These are the known numbers so far:
$$
S = \left(\begin{matrix}
0 & 120 \\
1 & 106 \\
2 & 112 \\
3 & 108 \\
4 & 80 \\
5 & 92 \\
6 & 84
\end{matrix}\right)
$$
When investigating the numbers, I noticed a pattern in them:
$$
\Delta S = 112 - 106 = 6 \\
\Delta S = 112 - 108 = 4
$$
$$
\Delta S = 92 - 80 = 12 \\
\Delta S = 92 - 84 = 8
$$
These number seem to be related to each other in groups of 3 and their $\Delta$'s between each other seem to be doubling. Backwards engineering using this, leaves us with:
$$
\Delta S = a - 120 = 2 \\
a = 2 \\
S(-1) = 122
$$
Now let's compare the highest numbers of all the known triplets:
$$
\Delta S = 122 - 112 = 10 \\
\Delta S = 112 - 92 = 20
$$
Which confirms my theory that the pattern must be:
$$
\text{in }\mathbb{Z} \\[12pt]
S(x) = \begin{cases}
S(x+1) - \left(6 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) & \text{if } ((x-1) \text{ mod } 3) = 0 \\
132 - \left(20 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) & \text{if } ((x-2) \text{ mod } 3) = 0 \\
S(x-1) - \left(4 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) & \text{if } (x \text{ mod } 3) = 0
\end{cases}
$$
The next number in the sequence would be:
$$
\begin{align}
S(7) &= S(7+1) - \left(6 * 2^{\left\lfloor\frac{7-1}{3}\right\rfloor}\right) \\
&= \left(132 - \left(20 * 2^{\left\lfloor\frac{8-1}{3}\right\rfloor}\right)\right) - \left(6 * 2^{\left\lfloor\frac{7-1}{3}\right\rfloor}\right) \\
&= \left(132 - \left(20 * 4\right)\right) - \left(6 * 4\right) \\
&= \left(132 - 80\right) - 24 \\
&= 52 - 24 \\
&= 28
\end{align}
$$