Counting/Combinatoric Problem involving digits

Arcmed

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Dec 21, 2022
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How many integers with 13 digits are there, that start with 1 and for which every digit differs by at most 1 from the previous digit (e.g. 1234 is a valid number, but 1342 is not)?

The answer is: 742,900. However, I do not know how they came up with the answer.

My closes what I came up with is this:
1. The first digit there is only 1 possibility.
2. The second digit has 3 choices: 0 1 2
3. The third digit has: (2 + 3 + 3) = 8
... It will eventually lead to a number bigger than 742,900.
 
This is an interesting problem. I got an answer greater than the provided answer. Need to think about it some more.
 
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