TITLE: How many numbers of length N(N is the number of digits) are possible
QUESTION [7 upvotes]: Given $N \le 10^9$, determine how many numbers of length $N$ are possible with the following constraint: the adjacent digits of the number should have an absolute difference of $1$.
For eg, for $N = 3$, numbers starting with $1$ are $101,121,123$. For $N =4$, numbers starting with $1$ are $1210,1212,1232,1234$. 
My approach to this problem is: first digit could be anything from $1$ to $9$. The second digit could be firstDigit$+1$ or firstDigit $-1$. And so on, if we do, we get the total number of possible numbers are $9\cdot2^{N-1}$. But the catch is, we must subtract those numbers with digit less that $0$ or greater than $9$. 
I am stuck here. How to get a formula to subtract these numbers from the total.
Or is there a better approach to solve this problem?

REPLY [7 votes]: I try a different approach: I write a $10 \times 10$ transition matrix (one dimension per digit, like a markov chain, but this time it is not about the probabilities, but about counting the number of possibilities) where each entry $a_{ij}$ says whether we can go from digit $j$ to digit $i$. Lets say that the top left entry has indices $(0,0)$ just for this example (I know this is not usually done like this) This matrix looks as follows:
$A= \begin{pmatrix}
 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
\end{pmatrix}$
If we want to know what the next digit could be if we are at $n=4$ we just multiply the matrix $A$ with the vector $e_4 = (0,0,0,0,1,0,0,0,0,0)^t$ (the one is in the fourth position when beginning counting at 0. Then we get the vector $$Ae_4 = (0,0,0,1,0,1,0,0,0,0)^t$$.
After two steps (so for the third digit) we get the following vector:
$$A^2e_4 = (0,0,1,0,2,0,1,0,0,0)^t$$
and so forth. So the $n$th entry of this vector says how many numbers have the digit $n$ as their third digit (=after two steps).
So the total of the new possibilities of those numbers is just the sum of the entries of the new vector. This is if we just take only one step (transition from one to the next digit). So when we want to get the possibilities of the next step, we just have to multiply with $A$ one more time. 
As we want all combinations with all starting numbers we have to use the starting vector $v=(0,1,1,1,1,1,1,1,1,1)^t$ (since we cannot start with the digit 0 as @Mathmo123 mentioned (Thanks!))
In total we have $N-1$ transitions, so $v$ needs to be multiplied by $A^{N-1}$. For $N=9$ get:
$$A^8 v = (56,126,153,208,208,228,201,181,125,70)^t$$
Which makes a total of 1556. Or for $N=3$ we get
$$A^2 v = (1,3,3,4,4,4,4,4,3,2)$$
which makes a total of $32$.
Edit
If you want the total of all combinations from $N=1,...,10^9$ we can make use of the geometric series. 
We want to sum up all the solutions from all those $N$, then we get (let m=10^9$ be the maximal exponent):
$Iv+Av+A^2v+A^3v+...+A^mv = (I+A+A^2+A^3+...+A^m )v = (I-A)^{-1}(I-A^{m+1}) v$
Of course here we still need a fair amount of matrix multiplication, especially for calculating $A^{m+1}$.
Appendix
PS: Of course I did not calculate this by hand, I just quickly did it in Octave/Matlab: 
a=diag(ones(9,1),-1);
a=a+a'; %now we have our finished matrix a
e4 = [0,0,0,0,1,0,0,0,0,0]'; 
%the two explaining examples
disp(a*e4);   
disp(a^2*e4);
%our resulting vector:
 v = ones(10,1);
v(1)=0;
disp(a^8 * v);
%the total for N=9
disp( sum(a^8 * v));