Two different answers while doing Inverse Z-transform Given, $$X(z) = \frac{z}{3z^2 - 4z + 1}$$
Question 1

I need to calculate inverse z-transform for ROC $|z|&gt;1$

When I try to calculate inverse $z$-transform using partial fraction of $X(z)$ and using long division method, I seem to get two different answers.

Let us use partial fraction first. Since $X(z)$ is already in its proper form, the partial fraction evaluates to:
$$ X(z) = \frac{\frac 3 2}{z-1} - \frac{\frac 1 2 }{z-\frac 1 3}$$
On rearranging,
$$X(z) = \frac 3 2 \frac{z^{-1}}{1-z^{-1}}-\frac 1 2 \frac{z^{-1}}{1-\frac 1 3 z^{-1}}$$
Taking inverse z-transform,
$$x(n) = \frac 3 2 u(n-1) - \frac 1 2 (\frac 1 3)^{n-1}u(n-1) \tag{1}$$

Now on using long division method, the first few terms of the division are:
$$\frac 1 3 z^{-1} + \frac 4 9 z^{-2} + ...$$
This suggests that the signal in time domain is:
$$x(n) = \{0, \frac 1 3, \frac 4 9 , ...\}$$

But when I substitute $ n = 0,1,2,...$ in equation (1), I get the sequence:
$$x(n) = \{0,1,\frac 4 3,...\}$$

Why am I getting two different sequences for the same function $X(z)$?

Question 2

The notes my teacher provided me had this exercise done by finding partial fraction for $\frac{X(z)}{z}$ rather than $X(z)$ itself. As the given function is already in the proper form (order of denominator is greater than the order of numerator), is it incorrect to calculate partial fraction of $X(z)$ itself?

This has eaten almost three hours of my time. Any help would be appreciated. Thanks.

EDIT

For when the ROC of $X(z)$ is $|z| &lt; \frac 1 3$, $x(n)$ should be anti-causal. While shifting $x(n)$ due to presence of $z^{-1}$ in the numerator, which one among the following two is correct?

$$x(n) = - \frac 3 2 u(-n) + \frac 1 2 (\frac 1 3)^{n-1} u(-n) \tag{2}$$
or
$$ x(n) = - \frac 3 2 u(-n-2) + \frac 1 2 (\frac 1 3)^{n-1} u(-n-2) \tag{3}$$
Q1: You should ask yourself if

$$\frac{z}{3z^2 - 4z + 1}\stackrel{?}{=}\frac{\frac 3 2}{z-1} - \frac{\frac 1 2 }{z-\frac 1 3}$$

really holds. You'll find out that you forgot to scale correctly.

Q2: There is no reason to consider $X(z)/z$ instead of $X(z)$ in this case.

And concerning your question about the anti-causal sequence, a multiplication by $z^{-1}$ always corresponds to replacing $n$ by $n-1$, no matter if a sequence is right-sided or left-sided.