Let me first explain what I call a "super-palindrome":

Consider the number $99999999$. That number is obviously a palindrome. ${}{}{}{}$ The largest prime factor of $99999999$ is $137$. If you divide $99999999$ by $137$, you get $729927$. This number is also a palindrome.

The largest prime factor of $729927$ is $101$. $729927/101=7227$ which again is a palindrome.

The largest prime factor of $7227$ is $73$. $7227/73=99$ which again is a palindrome.

By further dividing by the largest prime factor, you get $9$, $3$ and finally $1$, which, being one-digit numbers, are also palindromes. Since $1$ has no prime factors, the procedure ends here.

Now generalizing this observation, I define a super-palindrome as a palindrome which is either $1$, or which gives another super-palindrome if divided by its largest prime factor.

Note that not all palindromes are super-palindromes. The smallest palindrome which is not a super-palindrome is $252$.

Now an obvious question is whether there are infinitely many super-palindromes. One obvious possibility would be if there are infinitely many palindromic primes (because each palindromic prime is a super-palindrome). However according to Wikipedia that question is still open (which of course means there won't be an affirmative "no" to my question, as that would imply an answer to that question as well).

However, since the prime factors of the super-palindromes (except for the smallest one) do not have to be palindromes themselves, there might be other ways to prove that there are infinitely many super-palindromes (assuming there are). Maybe there's a very easy argument to see that there must be infinitely many super-palindromes which I just don't see.

So does anyone have an idea?

BTW, is this already a known concept? At least OEIS doesn't seem to have the sequence.