For this to be an integer, $a$ must be of the form $kb(b+1)+b$ for some positive integer $k$. This makes the expression $k(b+1)+1+kb+1=k(2b+1)+2$. This means that any number $x$ such that $x-2$ is not a multiple of any odd number greater than $1$ (or a power of $2$) cannot be expressed, meaning the answer is the set $\{2^k+2: k\in \mathbb{N}_0\}$.