Consecutive Prime Powers

Let us define a set of 3 numbers $N-1, N, N+1$ as a sequence. Note that no matter what $N$ is, at least one of the numbers is divisible by 2. The same holds for 3. Since $N-1$ and $N+1$ have the same parity while prime towers of 2 and 3 do not, we know that $N$ must be a power of 2 or 3. We can now separate the problem into two cases:

Case 1: $N$ is a power of 2

If $N$ is a power of 2, then we know that either $N-1$ or $N+1$ is a power of 3.

If $N-1$ is a power of 3, then we have that $2^a-1=3^b$ for some $a$ and $b$ . Taking both sides mod 3, we get $2^a \equiv 1 \pmod{3}$ , which only occurs if $a$ is even. Letting $a=2r$ and substituting back into our original equation, we have $2^{2r}-1=(2^r-1)(2^r+1)=3^b$ , thus implying that both $2^r-1$ and $2^r+1$ are powers of 3. Because these are not equivalent mod 3, one of these must be 1, so we have $2^r-1=1 \rightarrow r=1$ . This yields $a=2$ and thus our sequence is $\boxed{3,4,5}$ , which works.

If $N+1$ is a power of 3, then we have that $2^a=3^b-1$ . For $a=1, b=1$ , the sequence becomes $1,2,3$ which does not work. Therefore, we have that $a \ge 2$ and thus $4|2^a$ . Taking both sides of our original equation mod 4, we get $3^b \equiv 1 \pmod{4}$ which occurs only when $b$ is even. Then let $b=2s$ . Proceeding similarly to our first case, we get $(3^s-1)(3^s+1)=2^a$ . Both multiplicands are thus powers of 2. The only two powers of 2 with a difference of exactly 2 are 2 and 4, so we then have that $3^s-1=2$ and $3^s+1=4$ , yielding $s=1$ and thus $b=2$ . We then have that our sequence for this case is $\boxed{7,8,9}$ which, again, works.

Case 2: $N$ is a power of 3

In this case, either $N-1$ or $N+1$ is a power of 2. If $N-1$ is a power of 2, then we have $3^a-1=2^b$ . For $b \ge 2$ we have already seen that the only solutions are $a=2$ and $b=3$ , which in this case yields the sequence $8,9,10$ , which does not work. For $b=1$ , we get $a=1$ and $\boxed{2,3,4}$ as a sequence, which does indeed work.

If $N+1$ is a power of 2, then we have $2^a-1=3^b$ . We have already seen that the only solution to this equation is $a=2, b=1$ . If this is the case, our sequence is then $2,3,4$ , which we have already counted.

Thus, our solutions are $[2,3,4], [3,4,5], [7,8,9]$ and our answer is then $3+4+8=\boxed{15}$