Selfy number couldn't exist? For a positive integer $x$, let $S(x)$ denote the sum of the digits of $x$, and $l(x)$ denote the number of digits of $x$ (in base $10$). Now given a positive integer $n \ge 2$, it seems that there always exists $x$ such that $n=x+S(x)$, or $n=x+l(x)$, or $n=x+S(x)+l(x)$.
A Selfy number is a positive integer that can not be expressed in such a way. For example number $19$ is not Selfy because $19=14+5$; $16$ is not Selfy because $16=14+2$; and $21$ is not Selfy because $21=14+5+2$. I've checked some numbers but I couldn't find any such number.
My question: Is there any Selfy number?
 A: Note that $\ell(x)$ is an exceptionally slow-growing increasing function, and never increases by more than $1$, so the sequence $(n+\ell(n))_{n\in\Bbb N}$ increases in steps of $1$ and occasionally $2$, and so already almost all numbers are covered except those skipped over in the gaps when it increases by $2$. These are the points at which $\ell(n)$ increases, namely at $n=10^k$, at which point you have $(10^k-1)+\ell(10^k-1)=10^k-1+k$ and $10^k+\ell(10^k)=10^k+k+1$, skipping over the number $x=10^k+k$. Thus any Selfy number is of this form.
By brute-force calculation, I have verified that there are no such numbers for $k\le4000$, so we can assume that $k$ is large and so any Selfy number will have a large number of $0$'s in it separating the $10^k$ part from the $k$ part, so let's attempt to decouple the $10^k$ piece from the calculation. Since $s(x)\le 9\ell(x)$, any $x$ satisfying  $x+s(x)=10^k+k$ or $x+s(x)+\ell(x)=10^k+k$ is obviously $\le n$, so $\ell(x)\le\ell(n)=k+1$ and so $s(x),s(x)+\ell(x)\le 10(k+1)$ implying that any solution lies in the region $[10^k-9k-10,10^k+k)$.
This search area also breaks into three main regions, two for $x<10^k$ and one for $x\ge10^k$, because in the first case you have $s(x)\approx 9k$ and in the second you have $s(x)\approx 0$ (the deviations being of order $\log_{10}k$), leading to search regions near $10^k-9k$ (for $x+s(x)+\ell(x)$ solutions), $10^k-8k$ (for $x+s(x)$ solutions), and  $10^k+k$ (for $x+s(x)$ solutions). There are no $x+s(x)+\ell(x)$ solutions in the upper region because these would be larger than the $x+\ell(x)$ solutions that we have already excluded. This is the means by which the above brute-force search was performed.
Assuming $x<10^k$, we have $\ell(x)=k$, so if $x=\sum_{i=0}^{k-1}a_i10^i$, then let $x'=\sum_{i=0}^{k-1}(9-a_i)10^i$. We have $x+x'=10^k-1$ and $s(x')=\sum_{i=0}^{k-1}(9-a_i)=9k-s(x)$, so if $x$ is a solution then we have one of
$$10^k+k=x+s(x)\implies x'+s(x')+1=8k$$
$$10^k+k=x+s(x)+k\implies x'+s(x')+1=9k$$
In the other region, $x\ge 10^k$, let $x'=x-10^k$. Then $s(x')=s(x)-1$, so
$$10^k+k=x+s(x)\implies x'+s(x')+1=k.$$
Thus we have replaced the original problem with the claim that for each (large enough) $k$ one of these three equations has a solution. Or, in the language of self numbers:

$10^k+k$ is a Selfy number iff $k-1,8k-1,9k-1$ are all self numbers.


Edit: With the help of the above reductions to eliminate the $10^k$ piece of the equation and speed up the calculation, I found a counterexample by brute force: $10^{46927}+46927$ is a Selfy number. There are also Selfy numbers for $k=47827,49728,55927,56827$ and more.
