<s>[INST] Liouville sequences We have $(1+2+\cdots+n)^2=1^3+2^3+\cdots+n^3$. We call a finite sequence $a_1,\ldots,a_n$ Liouville if it satisfies \begin{equation}(a_1+\cdots+a_n)^2=a_1^3+\cdots+a_n^3.\end{equation}
Liouville discovered that for all $m\in\mathbb{N}$ the sequence $\tau(d_1),\ldots,\tau(d_n)$ where $d_i$, $i=1,\ldots,n$, are the positive divisors of $m$ and $\tau(k)$ is the number of positive divisors of natural number $k$. $m=27$ has the divisors 1, 3, 9, 27, and we obtain the sequence 1, 2, 3, 4. For $m=10$ we obtain the sequence 1, 2, 2, 4.
$a_1,\ldots,a_n$ with $a_i=n$ for $i=1,\ldots,n$ and 2, 2, 2, 2, 2, 2, 2, 2, 8 are Liouville but are not obtained by Liouville's method. 
Are there other such examples? 
 [/INST] I've searched through sequences of length 4, 5, 6 and 7 with entries up to 10, and there appear to be many other examples.  I haven't yet spent time analysing the results to find any patterns, so I'll just list some examples for the moment to answer your question.
Firstly, one obvious observation: sequences which don't contain a $1$ cannot come from $ \tau(d_i) $, since $1$ is a divisor of any $ m \in \mathbb{N} $ and $ \tau(1) = 1 $.
Length 4 sequences:
\begin{array}{c}
\{2, 2, 4, 4\}
\end{array}
Length 5 sequences:
\begin{array}{c}
\{3, 3, 3, 3, 6\}, \{3, 3, 3, 4, 6\}, \{1,2,2,3,5\}
\end{array}
The third sequence also does not come from $ \tau(d_i) $.  If this was from $ \tau(d_i) $, it would have to come from $ m = p^4 $ since there are $ 5 $ divisors.  But this generates $ \{1, 2, 3, 4, 5\} $, so doesn't work.
Length 6 sequences:
\begin{array}{c}
\{1, 1, 1, 2, 2, 5\}, \{1, 1, 1, 4, 4, 5\}, \{1, 1, 2, 4, 5, 5\}, \{1, 1, 4, 5, 5, 5\}, \\
\{1, 2, 2, 4, 4, 6\}, \{1, 4, 4, 4, 6, 6\}, \{2, 2, 2, 2, 2, 6\}, \{2, 2, 4, 4, 6, 6\}, \\
\{2, 4, 4, 5, 5, 7\}, \{2, 4, 4, 6, 6, 6\}, \{3, 3, 3, 3, 5, 7\}, \{3, 3, 3, 6, 6, 6\}, \\
\{3, 4, 5, 5, 6, 7\}, \{3, 5, 5, 5, 6, 7\}, \{4, 5, 5, 6, 6, 7\}
\end{array}
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