First, you can convert the recursion to
y_(k+1) = y_k + y_(k-1)
where y_k = x_k - 4. Then you can write
[y_(k+2);y_(k+1)] = A * [y_(k+1);y_k]
where A = [1,1;1,0]. Then find the eigenvalues and eigenvectors of A.
[V,D] = eig(A);
Hence you would have
[y_13; y_12] = A^11*[y_2;y_1] = (V*D^11*V')*[y_2;y_1]
where D^11 is merely the eleventh power of the diagonal eigenvalues. This expresses the value of y_13 (and therefore x_13) directly in terms of y_2 and y_1 without any iterated steps.
The Wikipedia article at:
http://en.wikipedia.org/wiki/Fibonacci_number
describes this procedure in their section 5 - Matrix Form.
