Let's say that a given $n\in\mathbb{N}$ is writable as sum of cubes in $k$ consecutive ways if it can be written as sum of $j,j+1,\ldots, j+(k-1)$ nonzero cubes, for some $j\geqslant 1$.
For example, $26$ is writable as sum of squares in $4$ consecutive ways, for \begin{align} 26 &= 1+25 &&\text{($2$ squares)}\\ 26 &= 1+9+16 &&\text{($3$ squares)}\\ 26 &= 4+4+9+9 &&\text{($4$ squares)} \\ 26 &= 1+1+4+4+16 &&\text{($5$ squares)} \end{align}
I'm trying to show that there is $n\in\mathbb{N}$ such that $n$ is writable as sum of cubes in $9$ consecutive ways, i.e. that there is $j\geqslant 1$ such that $n$ is the sum of $j,j+1,\ldots,j+8$ cubes.
I'm using that every number can be written as sum of $9$ or fewer cubes. (Or at least that there is some $g(3)\in\mathbb{N}$ such that every number is the sum of $g(3)$ or fewer cubes. I prefer this one because this problem seems to ask to be generalized to Waring bases $P_k=\{n^k:n\in\mathbb{N}\}$ in general.)
The example I found for $26$ for the case of squares was calculated manually, but I believe that there must be some general strategy to tackle this problem. I'm not much interested in to find an specific $n$ that is sum of cubes in $9$ consecutive ways, but in the method of finding one (if there is such).
I believe it holds for any $k$ in any Waring basis, but the case of cubes for $k=9$ would be good enough haha.
Any hints and tips would be much appreciated!
$1072$ represented by the sum of $k$ cubes $k=2,...,30$