Palindromes in a base less than the square root

[David Wilson]

Every number n >= 0 is a palindrome in some base b >= 2.  0 has
the palindromic representation 0 (or the empty string) in every
base b >=2, while n >= 1 has the trivial 1-digit palindromic
representation n in every base b > n.  Additionally, every n >= 3
has the palindromic representation 11 in base b = n-1.

Your "unkempt" numbers n are nonpalindromic in bases
2 <= b <= sqrt(n).  In other words, unkempt numbers have no
palindromic representation of 3 or more digits in any base.

But there are numbers that are yet more palindrome-hostile
than the unkempt numbers.  These "strictly nonpalindromic
numbers", given by A016038, are nonpalindromic in every
base 2 <= b <= n-2.  In other words, their only palindromic
representation of 2 digits or more is 11 in base n-1.

The unkempt numbers include the strictly nonpalindromic
numbers, so if A016038 is infinite, then so are the unkempt
numbers.  I strongly suspect that A016038 is indeed infinite,
but I can't offhand prove this.  When I look for the smallest
member of A016038 >= 10^k, I get

1, 11, 103, 1019, 10069, 100333, 1000183, 10000189

This indicates to me that A016038 is not thinning out very
quickly for large n, which bodes well for its being infinite.
I'll have to think about it some more.

----- Original Message ----- 
From: "Dr. Gordon Hamilton" <hamiltonian at shaw.ca>
To: <ham>; <seqfan at ext.jussieu.fr>
Sent: Tuesday, March 01, 2005 9:56 PM
Subject: Palindromes in a base less than the square root


> A100563 reports the number of times that a number is a palindrome in a 
> base less than its square root.
>
> For example 100 is a palindrome in bases 3, 7, and 9 so F(100) =3.
>
> Are there an infinite number of "unkempt" numbers with F(n) = 0 ?
>
> Thanks for any insight...
>
> Gordon