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?
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 $6$.