<s>[INST] Why does this strategy sieve out composites without factors $2,3$? Take
$$
3 + 2 = 5 \\
3 + 4 = 7 \\
5 + 6 = 11 \\
5 + 8 = 13 \\
7 + 10 = 17 \\
7 + 12 = 19 \\
9 + 14 = 23 \\
9 + 16 = 25 \\
\cdots
$$
I'm having trouble writing down what the pattern is to start with.  
I think it's $2n + 1 + 2n + 2$ and $2n + 1 + 4n$ interwoven so the list essentially enumerates the two sets $4\Bbb{N} + 3$ and $6\Bbb{N} + 1$.  So how do I go from there to say that the list of right hand side is the ordered list of all numbers not divisible by $2,3$?
Secondly, how do we generalize this summation procedure to filter out more composites?
 [/INST] Working modulo $6$, we know that numbers divisible by $2$ or $3$ will have remainders $0, 2, 3$ or $4$. Therefore, we have to show that the numbers in this list have remainders $1$ or $5$.
We have:
$(3)+(2) \equiv 5 \pmod 6$   
$(3)+(2+2) \equiv 1 \pmod 6$  
$(3+2)+(2+2+2) \equiv 5 \pmod 6$   
$(3+2)+(2+2+2+2) \equiv 1 \pmod 6$ 
and so on. Notice that from line 1 to line 2, we add $2$ to $5$, which is $1$ modulo $6$. Then, from line 2 to line 3, we add $4$ to $1$, which is $5$ again, and the cycle repeats.
Therefore, the only remainders in the sequence are $1$ and $5$, which are also the remainders which are not divisible by $2$ or $3$ modulo $6$.
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