Numbers with the property that qp = 0, where p = (sum of digits of n) times (number of digits of n) (A110805) and q = (sum of {each digit of n} raised to the power {number of digits in n}) (A101337).
From David Consiglio, Jr., Jan 29 2017: (Start)
The sequence is finite and 210 is the last term. Proof:
1. Let Y = length(n).
2. The maximum value of digit_sum(n) is 9Y.
3. Since p = Y*digit_sum(n), p has a maximum value of 9Y^2.
4. 2^Y > 9Y^2 for all Y > 9. Therefore q > p for all numbers longer than 9 digits that contain any digits > 1.
a. Example: 1,000,000,002. q = 1^5 + 8*0^5 + 2^10 = 1025. The largest p value for a 10 digit number would be for 9,999,999,999 which has p = 10*(9*10) = 900. Since q > all possible p values at this size, any member of this sequence must either have fewer than 10 digits or contain only 0's and 1's as digits.
5. Now we can consider only numbers with 0 and 1 digits.
6. Let Z = number of 1 digits in a number n.
7. q = Z because q = Z*1^Y + (YZ)*0^Y = Z.
8. p = YZ because the sum of the digits is equal to the number of 1's.
9. Z = YZ only in the case of Y = 1. Thus, the only member of this sequence that contains only 0's and 1's has a length of only 1 digit. Thus, n = 1 is in this sequence.
10. Therefore a candidate number must have fewer than 10 digits if it contains a digit 2 or larger, and must have fewer than 2 digits if it does not. All numbers in this range have been checked, and no additional values of n with q = p have been found.
Thus the sequence is finite. (End)
