Let $\text{reverse}_{10}(n)$ be the digit reversal (in base 10) of $n$, and let $\text{sopfr}(n)$ be the sum of $n$'s prime factors, with repetition. What numbers $x$ exist such that $x = \text{sopfr}(\text{reverse}_{10}(x))$? To keep things interesting, let's also say that $x$ must be composite.
Through brute force, the only numbers that I've been able to find are of the form $5\times10^k-1$ such that $5\times10^k-3$ is prime. (4, 49, 4999999999999999...) Are these the only composite numbers that have the above property?
What if we were to use a different base? Using base 11, for example, seems to yield many more numbers. Is there some reason behind this? Is there some known way of enumerating those?