Palindromes Within Palindromes

I'm surprised that I've not covered this sequence before but checking through my previous posts it certainly seems as if I haven't. Here is the sequence in question:

A046351  Palindromic composite numbers with only palindromic prime factors.

The initial members of this sequence, up to 40000, are (permalink):

4, 6, 8, 9, 22, 33, 44, 55, 66, 77, 88, 99, 121, 202, 242, 252, 262, 303, 343, 363, 393, 404, 484, 505, 525, 606, 616, 626, 686, 707, 808, 909, 939, 1111, 1331, 1441, 1661, 1991, 2112, 2222, 2662, 2772, 2882, 3333, 3443, 3773, 3883, 3993, 4224, 4444, 5445, 5555, 5775, 6336, 6666, 6776, 6886, 7777, 7997, 8448, 8888, 9999, 10201, 12221, 13231, 14641, 15251, 15851, 18281, 19291, 20402, 20602, 22622, 22822, 23232, 24442, 24842, 25152, 25452, 26462, 26662, 28682, 30603, 30903, 31613, 33933, 34643, 35653, 36663, 37673, 37873, 38683, 39693, 39993

There are 94 terms in all. Let's just look at the numbers with two distinct prime factors but each of which is two digits or longer. It can be noted that these numbers have either 11 or 101 as factors.

  number   factors

  1111     11 * 101
  1441     11 * 131
  1661     11 * 151
  1991     11 * 181
  3443     11 * 313
  3883     11 * 353
  7997     11 * 727
  13231    101 * 131
  15251    101 * 151
  18281    101 * 181
  19291    101 * 191
  31613    101 * 313
  35653    101 * 353
  37673    101 * 373
  38683    101 * 383

From the above it can be noted that multiplying a palindrome by 11 or 101 seems to produce another palindrome. By extension, multiplying a palindrome by 1001, 10001, 100001 etc. will often produce another palindrome. For example:$$10001\times 1340431 = 13405650431$$Even a series of alternating 1's and 0's may also produce palindromes. For example:$$10101 \times 1340431 = 13539693531$$However, such products of palindromes are NOT always palindromic. For example:$$ \begin{align} 11 \times 1949999491 &= 21449994401 \\ 101 \times 1949999491 &= 196949948591\\1001\times1940491 &= 1942431491\\10101 \times 1940491 &= 19600899591 \end{align}$$If all the digits of the second palindrome are less than 5 then the multiplication will always produce palindromes.