Proof of Theorem 1.3

For $k \geq 3$ , let

$$ \begin{align*}\alpha_k = (6, 6, \underbrace{1, \ldots, 1}_{k-2}) \quad\text{and}\quad \beta_k = (9, 2, 2, \underbrace{1, \ldots, 1}_{k-3}).\end{align*} $$

Note that both are partitions of $k+10$ having k parts each. We have ${pre}_k(\alpha _k) = {pre}_k(\beta _k) = 36$ . More specifically, for $k=3$ , this gives the partitions $(6, 6, 1)$ and $(9, 2, 2)$ of 13, having the same image, 36, under $pre_3$ . This, in fact, readily generates an infinite family of pairs of partitions with identical $pre_3$ images, namely, $(6m, 6m, m)$ and $(9m, 2m, 2m)$ for each $m \geq 1$ . Each is a partition of $13m$ , with the $pre_3$ image $36m^3$ . We now give another family where the parts of each of the partitions are relatively prime, namely, $(3m, 2m-1, 2)$ and $(4m-2, m, 3)$ for $m \geq 3$ . For $(3m, 2m-1, 2)$ , clearly, 2 and $2m-1$ are relatively prime, and hence the triplet of numbers has this property as well. In addition, for $(4m-2, m, 3)$ , if 3 does not divide m, then 3 and m are relatively prime. Otherwise, $4m-2$ is not divisible by 3 and, hence, relatively prime to 3. In either case, the triplet of numbers has the greatest common divisor equal to one. Moreover, observe that both are partitions of $5m+1$ having $3$ parts, with the identical ${ pre}_3$ image $6m(2m-1)$ .

More generally, if $p> q$ are two primes, then the family of pairs of partitions given by $(q(1 + m(p-1)), \ 1 + m(q-1), \ p)$ and $(p(1 + m(q-1)), \ 1 + m(p-1), \ q)$ , where $m \geq {(p-1)}/{(q-1)}$ , are both partitions of $p+q+1+m(pq-1)$ having three parts with the same $pre_3$ image $pq(1 + m(p-1))(1 + m(q-1))$ . The condition on m is necessary to ensure that in the first named partition, we have $1 + m(q-1) \geq p$ . The second family alluded to in the previous paragraph, that is, $(3m, 2m-1, 2), (4m-2, m, 3)$ can be obtained from the special case $p=3, \ q=2$ . Moreover, for each $k \geq 3$ , we can extend the pair at the beginning of this paragraph to

$$ \begin{align*}& (q(1 + m(p-1)), \ 1 + m(q-1), \ p, \ \underbrace{1, \ldots, 1}_{k-3}), \\ & (p(1 + m(q-1)), \ 1 + m(p-1), \ q, \ \underbrace{1, \ldots, 1}_{k-3}),\end{align*} $$

with $m \geq {(p-1)}/{(q-1)}$ , giving an infinite three-parameter family of pairs of partitions of $p+q+k+m(pq-1)-2$ with k parts, having the same image under $\mathit{pre}_k$ , namely, $pq(1+m(p-1))(1+m(q-1))$ . Thus, for infinitely many n, the map ${pre}_k$ fails to be injective on partitions of n with k parts.

Proof of Theorem 1.5

Let $\lambda = (\lambda _1, \ldots , \lambda _{\ell }), \ \mu = (\mu _1, \ldots , \mu _{\ell })$ with ${pre}_{\ell - k}(\lambda ) = {pre}_{\ell - k}(\mu )$ . Then, the products of the parts of the partitions $pre_{\ell - k}(\lambda ), \ {pre}_{\ell - k}(\mu )$ are also the same. Since each $\lambda _i$ occurs once in exactly $\binom {\ell - 1}{\ell - k - 1}$ parts of ${pre}_{\ell - k}(\lambda )$ , the product of the parts of $pre_{\ell - k}(\lambda )$ is therefore $(\prod _{i=1}^{\ell } \lambda _i)^{\binom {\ell - 1}{\ell - k - 1}}$ and that of $pre_{\ell - k}(\mu )$ is, similarly, $(\prod _{i=1}^{\ell } \mu _i)^{\binom {\ell - 1}{\ell - k - 1}}$ . Thus, $(\prod _{i=1}^{\ell } \lambda _i)^{\binom {\ell - 1}{\ell - k - 1}} = (\prod _{i=1}^{\ell } \mu _i)^{\binom {\ell - 1}{\ell - k - 1}}$ , giving $\prod _{i=1}^{\ell } \lambda _i = \prod _{i=1}^{\ell } \mu _i$ . The fact that ${pre}_{\ell - k}(\lambda ) = {pre}_{\ell - k}(\mu )$ means that every product of $\ell - k$ numbers from the partition $\lambda $ equals a similar product of the parts of $\mu $ . Combined with $\prod _{i=1}^{\ell } \lambda _i = \prod _{i=1}^{\ell } \mu _i$ , this implies that there is a correspondence between the product of any k parts of $\lambda $ with those of $\mu $ . Therefore, $pre_k(\lambda ) = {pre}_k(\mu )$ . Since $pre_k$ is injective, this finally tells us that $\lambda = \mu $ . Hence, ${pre}_{\ell -k}$ is also injective.

