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Hі,

Wikipedia states that 2201 is the "only known non-palindromic number whose cube is palindromic", and lists no reference. It is in fact true that $2201^3=10662526601$, which is a palindrome. But to say there isn't any other number with this property seems a rather bold statement. Is this provable?

Thanks,

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  • $\begingroup$ @AustinMohr, sorry for the wrong quoting, I'll edit my question. But nbubis is right (the comment he made), it is a bold claim nonetheless. What is strange is that huge primes have been found (and those are computationally expensive to find), but there hasn't been found any other number like this. It's just so awkward. $\endgroup$
    – JMCF125
    Jul 1, 2013 at 22:49
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    $\begingroup$ You should not expect many. All else being equal, you'd expect the probability that $n^3$ is a palindrome to be around $n^{-3/2}$ and $\sum_{n=1}^\infty n^{-3/2}\approx 2.6$. Considering that the numbers less than $10^8$ have been searched the expected number remaining is $\sum_{n=10^8}^\infty n^{-3/2}\approx 0.0002$. $\endgroup$
    – deinst
    Jul 1, 2013 at 23:46
  • $\begingroup$ @deinst, why is the probability of $n^3$ being a palindrome $n^{-\frac 3 2}$? I didn't get that part (although the rest makes sense if that's true). $\endgroup$
    – JMCF125
    Jul 2, 2013 at 11:53
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    $\begingroup$ To be a palindrome, a number has half of its digits specified, so the probability that an arbitrary number is a palindrome is about $n^{-1/2}$. For example there are about $10^{10}$ 20 digit palindromes ($9\times 10^9$ if we disallow leading 0's). This is not exact, but is only off by at most a small constant factor, and so is good enough to indicate that one is unlikely to find another non palindrome whose cube is a palindrome. $\endgroup$
    – deinst
    Jul 2, 2013 at 14:26
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    $\begingroup$ @deinst, thanks for explaining, I get it now. However, your argument is not conclusive. For example, as $n\to\infty$, the probability $n$ is a prime decreases logarithmically. Yet, there's an infinite number of primes. And I'm not claiming there's an infinite amount of non-palindromic numbers whose cube is palindromic. I say there may be another palindromic number ($\forall x \in \mathbb R, x^{-\frac3 2}\not=0$). Maybe the ceiling (function) of googolplex times pi is that next number with the 2201 property. We may never know... $\endgroup$
    – JMCF125
    Jul 2, 2013 at 22:02

1 Answer 1

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A short computer run will verify this is true for all $n$ smaller than (updated) $10^{11}$.

import math,sys

def isPali(n): return str(n)[::-1] == str(n)
n = 0 
try:    
  while True:       
  n += 1        
  if not isPali(n) and isPali(n*n*n): print 'Found one!',n, n*n*n 
except KeyboardInterrupt:
  print 'Searched until', n 
  sys.exit()

Anyone with more computing time is welcome to add some powers to that $10$.

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  • 1
    $\begingroup$ That is indeed useful, although I was looking for some sort of proof by contradiction (I tried, but got nowhere). BTW, is that Python? $\endgroup$
    – JMCF125
    Jul 1, 2013 at 23:09
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    $\begingroup$ @JMCF125 - yes it is. by saying "the only" known, whomever wrote that, excluded the existence of a known proof, so I wouldn't place my bets on it being an easy proof. I'll update the answer later with some higher bounds. $\endgroup$
    – nbubis
    Jul 1, 2013 at 23:25
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    $\begingroup$ G.J.Simmons verifies this up to $2.8\cdot 10^{14}$ in 1970, in this article, question 2. $\endgroup$
    – Vepir
    Oct 25, 2019 at 18:52
  • $\begingroup$ @Vepir - You should add that as an answer - this sounds like a canonical source any future papers would reference. $\endgroup$
    – nbubis
    Oct 27, 2019 at 1:12
  • $\begingroup$ Verified up to $10^{15}$ according to oeis.org/A002780 $\endgroup$
    – Gerry Myerson
    Jan 1, 2021 at 11:06

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