(Note: This post is a bit related to this earlier MSE question.)
The title says it all.
Is $\sigma(2^r)$ a palindrome (in base $10$) for some $r > 2$, where $\sigma$ is the sum-of-divisors function?
If $r = 1$, then $\sigma(2^r) = \sigma(2) = 3$ is trivially a palindrome.
Similarly, if $r = 2$, then $\sigma(2^r) = \sigma(4) = 7$ is also trivially a palindrome.
I noticed that $$\sigma(2^r) \equiv \begin{cases} { 1 \pmod{10}, \hspace{0.1in} r \equiv 0 \pmod 4 \\ 3 \pmod{10}, \hspace{0.1in} r \equiv 1 \pmod 4 \\ 7 \pmod{10}, \hspace{0.1in} r \equiv 2 \pmod 4 \\ 5 \pmod{10}, \hspace{0.1in} r \equiv 3 \pmod 4 } \end{cases} $$
and that the leading decimal digit of $\sigma(2^r)$ does not coincide with the trailing decimal digit, for almost all $r > 2$. (I arrived at this observation by coding the appropriate cell formulas in a spreadsheet.)