The formula $c(n)=a(n+1)$ is pretty much straightforward, noticing that
$$\lfloor \log_{i+1}(n-i)\rfloor -  \lfloor\log_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$
The latter condition means that $n+1=k+k^m$ with $k:=i+1$.