Does there exist a positive integer solution to $1^{m}+2^{m}+3^{m}+\cdots+l^{m} = 123\cdots n$, where $l,n \geq2$?
Note: $123\cdots n$ is the concatenation of first $n$ positive integers.
I know that when $m=1$, then this is equivalent to asking "Is $123\cdots n$ ever a Triangular Number?". I did check all values of $n$ up to $5000$ and no number listed there is a Triangular Number.
If $m=3$, then it is same as asking is $123\cdots n$ ever a square number?, which is already discussed here. So we can rule out when $m=3$.
For $m=5$, the last two digits of $1^{5}+2^{5}+3^{5}+\cdots+l^{5}$ can only be $00,01,08,25,33$ or $76$. So that means the last two digits of $123\cdots n$ must be one of them. In addition, $123\cdots n$ can only be equal to $1\pmod{9}$ if $3\nmid n$, while $1^{5}+2^{5}+3^{5}+\cdots+l^{5}$ can be only $1,4,6\pmod{9}$, so if there is a integer solution to $1^{5}+2^{5}+3^{5}+\cdots+l^{5}=123\cdots n$, then $n$ must be either not divisible by $3$ or $n\equiv3\pmod{9}$.
If $m=2$, then there is no integer solution to $1^{2}+2^{2}+3^{2}+\cdots+ l^{2}=123\cdots n$ if $l\not\equiv0,1,4,5,6,8,9,13,17,18,26\pmod{27}$.
If $m$ is a prime number $\geq7$, then $1^{m}+2^{m}+3^{m}+\cdots+l^{m}=123\cdots n$ is not solvable when $l\equiv4\pmod{9}$.
As for other values of $m$, I have no idea.
I checked all values of $l,m,n$ up to $5000$, and it yields no solution.
