OFFSET
1,2
COMMENTS
The sequence is infinite. One can always multiply a k-digit palindrome by the palindromic factor 10^k+1 to get another palindrome. - Franklin T. Adams-Watters, Jun 06 2006
Apart from a(1), identical to A088114. - Giovanni Resta, Oct 08 2025
From David A. Corneth, Oct 18 2025: (Start)
a(27), a(28), a(29) <= 61400579344397500416 via (for example) 2^19 * 3^1 * 4^2 * 11^3 * 25852^1 * 70907^1, 2^21 * 3^1 * 4^1 * 11^3 * 25852^1 * 70907^1, 2^23 * 3^1 * 11^3 * 25852^1 * 70907^1 respectively.
When searching the factorizations of some palindrome P into smaller palindromes (>1) for this sequence one can ease the search by filtering out some palindromes from possible factors. If a palindrome can be written as the product of two palindromes > 1 we do not have to consider that palindrome as a factor. Also we can remove P to be considered as well except for n = 1. These would not be in a factorization into as much palindromes as possible.
For example, the palindrome divisors (1 < p < 252) of 252 are {2, 3, 4, 6, 7, 9}. However 4, 6 and 9 can be written as a product of smaller palindromes > 1 so we can leave them out and just consider factorizations of 252 into 2, 3 and 7 which eases the search. (End)
EXAMPLE
a(4) = 88 = 2*2*2*11.
a(5) = 252 = 2*2*3*3*7.
a(26) = 8463613290923163648 = 2^21 * 11 * 101 * 111 * 878 * 37273.
CROSSREFS
KEYWORD
base,nonn,more
AUTHOR
Amarnath Murthy, Jun 07 2002
EXTENSIONS
Corrected and extended by Sascha Kurz, Jan 02 2003
Name corrected by Robert Israel, Jan 08 2025
a(21)-a(26) from Giovanni Resta, Oct 08 2025
STATUS
approved