Proof of Theorem 1.6

Let $\lambda = (\lambda _1, \lambda _2, \lambda _3, \lambda _4)$ and $\mu = (\mu _1, \mu _2, \mu _3, \mu _4)$ be partitions of n with ${pre}_2(\lambda ) = { pre}_2(\mu ) = \nu $ . Then, $\nu = (\lambda _1 \lambda _2, \lambda _1 \lambda _3, \ldots , \lambda _2 \lambda _4, \lambda _3 \lambda _4) = (\mu _1 \mu _2, \mu _1 \mu _3, \ldots , \mu _2 \mu _4, \mu _3 \mu _4)$ . Therefore,

(3.1) $$ \begin{align} \lambda_1 \lambda_2 = \mu_1 \mu_2, \quad \lambda_1 \lambda_3 = \mu_1 \mu_3, \quad \lambda_2 \lambda_4 = \mu_2 \mu_4, \quad \lambda_3 \lambda_4 = \mu_3 \mu_4. \end{align} $$

In other words,

(3.2) $$ \begin{align} \frac{\lambda_1}{\mu_1} = \frac{\mu_2}{\lambda_2} = \frac{\mu_3}{\lambda_3} = \frac{\lambda_4}{\mu_4} = k \quad (\text{say, a positive rational number}). \end{align} $$

Since $\lambda _1 + \lambda _2 + \lambda _3 + \lambda _4 = n = \mu _1 + \mu _2 + \mu _3 + \mu _4$ , it follows from (3.2) that $k \mu _1+ \mu _2/k + \mu _3/k + k\mu _4 = \mu _1 + \mu _2+ \mu _3 + \mu _4$ . This implies that

(3.3) $$ \begin{align} \mu_1 + \mu_4 = \frac{\mu_2 + \mu_3}{k}. \end{align} $$

Note that the ordering among the four indicated parts of $\nu $ is fixed, as shown in (3.1), but the order of the remaining two parts is unrestricted. We now consider two cases depending on two possible equalities.

Case 1: $\lambda _1 \lambda _4 = \mu _1 \mu _4$ and $\lambda _2 \lambda _3 = \mu _2 \mu _3$ . Then, $k = {\lambda _1}/{\mu _1} = {\mu _4}/{\lambda _4}$ . However, from (3.2), ${\mu _4}/{\lambda _4} = 1/k$ . Thus, $k = 1/k$ , implying $k=1$ (since k is positive). Again, (3.2) implies, for $k=1$ , that $\lambda _i = \mu _i$ for all $i=1,2,3,4$ , which of course means that $\lambda = \mu $ .

Case 2: $\lambda _1 \lambda _4 = \mu _2 \mu _3$ and $\lambda _2 \lambda _3 = \mu _1 \mu _4$ . Then, $k \mu _1 \times k \mu _4 = \mu _2 \mu _3$ , implying that

(3.4) $$ \begin{align} \mu_1 \mu_4 = \frac{\mu_2 \mu_3}{k^2} = \frac{\mu_2}{k}\frac{\mu_3}{k}. \end{align} $$

From (3.3) and (3.4), $\mu _1, \mu _4$ are the roots of the quadratic equation

$$ \begin{align*}x^2 - \bigg(\frac{\mu_2}{k} + \frac{\mu_3}{k}\bigg)\,x + \frac{\mu_2}{k}\frac{\mu_3}{k} = 0.\end{align*} $$

However, the roots of this equation are clearly ${\mu _2}/{k}$ and ${\mu _3}/{k}$ , and ${\mu _2}/{k} \geq {\mu _3}/{k}$ . Since $\mu _1 \geq ~ \mu _4$ , we deduce that $\mu _1 = {\mu _2}/{k}$ and $\mu _4 = {\mu _3}/{k}$ . Moreover, $\mu _1 = {\mu _2}/{k} \leq {\mu _1}/{k}$ , so $k \leq 1$ . Similarly, $\mu _4 = {\mu _3}/{k} \geq {\mu _4}/{k}$ , so that $k \geq 1$ . We thus conclude that $k=1$ , which implies that $\lambda _i = \mu _i$ for all i from (3.2). Hence, $\lambda = \mu $ .

