Maximal $n$ such the the additive partition with a given product is unique. Given $n$, there are many tuples with $a + b + c = n,0 < a < b < c$. For large $n$, different tuples may give the same products. E.g. $2+8+9=19=3+4+12,2\times8\times9=144=3\times4\times12$.
What is the largest value of $n$, such that there is no tuples with the same product? The computer tells me it's $22$. Maybe directly proving from $22$ is hard. We can just prove that for sufficiently large $n$, there are always tuples with the same product.
Any ideas?
 A: 
For sufficiently large $n$, there are always unique tuples $a,b,c, \ 0<a<b<c$ and $a',b',c', \ 0<a'<b'<c'$ such that $a+b+c=n=a'+b'+c'$ and $a \times b \times c = a' \times b' \times c'$.

Proof. Say we have $b$ divisible by $3$ and $c$ divisible by $2$. This assumption will be justified later. Relate $a,b,c$ with $a',b',c'$ as follows:
$$
a' = 2a \qquad b' = \frac{b}{3} \qquad c' = \frac{3c}{2}
$$
Clearly these are all integers, and clearly we have $a' \times b' \times c' = a \times b \times c$. The difference between the new sum and the original sum, which we need to be $0$, is
$$a-\frac{2b}{3}+\frac{c}{2} = 0$$
which has solutions $0<a<b<c$ of the form
$$b = 3x, \quad c=4x-2a \quad | \quad a,x \in \mathbb{Z}^+ \text{ and } x>2a$$
which also justifies our divisibility assumptions about $b$ and $c$.
So if we can write $n=a+b+c=a + (3x) + (4x - 2a) = 7x - a$ where $x$ is a positive integer greater than $2a$, then we can certainly find a working solution. 
It is easy to see that any sufficiently large $n$ can be written in this form. Given some $n$, we choose $x$ so that $7x$ is the smallest multiple of $7$ greater than $n$, and we pick and $a=7x-n$ to correct for the error. Then $a$ will always be between $1$ and $7$. So if $x \geq 15$, we can ensure that $x>2a$ holds.
The last step is to ensure that the inequality $0<a'<b'<c'$ holds. We have positive $a$, so certainly $0<2a=a'$. Since $x>2a$, we have that $a'=2a<x=\frac{3x}{3}=\frac{b}{3}=b'$. Finally, $b'=\frac{b}{3}<b<c<\frac{3c}{2}=c'$. 
Hence, for all sufficiently large $n$ (specifically $n \geq 98$), we can find a solution pair $a,b,c$ and $a',b',c'$ as described above. Q.E.D.

Summary. Given $n \geq 98$, calculate $x=\lceil \frac{n+1}{7} \rceil$. Then the following is a working pair of tuples:
$$\boxed{(a,b,c)=(7x-n, \ 3x, \ 2n-10x)} \\
\boxed {(a',b',c')=(14x-2n, \ x, \ 3n-15x)}$$
This follows from the proof and the fact that $\lceil \frac{n+1}{7} \rceil$ is the $x$ such that $7x$ is the smallest multiple of $7$ greater than $n$. The rest is substitution.
