Does $ \sum_{i=1}^n n^{k_i} $ determines $(k_1,...,k_n)$?

Question

Let $k_1,...,k_n\in\mathbb{N}$. Does the power sum $$ \sum_{i=1}^n n^{k_i} $$ uniquely determines the $n$-tuple $(k_1,...,k_n)$?

Remark: In the case $n=2$, this is true. However, when trying to generalize to an arbitrarily sized sum, it doesn't hold. For example, $$ 2^0+2^0+2^2=2^1+2^1+2^1. $$ I then thought of a fixed sum size, equal to the considered base, but I don't know exactly how to argue or build this bijection. The motivation behind this question comes from trying to determine the $n$-tuple $(k_1,...,k_n)$ from the sum $$ \sum_{i=1}^n g(k_i), $$ where $g$ is some constructible function.

Answer 1

Obviously you can only hope to determine the $n$-tuple up to permutation, and you can only hope to determine even that when $n>1$. So I'll assume that's what you want. And as you see, it doesn't work in general when the number of $k_i$ is not equal to the base of the exponent.

Then the answer is yes, if you know $n$. It's a corollary of the uniqueness of base-$n$ representations. You can read off the $k_i$ from the digits of the number $s = \sum_{i=1}^n n^{k_i}$ written base $n$. The $j^{\small\text{th}}$ digit ($j=0$ is the first digit) will tell you how many $k_i$ are equal to $j$, unless there is exactly one $1$ and no other nonzero digits in $s$, in which case all the $k_i$ are equal to $j-1$ where $j$ is the place where the $1$ occurs.

Answer 2

Adding more details.

Let $a_j$ be the number of the exponents that are equal to $j$. I.e. $$ a_j = \#\{i \mid k_i=j\}. $$ We have $$S:=\sum_{i=1}^n n^{k_i} = \sum_{j\ge0}a_jn^j.$$ Case 1 - Not all of the exponents are equal: Then $0\le a_j<n$ for all $j$. Thus the sequence $(a_0, a_1, \ldots)$ is the base $n$ representation of $S$.

Case 2 - All of the exponents are equal: Say $k_i=j$ for all $i$. Then $S=n^{j+1}$.

Note crucially this: Numbers in case 1 have different base $n$ representations (by uniqueness of such representations). Also, no number can be in both cases, again by uniqueness: If $S$ is a perfect power of $n$ (case 2), then it only has one non-zero digit base $n$, so it can't be in case 1.

Therefore we have a strategy: Write $S$ in base $n$. If there is more than one non-zero digit, read off the digits as the unique answer for the exponents. Otherwise, $S$ must be a perfect power of $n$, say equal to $n^{j+1}$. Then all exponents are equal to $j$.