Reversing the digits of $2^n$ to yield a prime power

Question

This question asks about the existence of an $n\in\Bbb N$ such the number obtained reversing de digits of $2^n$ is a power of $7$. The same question with $5$ instead of $7$ was asked by Freeman Dyson, according to this post. So I asked myself: for which $n\in\Bbb N$ is the number obtained reversing de digits of $2^n$ a prime power? Clearly $n=1,2,3$ work, as do $n=4,5$, since $61$ and $23$ are prime. A computer search up to $n\le 10\,000$ found $$ 1,2,3,4,5,17, 24, 37, 45, 55, 70, 77, 107, 137, 150, 271, 364, 1157, 1656,\\ 2004, 2126, 3033, 3489, 3645, 4336, 6597, 7279 $$ Except for $n=2,3$, the number obtained reversing de digits of $2^n$ is prime (sequence A057708). So, my question is:

Is there an an $n\in\Bbb N$, $n\ne2,3$, such that the number obtained reversing de digits of $2^n$ is a prime power with exponent at least $2$?

Do you restrict your attention to base $10$, or are other bases (besides powers of $2$) allowed? — Servaes, Aug 25 '18 at 20:14
At this moment, base 10 is my main focus. But of course other bases and powers are posible. — Julián Aguirre, Aug 26 '18 at 08:06
No further prime powers upto $n=10^4$, hence a further example would be a miracle , but I do not see a way to rule it out. — Peter, Aug 27 '18 at 17:34
@JuliánAguirre, I was reacting to the first sentence and I quote "...such the number obtained reversing de digits of 2^n is a power of 7." — user25406, Aug 27 '18 at 21:09
I also quote:for which $n\in\Bbb N$ is the number obtained reversing de digits of $2^n$ a prime power? — Julián Aguirre, Aug 27 '18 at 22:31

score 0 · Answer 1 · answered Mar 08 '19 at 00:28

Let $r$ be the reversal function.
$2^{10} = r(7^4)-r(3^4)$
$6^{3} = r(2^9)+1$
$19^{2} = r(7^3)+r(3^4)$
$54^{2} = r(2^{13})-2$
$252^{2} = r(2^{16})-r(5^2)$
$973^{2} = r(7^6)-r(3^4)$

More interesting is to ask how close one can get, setting up a reverse-digit ABC conjecture equivalent.

Reversing the digits of $2^n$ to yield a prime power

1 Answers1