The answer are all $c$ that are of form $u^2$ or $t^2+t$, with $u,t$ being arbitrary. Suppose $a_n$ is a perfect square for $n=2k$. Then, $a_{2k} = 4^{2k}c + (4^{2k}-1)/5 = u^2$, yielding $u^2\equiv -1\pmod{4}$ if $k\ge 1$, a contradiction. On the other hand, $a_0 = c$ so $c=u^2$ works. Next, let $a_n$ is a perfect square for $n=2k+1$, so $(5c+1)4^{2k+1} +1 = 5u^2$ for some $u$. If $k\ge 1$, then $4^{2k+1}\equiv 0\pmod{8}$, so $5u^2\equiv 1\pmod{8}$, a contradiction. So, $k=0$ and $n=1$. If $a_1 = 4c+1$ is a square, then $4c+1 = u^2-1$, thus $u$ is odd. Set $u=2t+1$ and arrive at $c=t^2+t$.