Unique Partition

First, note that $w(10) = 10$ :

• 10
• 9 + 1
• 8 + 2
• 7 + 3
• 7 + 2 + 1
• 6 + 4
• 6 + 3 + 1
• 5 + 4 + 1
• 5 + 3 + 2
• 4 + 3 + 2 + 1

This is enough to answer the problem. (It probably should have asked the sum of all $a$ such that $w(a) = a$ to make it not so simple.) But we will show that this is indeed the only solution.

It can be verified manually (if with difficulty) that $w(n) < n$ for all $n = 2, 3, \ldots, 9$ . Thus there's no solution for $n < 10$ .

For $n > 10$ , observe that the following are partitions with no repeated parts:

• $n$
• $(n-k) + k$ for $k = 1, 2, \ldots, \lceil \frac{n}{2} \rceil - 1$
• $(n-k-1) + k + 1$ for $k = 2, 3, \ldots, \lfloor \frac{n}{2} \rfloor - 1$

They account for a total of $1 + \left( \lceil \frac{n}{2} \rceil - 1 \right) + \left( \lfloor \frac{n}{2} \rfloor - 2 \right) = n-2$ partitions. Thus it's enough to prove that for $n > 10$ , there exists three other partitions with no repeated parts to claim $w(n) > n$ . Sure, this is easy enough: $(n-5) + 3 + 2$ , $(n-6) + 4 + 2$ , $(n-6) + 3 + 2 + 1$ suffice. This completes the proof.

We claim that for $a \geq 10,$ , we have $w(a+1)-w(a) \geq 1$ .

Consider any expression of $a= x_{1} + x_{2} + ... + x_{n}$ , where $x_{1}>x_{2}>...>x_{n}$ . Then $a+1= (x_{1}+1) + x_{2}+...+x_{n}$ . Thus $w(a+1)>=w(a)$ .

However, whenever we have a partition of $a+1$ , in which the largest term is exactly one more than the second largest term, there won't be a corresponding expression for $a$ .

As an example, consider $6= 5+1 = 4+2= 3+2+1$ . Adding one to the largest term in each such expression, we get

$7= 6+1= 5+2= 4+2+1$ , but we MAY get additional expressions, such as $4+3$ , where there is no corresponding $3+3$ as an expression for $6$ . So if we can prove that for all $a \geq 10$ , we can find such an expression, our claim is proved. This is easy , because

$3k= (k+1)+k+(k-1)$ $3k+1= (k+2)+(k+1)+(k-2)$ $3k+2= (k+2)+(k+1)+(k-1)$

Notice, $w(11)=12$ . This means $w(a)\geq a+1$ for all $a\geq 11$ . It suffices to evaluate $w(a)$ for all $a \leq 10$ . A quick check shows that $a=10$ is the only solution.

If we check the number of ways to express a certain integer, a pattern will be observed.The pattern is ,

$ω$ ( $1$ ) = $1$

$ω$ ( $2$ ) = $1$

$ω$ ( $3$ ) = $2$

$ω$ ( $4$ ) = $2$

$ω$ ( $5$ ) = $3$

$ω$ ( $6$ ) = $4$

$ω$ ( $7$ ) = $5$

$ω$ ( $8$ ) = $6$

$ω$ ( $9$ ) = $8$

$ω$ ( $10$ ) = $10$

$ω$ ( $11$ ) = $12$

$ω$ ( $12$ ) = $15$

You can see that difference between the numbers increases by 1 after every set of 4 numbers.The pattern is not clear for the first 4 numbers but after that the pattern can be identified clearly. So the number of ways increases continuously and after $10$ , $ω$ (a) > a