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(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.)

Vepir
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Jose Arnaldo Bebita Dris
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    Note $\sigma(2^r) = 2^{r+1}-1$. – Daniel Fischer Feb 13 '16 at 15:15
  • @DanielFischer, is it possible to compute the leading decimal digit of $2^{r+1} - 1$ using some formula? – Jose Arnaldo Bebita Dris Feb 13 '16 at 15:22
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    Sort of. For $k \geqslant 4$, the leading decimal digit of $2^k - 1$ is the leading decimal digit of $2^k$. Then you have $\log_{10} 2^k = k\cdot \log_{10} 2$, and from the fractional part of that you can read off the leading digit(s). For large $k$, however, computing the logarithm with sufficient accuracy becomes complicated. – Daniel Fischer Feb 13 '16 at 15:27
  • Heuristically, Benford's law applies so that we can say that the leading digit is $1$ quite often ... – Hagen von Eitzen Feb 13 '16 at 15:28

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