Consider the sequence of palindromic numbers where each term is defined as the smallest palindromic number with exactly k distinct palindromic prime factors (to avoid ambiguity, here I mean a palindromic number whose prime factors are ALL also palindromic numbers.). The sequence begins: $2$,$6$,$66$,$6666$:

  • $2$
  • $6$=$2$.$3$
  • $66$=$2$.$3$.$11$
  • $6666$=$2$.$3$.$11$.$101$

What is the next term after $6666$,does the next term exist ?. I have checked palindromes up to $10^9$( But I know that if there are infinitely many primes of the form $10000....00001$,then this sequence is absolutely infinite.). So what is the next term ?

  • $\begingroup$ Where does this come from? It looks like a contest, maybe ProjectEuler? $\endgroup$ – Jean-Claude Arbaut Oct 24 '15 at 11:01
  • $\begingroup$ @Jean-ClaudeArbaut,no, it just come to my head : D $\endgroup$ – Orange Diamond Oct 24 '15 at 11:04

From a relatively quick test, the recurrent sequence
$$a_0=6,\quad a_{n+1}=a_n\cdot (10^{2^{n-1}}+1), \quad n\ge3$$ are of numbers with $2^n$ repeated 6's.

I have no answer if there something that can generate other lenghts, like 666 or 66666, etc...

  • $\begingroup$ I mean any palindrome, not just restricted to string of digit $6$ $\endgroup$ – Orange Diamond Oct 24 '15 at 10:57

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