In both cases, we have therefore shown that $pre_2$ is injective on partitions of n with four parts.

We move on to partitions with five parts. Assume $\lambda = (\lambda _1, \lambda _2, \lambda _3, \lambda _4, \lambda _5)$ and $\mu =~(\mu _1, \mu _2, \mu _3, \mu _4, \mu _5)$ are such that ${pre}_2(\lambda ) = {pre}_2(\mu ) = \nu $ . From the ordering relations among the parts of $\nu $ , as in the proof for partitions with four parts above, we have

(3.5) $$ \begin{align} \lambda_1 \lambda_2 = \mu_1 \mu_2, \quad \lambda_1 \lambda_3 = \mu_1 \mu_3, \quad \lambda_3 \lambda_5 = \mu_3 \mu_5, \quad \lambda_4 \lambda_5 = \mu_4 \mu_5. \end{align} $$

We note that the product of parts of ${pre}_2(\lambda )$ and ${pre}_2(\mu )$ are equal since the partitions are themselves equal, giving $(\lambda _1 \times \cdots \times \lambda _5)^4 = (\mu _1 \times \cdots \times \mu _5)^4$ . Thus, $\prod _{i=1}^{5} \lambda _i = \prod _{i=1}^{5} \mu _i$ . However, $\lambda _1 \lambda _2 = \mu _1 \mu _2$ and $\lambda _4 \lambda _5 = \mu _4 \mu _5$ , which implies that $\lambda _3 = \mu _3$ . From each of the relations in (3.5), we can then deduce that $\lambda _i = \mu _i$ for the other values of i as well. Therefore, $\lambda = \mu $ and ${pre}_2$ is injective on partitions of n with five parts.

Finally, for partitions with six parts, assume $\lambda = (\lambda _1, \ldots , \lambda _6)$ and $\mu = (\mu _1, \ldots , \mu _6)$ satisfy ${pre}_2(\lambda ) = {pre}_2(\mu ) = \nu $ . As before, we have the ordering relations

(3.6) $$ \begin{align} \lambda_1 \lambda_2 = \mu_1 \mu_2, \quad \lambda_1 \lambda_3 = \mu_1 \mu_3, \quad \lambda_4 \lambda_6 = \mu_4 \mu_6, \quad \lambda_5 \lambda_6 = \mu_5 \mu_6. \end{align} $$

We also have the product of parts of $\nu $ equal to $(\prod _{i=1}^{5}\lambda _i)^5 = (\prod _{i=1}^{5} \mu _i)^5$ . This implies that $\prod _{i=1}^{5}\lambda _i = \prod _{i=1}^{5} \mu _i$ . From (3.6), $\lambda _1 \lambda _2 \lambda _5 \lambda _6 = \mu _1 \mu _2 \mu _5 \mu _6$ . Now, $\prod _{i=1}^{5}\lambda _i = \prod _{i=1}^{5} \mu _i$ implies that $\lambda _3 \lambda _4 = \mu _3 \mu _4$ . In a similar vein, we get the other order relations listed as follows (we include $\lambda _3 \lambda _4 = \mu _3 \mu _4$ too for convenience):

(3.7) $$ \begin{align} \lambda_3 \lambda_4 = \mu_3 \mu_4, \quad \lambda_3 \lambda_5 = \mu_3 \mu_5, \quad \lambda_2 \lambda_5 = \mu_2 \mu_5, \quad \lambda_2 \lambda_4 = \mu_2 \mu_4. \end{align} $$

The relations in (3.6) and (3.7) yield

(3.8) $$ \begin{align} \frac{\lambda_1}{\mu_1} = \frac{\mu_2}{\lambda_2} = \frac{\mu_3}{\lambda_3} = \frac{\lambda_4}{\mu_4} = \frac{\lambda_5}{\mu_5} = \frac{\mu_6}{\lambda_6}. \end{align} $$

From (3.8), we also get the additional relation $\lambda _1 \lambda _6 = \mu _1 \mu _6$ . Collecting all the known relations so far, we have

