Inverse Z-transforms
$ x[n] = \begin{cases} 1, & n = 4 \\ 2, & n = 5 \\ 3, & n = 2 \\ 0, & \mbox{else} \end{cases} $
This is equivalent to
$ \begin{align} x[n] &= \delta[n-4] + 2\delta [n-5] + 3\delta [n-2], \\ X(z) &= z^{-4} + 2z^{-5} + 3z^{-2}, (\text{converges for all } z \neq 0) \\ &= \left[z^{-4} + 2z^{-5} + 3z^{-2}\right] \sum_{n=-\infty}^{\infty} \delta[n]z^{-n} \\ &= \sum_{n=-\infty}^{\infty} \delta[n]z^{-n-4} + 2\sum_{n=-\infty}^{\infty} \delta[n]z^{-n-5} + 3\sum_{n=-\infty}^{\infty} \delta[n]z^{-n-2} \end{align} $
Using a change in variables to bring equation to the right form,
$ \begin{align} j = n+4 \\ k = n+5\\ l = n+2 \\ X(z) &= \sum_{j=-\infty}^{\infty} \delta[j-4]z^{-j} + 2\sum_{k=-\infty}^{\infty} \delta[k-5]z^{-k} + 3\sum_{l=-\infty}^{\infty} \delta[l-2]z^{-l} \\ x[n] &= \mathcal{Z}^{-1}(X(z)) \text{ and } X(z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n}\\ x[n] &= \delta[n-4] + 2\delta [n-5] + 3\delta [n-2] \end{align} $
