Given occasional LFSR samples can the next sample be computed?
It is very easy to predict and should not be used as a main cryptographic component on its own.

Firstly, many $k-$decimations $s_{kt}$ of LFSR sequences (those with $gcd(k,2^n-1)=1$) are shifted LFSR sequences themselves so this can be used for prediction.

If you can find a template $T=\{0,t_1,\ldots,t_{n-1}\}$ (with $0&lt;t_1&lt;t_2&lt;\cdots&lt;t_{n-1}$ such that the samples you have of the LFSR sequence $(s_t)$ cover $n+1$ shifts of this template you can set up a linear system and solve for the recurrence. In fact assuming $n=3$ one can (Berlekamp Massey is more efficient of course, this would need Gaussian elimination) set up the system

$$c_0 x_t + c_1 x_{t+1} + c_2 x_{t+2}=x_{t+3}\\
c_0 x_{t+1} + c_1 x_{t+2} + c_2 x_{t+3}=x_{t+4}\\
c_0 x_{t+2} + c_1 x_{t+3} + c_2 x_{t+4}=x_{t+5}$$

and solve for the LFSR coefficients $c_i$. For the template I mentioned, a similar system needs to be solved. This case is the template $T=\{0,1,2\}$ with shifts also $\{0,1,2\}.$