1
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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...

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.