The search can be optimized, and I have been running one for the last two hours or so. There are no lucky palindromes with 16 digits. There are a couple with 17 digits, and I just found one with 18 digits:

294507705507705492 == 2231118981118981 11 3 2 2

but no more than five factors so far.

Wow, Roman, great, thank you! It would be really interesting to see the optimized code.

"...no lucky palindromes with 16 digits"

-- do you mean in both ascending and descending order?

294507705507705492 == 223111898111898111322

-- remarkable! And again 5 factors tops - very interesting. I wonder if 7-factors even possible.

Great question. Because there is no asymmetry. It is easy to get a palindrome from 2 or more identical palindromes. But it is much harder for not-identical non-palindromic factors to line up to form a palindrome. I am just curious about these not-trivial cases, if that makes sense.

There are not that many. There are just two examples with 17 digits,

10022212521222001 == 100111001 100111001
12323244744232321 == 111010111 111010111

none with 18 digits (there's just a single, very lucky, palindrome with 18 digits) and none so far with 19 digits (search still ongoing):

There is one construction method that leads to the majority of lucky palindromes so far. Take a prime whose digits are all ones or zeroes, with at most 9 ones, such that its digit reversal is also prime. Their product will not generate any carries, and is a (doubly lucky) palindrome, such as

121121010242121121383121121242010121121 == 11001000011000001011*11010000011000010011

There can also be a two and up to five ones, that just fits (note the central 9):

10201112123150905132121110201 == 100001001010201*102010100100001

And then there are the repunit primes. Combined with 11 they form palindromes of the form 122...221.

12222222222222222221 == 1111111111111111111*11 == 11*1111111111111111111

This happens to be the only lucky palindrome of length 20. I'll wrap up my search and report back with a followup post.