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.

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    If there are infinitely many primes of the form $10^k + 1$, then there are infinitely many superpalindromes. Start with a prime number $p_0$, then choose a prime $p_1 = 10^{k_1} + 1$ where $k_1$ is at least as big as the number of digits of $p_0$. Then $p_0p_1$ is a superpalindrome. Then choose $p_2 = 10^{k_2} + 1$ with $k_2$ at least as big as the number of digits of $p_0p_1$. Then $p_0p_1p_2$ is a super palindrome... Having said all that, I have no idea whether the statement "There are infinitely many primes of the form $10^k + 1$" is true or false. – Michael Albanese Sep 22 '12 at 19:41
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    @MichaelAlbanese: If there are infinitely many primes of the form $10^k + 1$, then automatically there are infinitely many palindromic primes, which is unknown :) – kennytm Sep 22 '12 at 19:45
  • @MichaelAlbanese: Since all primes of the form $10^k+1$ are palindromic primes, this would imply that there are infinitely many palindromic primes (which as I mentioned is an open question). You are right that it would be sufficient for there to be infinitely many palindromic primes, though. – celtschk Sep 22 '12 at 19:45
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    BTW, for the primes of the form $10^k+1$ there's already another [math.SE question](http://math.stackexchange.com/q/34877/34930). Note that according to the discussion linked to by one of the answers, there are no primes for $2 – celtschk Sep 22 '12 at 19:52
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    Hmm... that's one of the most interesting ways I've proved a theorem by assuming it's true. I think I need sleep :) – Michael Albanese Sep 22 '12 at 19:56
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    the only obvious way (to me) to approach this question without referring to the conjecture about palindromic primes would be to exhibit infinitely many super-palindromes from some finite suite of primes. A quick script shows that there are only finitely many super-palindromes with prime divisors no larger than, say, 200,000, which of course doesn't prove that the approach won't work, but certainly you'll need to use some big primes. – Dustan Levenstein Sep 22 '12 at 20:19
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    Your example 99999999 works because $137\cdot 101\cdot 73 = 10^6+1$ and the smallest factor 73 has two digits and $10^2-1$ is a superpalyndrome, and 101 makes palindromes from palindromes if there is no carry. This may look like a bit of luck, but maybe also inspire something? – Hagen von Eitzen Sep 27 '12 at 19:30
  • Observation: as palindromes are base dependent, any number $n < b$, where $b$ is base of the numeral system, is a superpalindrome. $b\rightarrow\infty$. Hence the question is: do you have any specific constraints on $b$? Does it necessarily have to be $10$? – peterph Nov 08 '12 at 15:21
  • You might want to rephrase that @peterph: http://stackoverflow.com/a/364904/983722 – Dennis Jaheruddin Nov 16 '12 at 17:34

1 Answers1


Maybe it would be more productive to think about this the other way around. Instead of breaking down super-palindromes, build them up. Then the question becomes one of testing whether products of known super-palindromes are palindromes.

Victor Engel
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    Actually the question becomes whether products of a known super-palindrome with a prime larger than the largest prime factor of that super-palindrome are palindromes. – celtschk May 26 '13 at 13:36
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    This seems to be the correct way to approach the problem, but a proof might be difficult to nail down. – recursion.ninja Jun 16 '13 at 00:01