

Question

Here is a hexagonal tiling, borrowed from Wikipedia.

I start in any hexagon on the left hand side. I end at any hexagon on the right hand side. I can only travel to the right, not up, down or backwards.

In how many ways can this be done?

Answer 1

The answer is

74280

Method

Every cell in the first column can be reached 1 way.
Other cells can be reached as many different ways, as the sum of their two left neighbours, giving:

The sum of the numbers in the last column is the answer to the question posted.

Answer 2

The correct answer might be:

73392?

Reasoning:

Well, I made a Pascal's Triangle-like chart with alternating 11 and 12 columns, and 14 rows, where each number was the sum of the two numbers northwest and northeast of it. I added all of the numbers on the bottom row. It's quite likely I messed up some addition, though.

Answer 3

Answer:

$83,746$

Method:

Consider the first 2 columns of this diagram. There are 11 starting positions, and 12 destinations. Each starting position can go to one of 2 destinations, so the number of possible paths is equal to $11 \cdot 2 = 22$.

Now, consider columns 2 and 3 of the diagram. There are 12 starting positions, and 11 destinations. 10 of the starting positions have 2 possible destinations, and 2 of them only have 1 possible destination (the top and bottom ones). This means the total number of possible paths is $(10 \cdot 2) + 2 = 22$.

If we now consider the first 3 columns, it can be seen that each of the internal starting positions has 4 possible paths, and the top and bottom ones have 3. Therefore, in total, there are $(9 \cdot 4) + (2 \cdot 3) = 36 + 6 = 42$ possible paths across the first 3 columns.

In the initial consideration, the inner hexes were each finished on twice, and the outer ones once. Therefore, a second way to work out the number of paths in the third consideration, would be to edit the calculation for consideration 2 to be $((10 \cdot 2) \cdot 2) + (2 \cdot 1) = 40 + 2 = 42$.

Next, considering the first 4 columns, each of the paths already determined for the first 3, has 2 more possible destinations, so the number of paths would total $42 \cdot 2 = 84$.
Using these values, I can determine a formula for calculating the number of paths, where n increases by 1 for every repeating pair of columns.
$\Sigma_n = ((\Sigma_{n-1} - n) \cdot 4) + 2n$

Therefore, we can step through this calculation to determine the results of:
$\Sigma_1 = 22$
$\Sigma_2 = 84$
$\Sigma_3 = 330$
$\Sigma_4 = 1312$
$\Sigma_5 = 5238$
$\Sigma_6 = 20,940$
$\Sigma_7 = 83,746$
As there are 7 pairs of columns in the above image, (if my formula is correct), I believe there to be $83,746$ possible paths.

Bonus Question:

$\Sigma_n = ((\Sigma_{n-1} - n) \cdot 4) + 2n$ where $n=2k$