Self and Earl Bird numbers: are there infinitely many natural numbers that are both?

Question

Self-numbers (A003052 in the OEIS) are natural numbers which are not the sum of a smaller number and the sum of its digits. Early Bird numbers (A116700) are numbers that occur ahead of their natural place in the string 12345678910111213....

Are there infinitely many numbers, such as 110, that are both a Self-number (because no number smaller than 110 exists such that added to the sum of its digits equals 110) and an Early Bird number (because in the string 12345...99100101102103... 110 appears earlier than its natural place, just a few digits further)?

Answer 1

Yes, there are infinite such numbers. My proof is based on this reduction test that can found here.

Take $a = 1$ and $c = 10$. Then, $m_1 = 10 - 1 = 9$ which is a self-number and $m_2 = 9b - 11$.
So let us pick any $b$ such that $9b-11$ is also a self-number.
Note the self-number generating recurrence relation (from wiki) $C_k = 8 \cdot 10^{k - 1} + C_{k - 1} + 8, C_1 = 9$.
This implies $C_k = (C_{k-1} - 2) \mod 9$, which means every value congruent to modulo $9$ repeats after every $9$ terms. Thus, there are infinite self-numbers that are congruent to $7 \mod 9\;$ so there will be infinite possible values for $b$ as well.

There is a $10$ at both the beginning and the end of this number so this is also an Early Bird number.

For example, take $b=12$ then $m_2 = 9 \cdot 12 - 11 = 97$ which is a self-number so $10^{12} + 10 = 1000000000010$ is also a self-number. This is also an Early bird number because it's a prefix of $10000000000, 10000000001$.

Answer 2

Yes, there are:

Proof: Given a number which is both, we'll construct a larger number that is also both.

Assuming we have such a number n we'll in addition assume that n+10 has k digits and that 10 + k x 9 < n which is e.g. satisfied by OP's example n = 110 and for any larger number. We'll also assume that n starts with a 1, a property that will trivially be preserved by the construction below.

Let J be the smallest positive number such that the last digit of n+J is 1. We claim that N := n+J + 10^k + 10^(k+1) + ... + 10^(k+J-1) is also both. Indeed, if N were non self, let M be the number that makes it so, i.e. N = M + digitsum(M). M must have the same number of digits as N with J leading digits 1. Removing the J leading 1s would yield a number m that renders n non self, a contradiction. As N starts and ends with a 1 (and its second digit is either 1 or 2) it is also early bird.