This is derived from $S_a=\sum_{i=0}^m(n+ai)=\frac12(m+1)(2n+am)$, $(n,m)\ge1$ when $a=3$
A number that works is some number where some $n,m$ Plugged into $S_a$ will result in that number. So $[n,m]=[1,2]$ plugged into $S_1=6$, so 6 works for $S_1$
Note: if $a=1$, $S_1$ can be every number that is not a power of $2$, and will never be a power of $2$. If $a=2$, $S_2$ can be every composite number but never a prime.
I found that for the same reasons $S_1$ can never be a power of $2$, $S_3$ can never be a power of $2$
I also found that $S_3$ can be every odd number greater than $4$($5$ being the smallest number $S_3$ can ever be) because $(n)+n+3=2n+3$, which if n is positive, will generate every odd number greater than $4$.
I also found that the numbers generated by $S_3=\{5+2x\}\cup\{12+3x\}\cup\{22+4x\}\cup\{35+5x\}\cup\cdots\cup\{\frac12(3k^2-k)+kx\}\cup\cdots$, $x=\{0,1,2,3,\cdots\}$
I then used that to find numbers that didn't work: https://www.desmos.com/calculator/ebklyzijdy $S_3\ne1,2,3,4,6,8,10,14,16,20,28,32,44,52,64,68,76,88,104,128,136,152,184,208,232,248,256,272,296,304,328,344,368,464,496,512,592,656,688,736,752,848,928,944,976,992,1024,1072,1136,1168,1184,1264,1312,1328,1376,1424,\cdots$
I do see a pattern though. Only primes times powers of $2$ don't work, for instance $496=2^4\cdot31$, $1376=2^5\cdot43$.
So that's where I am in finding the pattern. I think it may have to do with every number with $2$ odd factors working but I don't know.
First let's properly define $S_a$:$$S_a=\left \{ x \mid (\exists n,m \in \mathbb{N}) \left [x = \frac12(m+1)(2n+am) \right] \right \}$$
We are interested in the set $S_3$. I will separate the numbers into two classes based on the parity of $m$.
Say that $m$ is even. Let's write $m = 2k$. Then we can write our expression as $\frac12(2k+1)(2n+6k) = (2k+1)(3k+n)$.
Say that $m$ is odd. Let's write $m = 2k+1$. Then we can write our expression as $\frac12(2k+2)(2n+6k+3) = (k+1)(2n+6k+3)$.
What can we conclude from this? Well in case one $2k+1$ is a factor, and in case 2 $2n+6k+3$ is a factor, both of which are always odd. Therefore we can conclude that in all cases any $x \in S3$ must contain an odd factor. In other words, powers of two can't occur, and that observation of yours is correct.
When $k = 0$, case one is impossible (because then we'd have $m = 0$), but we see that case two simply becomes $2n + 3$. So every odd number $\geq 5$ is in $S_3$.
The only category of numbers left to explore is those of the form $2^s \cdot d$ with $d$ odd. In fact, I will factor further and write $x = 2^s \cdot p \cdot r$ where $p$ is the smallest odd factor of $x$, and $r$ is odd (and may be $1$).
$(2k+1)(3k+n) = 2^s\cdot p\cdot r$. To maximize the amount of numbers for which this is valid we find that $2k+1 = p$ and $3k+n = 2^s\cdot r$. This is possible when $3k < 2^s\cdot r$, and thus $p < \frac{2}{3}\cdot 2^s\cdot r + 1$ is the criterion for case one that allows the most numbers.
$(k+1)(2n+6k+3) = 2^s\cdot p\cdot r$. Here the most permissive strategy is to set $k+1 = 2^s$, and $2n + 6k + 3 = p\cdot r$. This is possible if $6k + 3 < p\cdot r$, which simplifies to $\frac{6\cdot 2^s - 3}{r} < p$.
Combining both cases we find that any number where
$$p < \frac{2}{3} \cdot 2^s \cdot r + 1 \quad \vee\quad \frac{6\cdot 2^s - 3}{r} < p $$
is in $S_3$ (in addition to any odd number $\geq 5$). Using DeMorgan's law we can invert this so find the criterion for numbers not in $S_3$:
$$\frac{2}{3} \cdot 2^s \cdot r + 1 \leq p \leq \frac{6\cdot 2^s - 3}{r}$$
But if $r \neq 1$ this inequality is impossible, thus we can conclude that $p$ is the only prime factor of $d$. Thus the only numbers bigger than $5$ not in $S_3$ (in addition to the powers of two) are those of the form $p \cdot 2^s$ where $p$ is prime and $s \geq 1$ and
$$\frac{2}{3} \cdot 2^s + 1 \leq p \leq 6\cdot 2^s - 3$$