A property of 47 with respect to partitions into five parts

Question

Is 47 the largest number which has a unique partition into five parts (15, 10, 10, 6, 6), no two of which are relatively prime?

Answer 1

Yes. Suppose $n>47$.

If $2\mid n$, we can take $(n-8,2,2,2,2),(n-10,4,2,2,2)$, which are distinct partitions for $n\geq 14$.

If $3\mid n$, we can take $(n-12,3,3,3,3),(n-15,6,3,3,3)$, which are distinct partitions for $n\geq 21$.

If $n\equiv 1\pmod 6$, we can take $(n-37,15,10,6,6),(n-43,15,12,10,6)$, which are distinct partitions for $n\geq 55$, and $(21,7,7,7,7),(14,14,7,7,7)$ for $n=49$.

If $n\equiv 5\pmod 6$, we can take $(n-41,15,10,10,6),(n-47,15,12,10,10)$, which are distinct partitions for $n\geq 59$, and $(20,15,6,6,6),(15,12,10,10,6)$ for $n=53$.

(Thanks to Gerhard for pointing out we can finish the argument quickly from what I wrote before)

Answer 2

Here is a quick demonstration that it is effectively solvable for any number of parts.

A sufficiently large number that has exactly one partition with the property must be prime. Otherwise we can write it as $n = a \cdot b$ with $1 < a \leq b$ and partition $b$ into $k$ parts in two different ways, then multiply them by $a$ to get two different partitions of $n$ with the property.

Now suppose $n = 6 \cdot m + 35$. A sufficiently large $m$ has two different partitions into $k - 1$ parts such that each part of each partition is divisible by either $5$ or $7$. Multiplying by $6$ and appending $35$ then gives two different partitions of $n$ with the property.

On the other hand, suppose $n = 6 \cdot m + 55 = 6 \cdot m + 5 \cdot 11$. The same logic above applies, and that covers all the cases. Filling in some more details results in a simple formula for an upper bound in terms of $k$: the first part says if $n$ is composite then $n \lt (k+2)^2$, and the second part chases down the primes relatively more quickly placing the largest one in $6 \cdot k + O(1)$.

More generally still, we can require that the pairwise $\text{gcd}$ of the parts be greater than or equal to some $d \ge 2$ (so the question of $47$ corresponds to the case $d = 2$ and $k = 5$). An effective upper bound on unique solutions can still be obtained: we just have to carry out the above argument for all congruence classes mod $d$, in each case representing the remainder as a semiprime with prime factors larger than $d$.

EDIT: That led me to the following even simpler version, working mod $2$ instead of mod $6$: if $m$ has two different partitions into $5$ parts, then $n = 2 \cdot m$ has two different partitions into $5$ parts, no two parts of which are relatively prime. So $n < 2 \cdot (5+2) = 14$ if $n$ is even. Otherwise, suppose $n = 2 \cdot m + 15$, and observe that

$18 = 3+3+6+6 = 3+3+3+9$,

$19 = 3+5+5+6 = 3 + 3 + 3 + 10$, and

$20 = 3+3+5+9 = 3+5+6+6$

to conclude $m \le 17$ and $n \le 49$.