How to prove this set is infinite? Suppose there is a set $\{o_n\} \subset \Bbb{N} : 3o_n + 1 = 2^k$.
How do I prove that there are infinitely many numbers in $o_n$? It seems like there should be because there are infinitely many powers of $2$, but I don't know how to formally prove it.
The first few values are $o_n = \{1, 5, 21, 85, 341, 1365, 5461...\}$
 A: If I have understood your question well, you are asking if there are infinitely many powers of two that are congruent to $1$ mod $3$.
Indeed there are, since
$$2^{2r}=4^r\equiv 1^r=1\pmod 3$$
for any positive integer $r$.
A: I put your sequence in the OEIS and got this result: $$\frac{4^n - 1}{3}$$ (but of course they write it as (4^n - 1)/3.)
That's not enough to prove the sequence is infinite, of course. Instead look at the powers of $4 \pmod 3$ to get $1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...$ This is because $4 \equiv 1 \pmod 3$, and since $1$ is a multiplicative identity in whatever number system it occurs, multiplying it by itself however many times you want just keeps giving you the same $1$ over and over again. If you subtract $1$ from each of those that gives you a sequence of $0$s, thereby proving that $3o_n + 1 = 4^k$ (or $2^k$ with the proviso that $k$ is even).
If you look at the powers of $2 \pmod 3$ the sequence is a little more interesting: $2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...$ But the previous sequence is enough to prove.
One more tidbit of assurance, look at the powers of $4$ in ternary (which is a lot like decimal but based around $3$ rather than $10$): $11, 121, 2101, 100111, 1110221, ...$
