4
$\begingroup$

Define $\text{sdig}(n)$ to be the sum of the decimal digits of $n$, where $n$ is a positive integer. My question is as follows:

Does there exist a function $h:\mathbb Z^+\mapsto\mathbb Z^+$ such that $$(h\circ h)(n)=\text{sdig}(n)$$ for all $n\in\mathbb Z^+$? That is, does there exist a functional square root of the digit sum function?

So far, this problem has stumped me. I've shown that if such $h$ exists, it must commute with the digit sum function. I've also shown that if $S_a$ is the set of numbers whose digit sum is $a$, then for all $a$, there exists some $b$ such that the function $h$ maps all elements of $S_a$ to elements of $S_b$. However, whether these discoveries lead to the construction of such a function or the proof of its nonexistence I do not know.

Any ideas?

$\endgroup$
2
$\begingroup$

I think you can by building $h(n)$ up by digit sums. It seems natural to set $h(n)=n$ for $n \lt 10$. Now you can let $h(n)=\operatorname{sdig}(n)$ if $10 \le 10\le 18$. Then we can let $h(n)=\operatorname{sdig}(n)+9$ if $\operatorname{sdig}(n)\le 9$ The first value we have not yet accounted for is $h(19)$. If $h(19)=a,$ we need $h(a)=10$ and then we need $h(h(a))=1$, so we pick a convenient $a$ like $100$ and set $h(19)=100$. Now we can set $h(n)=100$ if $\operatorname{sdig}(n) = 10$. The next one we need to handle is $h(29)$. Again, if $h(29)=b,$ we need $h(b)=11$ and $h(h(b))=2$, so $\operatorname{sdig}(b)=2$. We can set $h(n)=101$ if $\operatorname{sdig}(n) = 11$. I think you can just keep building up $h(n)$ from the digit sums this way. We have two tables $$\begin {array} {r |r}n&h(n)\\ \hline 1-9&n\\10-18&n-9 \end {array}$$
$$\begin {array} {r |r}\operatorname{sdig}(n)&h(n)\\ \hline 1-9&\operatorname{sdig}(n) \\10-19&\operatorname{sdig}(n)+90 \\20-99&\operatorname{sdig}(n)+900 \end {array}$$ Now if we have a number $k$ with sum of digits $19$ we need $h(h(k))=19$ and $h(k)$ needs a digit sum of $10$. $109$ works well and we can extend the bottom row of the table to cover $\operatorname{sdig}(n)$ of $19$. Because $h(20)=11$ wherever we send digit sums of $20$ has to have a digit sum of $11$. $920$ looks like a nice number and we can add a line to the table. For a sum of digits of $30$ we have $h(30)=12$ so we need to send it somewhere that has a sum of digits of $12$. We observe that we can extend the bottom line to cover digit sums up to $99$. It seems clear we can keep going.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.