(3.9) $$ \begin{align} \lambda_1 \lambda_2 = \mu_1 \mu_2, \quad \lambda_1 \lambda_3 = \mu_1 \mu_3, \quad \lambda_1 \lambda_6 = \mu_1 \mu_6, \quad \lambda_2 \lambda_4 = \mu_2 \mu_4, \quad \lambda_2 \lambda_5 = \mu_2 \mu_5. \end{align} $$

(3.10) $$ \begin{align} \lambda_3 \lambda_4 = \mu_3 \mu_4, \quad \lambda_3 \lambda_5 = \mu_3 \mu_5, \quad \lambda_4 \lambda_6 = \mu_4 \mu_6, \quad \lambda_5 \lambda_6 = \mu_5 \mu_6. \end{align} $$

From the remaining parts of $\nu = {pre}_2(\lambda )$ , the largest can be either $\lambda _1 \lambda _4$ or $\lambda _2 \lambda _3$ . We have a similar situation with ${pre}_2(\mu )$ . So, we will have either $\lambda _1 \lambda _4 = \mu _1 \mu _4, \ \lambda _2 \lambda _3 = \mu _2 \mu _3$ or $\lambda _1 \lambda _4 = \mu _2 \mu _3, \ \lambda _2 \lambda _3 = \mu _1 \mu _4$ .

Case 1: $\lambda _1 \lambda _4 = \mu _1 \mu _4, \ \lambda _2 \lambda _3 = \mu _2 \mu _3$ . This implies ${\lambda _2}/{\mu _2} = {\mu _3}/{\lambda _3} = {\mu _2}/{\lambda _2}$ , the latter equality following from (3.8). However, this gives $\lambda _2^2 = \mu _2^2$ and so $\lambda _2 = \mu _2$ as both are positive numbers. From (3.8), we then conclude that $\lambda = \mu $ .

Case 2: $\lambda _1 \lambda _4 = \mu _2 \mu _3, \ \lambda _2 \lambda _3 = \mu _1 \mu _4$ . Note that, after considering all the order relations in (3.9) and (3.10), the largest parts among the remaining parts of ${pre}_2(\lambda )$ and ${pre}_2(\mu )$ are $\lambda _1 \lambda _5$ and $\mu _1 \mu _5$ , respectively. Therefore, we deduce that $\lambda _1 \lambda _5 = \mu _1 \mu _5$ . Thus, ${\lambda _1}/{\mu _1} = {\mu _5}/{\lambda _5} = {\lambda _5}/{\mu _5}$ , the last equality coming from (3.8). So, with $\lambda _5^2 = \mu _5^2$ , we deduce again that $\lambda _5 = \mu _5$ and, hence, $\lambda = \mu $ .

Proof of Theorem 1.7

We consider two cases.

Case 1. Let $n+1$ be a composite number that is not a perfect square. Then, for each factorisation $n+1 = (a+1)(b+1)$ , with $a, b \geq 1$ , we obtain $n = ab + a +b = ab + b + a = {pre}_2(a + b +1)$ or ${pre}_2(b + a +1)$ . That is, for each factorisation of $n+1$ into two numbers greater than $1$ , we get a partition for n that is in the image of $pre_2$ . Moreover, interpreting $n+1 = (n+1)(0+1)$ with $b = 0$ gives $n = n = {pre}_2(n+1)$ . Thus, every pair of divisors of $n+1$ gives a partition of n that is in the image of $pre_2$ . Therefore, ${pre}_2(n) \geq \tfrac 12\tau (n+1)$ . If $n+1$ is a prime, then $\tfrac 12\tau (n+1) = 1$ and, again, we have ${pre}_2(n) \geq \tfrac 12\tau (n+1)$ .

Case 2. If $n+1$ is a perfect square, then in addition to its factorisations considered in Case 1, we also have $n+1 = (a+1)^2$ for some $a \geq 1$ . This gives $n = a^2 + a +a = {pre}_2(a+a+1)$ . Thus, using Case 1, ${pre}_2(n) \geq \tfrac 12(\tau (n+1) - 1) + 1 = \tfrac 12(\tau (n+1) + 1)$ .

Problem 5.1. Are there only finitely many n such that ${pre}_3$ is injective on partitions of n with three parts? In particular, do we have non-injectivity for all $n> 18$ , as the observed data seem to suggest? What about this phenomenon for ${pre}_k$ for $k \geq 4$ ? For sufficiently large n (depending on k), is ${pre}_k$ non-injective on partitions of n with k parts?

Problem 5.2. Can a proof by lattice analysis be given for the injectivity of ${pre}_2$ on partitions of n with more than six parts?

Problem 5.3. Does there exist an $n \geq 1$ such that $pre_2(n) = 1$ ?