This sequence is finite. For any j, the largest digit sum possible is 9*
A055642(j). Let j contain M digits. In order for j to be a palindrome and j + digsum(j) to be a palindrome, if digsum(j) affects the k-th digit of j, it must also affect the (M-k)-th digit of j for k = 1, 2, ..., M-1. For example, if j is 5 digits long and digsum(j) is 2 digits long, then j + digsum(j) and j - digsum(j) must affect the digit in the thousands place of j in order to produce a palindrome. This means that digsum(j) must be at least 4 digits long. Generally, we can say that
A055642(digsum(j)) >=
A055642(j) -
A055642(digsum(j)) + 1 and thus,
A055642(digsum(j)) >= (1/2)*(
A055642(j)+1). This, however, fails when j > 3 digits. When j is 4 digits, the maximum that the digit sum could be is 36, a 2-digit number. Since it is a 2-digit number, it must affect the digit in the hundreds place of j when it is added to or subtracted from j. However, this is not possible since digsum(j) is only 2 digits long. For j > 4 digits the argument is similar.
