## Motzkin numbers as a convergent of iterative matrix inversion

I have here left out the indices of the elements for easier typing but hopefully this can still be understood.

${A=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\\ x&x&x&x&x&x&1&0\\ x&x&x&x&x&x&x&1\end{array}\right)$

Shift down the elements in matrix ${A}$ two steps to get matrix ${B}$.

${B=}\left(\begin{array}{cccccccc}0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\end{array}\right)$

Add two diagonals of ones to the main diagonal (where row index = column index) and to the diagonal below (where row index = column index + 1) in matrix ${B}$ to get matrix ${C}$.

${C=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 1&1&1&0&0&0&0&0\\ x&1&1&1&0&0&0&0\\ x&x&1&1&1&0&0&0\\ x&x&x&1&1&1&0&0\\ x&x&x&x&1&1&1&0\\ x&x&x&x&x&1&1&1\end{array}\right)$

Change the sign of all the elements below the main diagonal in matrix ${C}$ so that you get matrix ${D}$.

${D=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -1&-1&1&0&0&0&0&0\\ -x&-1&-1&1&0&0&0&0\\ -x&-x&-1&-1&1&0&0&0\\ -x&-x&-x&-1&-1&1&0&0\\ -x&-x&-x&-x&-1&-1&1&0\\ -x&-x&-x&-x&-x&-1&-1&1\end{array}\right)$

Calculate the matrix inverse of matrix ${D}$ to get matrix ${E}$

Iterate by replacing matrix ${A}$ with matrix ${E}$.

Matrix ${E}$ should then converge to the Motzkin numbers in all columns.

Keywords: Matrix inversion, Motzkin numbers, iteration.