# Factoring digital IIR filter as a causal and anticausal pass?

Background: I'm new to signal processing and I'm reading some papers on B-spline interpolation of digital signals and trying to understand how a computation is derived. If I'm given samples from some function $$f_k = f(k)$$ as input and I would like to reconstruct it as

$$f(x) = \sum_k f_k \beta^3(x-k)$$

then we need to prefilter the sequence $$(f_k)$$, since $$\beta^3$$ is not an interpolating filter. In other words, I need to convolve $$\beta^3$$ with some other sequence $$c_k$$ which to satisfies

$$f_k = c_k * \beta^3(k),$$

where $$\beta^3(k)$$ is the sequence we get by sampling $$\beta^3$$ at integer points in order to get a reconstruction that passes through my given samples at integer points. To calculate the sequence $$(c_k)$$, I need find a sequence $$\beta^3(k)^{-1}$$ that is the inverse of the sequence $$\beta^3(k)$$ in terms of discrete convolution, i.e.

$$\beta^3(k) * \beta^3(k)^{-1} = \delta_k,$$

so that

$$c_k = f_k * \beta^3(x)^{-1}.$$

One way to find this inverse is to calculate the $$z$$-transform of $$\beta^3(k)$$, which I'm denoting below by $$B^3(z)$$:

$$B^3(z) = z^{-1} + 4z^{-2} + z^{-3}$$

Then the $$z$$-transform of this "convolution inverse" is:

$$H(z) = 1 / B^3(z) = \frac{1}{z^{-1} + 4z^{-2} + z^{-3}}.$$

My question: Somehow given just the $$z$$-transform representation above of the inverse filter, it's possible to calculate $$c_k$$. This can be done in two passes by first doing a causal pass over $$f_k$$ followed by an anticausal pass to get $$c_k$$. I've yet to see anywhere explained how from the $$z$$-transform above we can derive these two steps. Anyone know how this is done?