I have convinced myself (just with a calculator) that the next two entries are 10001 and 33. I am pretty sure that the sequence is infinite. 10^n + 1 = 100...01 is always a palindrome, and no matter what the last entry K is, a big enough n will avoid carries to produce a palindrome of the form K0...0K. This trick alone lets us produce an infinite sequence of distinct palindromes whose pairwise products are also palindromes, and the existence of at least one infinite sequence guarantees that there is a lexicographic minimum one. On Sun, Jan 12, 2020 at 1:22 PM Éric Angelini <eric.angelini@skynet.be> wrote:
Hello Math-Fun, is this seq finite (and correct)? S = 1,2,3,11,4,22,101,5,111,6,1001,7,88,77,8,1111,9,... (my knowledges about repunits and palindromes are close to the palindrome 0 themselves) Definition: Lexicographically earliest seq of distinct palindromes such that the product of two successive palindromes is a palindrome. Best, É.
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