I'm interested in knowing whether there is a non-trivial upper bound to the number $1234....n$ (a number consist of the digits, MSDigit-->LSDigit: $1,2,3,4,5,6,.....,n-1,n$), in base $n+1$.
By non-trivial I mean not $n^n$. if one found - what is the lowest one?
For example, look at 12345(base 6), which is 1865.
My guess is that it's indeed about $n^n$, but I have no proof or anything..
I'm searching for a good upper-bound in order to know how many bits are needed in order to save a number between ($0-1234....n$) in terms of $n$, when $n$ is already known.
If you can find a good upper-bound to the same number as above $+1$ (its digits are: $1,2,3,4,....,n-2,n,0$, or $12350_{n+1}=1866$ if we continue the last example), still in $base_{n+1}$. This is the number of possible numbers between $0-1234....n$) and it's what I'm looking for.
The exact value is $\dfrac{\dfrac{(n+1)^{n+1} - 1}{n}-1}{n}-1 = \dfrac{(n+1)^{n+1} - 1-n-n^2}{n^2}$
which is $\dfrac{(n+1)^{n+1}}{n^2} - 1$ rounded down when $n \ge 2$
This is between $n^{n-1}$ and $n^n$ when $n \ge 4$
It is almost OEIS A023811, which instead uses base $n$ and $1234\ldots(n-1)_{(n)}$, if you want more information