Given a natural number $N = a_{n}a_{n-1}\ldots a_{1}$ and its digits' multiset $D_{N} = \{a_{1},a_{2},\ldots,a_{n}\}$, we shall call it a self-written number if, and only if, it exists multisets $B = \{b_{1},b_{2},\ldots,b_{k}\}\subseteq D_{N}$ and $E = \{e_{1},e_{2},\ldots,e_{j}\}\subseteq D_{N}$ such that $B\cap E = \varnothing$ which satisfy the next condition: $$N = b^{e}\quad\text{where}\quad b = b_{k}b_{k-1}\ldots b_{1}\,\,\,\text{and}\,\,\, e = e_{j}e_{j-1}\ldots e_{1}$$ Once the definition has been made clear, we may now expose some examples. For instance, $121$ is self-written since $121 = 11^2$. The same happens to $1331$ once it can be rewritten as $1331 = 11^3$. Moreover, so it does to $14641$ given that $14641 = 11^4$. Thence it comes my first question: is there infinitely many self-written numbers? Secondly, is there any criteria to identify them quickly?
I think this is it. Thank you in advance for any contribution.
There's a neat trick for squaring numbers that end in $5$. Suppose we have a number $n = 10k+5$ for some $k \in \mathbb{N}$. Then it's not too difficult to show that $n^2 = 100(k)(k+1)+25.$
For example, for $25^2$, just take the non-five part $2$, multiply by the next natural number $3$ to obtain $6$, and slap that in front of $25$ to obtain $25^2 = 625$.
With that said, I think considering $N$ of the form $N = (100\dots 005)^2$ might do the trick.