Let $A=\{mn(m+n)\mid n,m\in \mathbb{N}_0\}$. Sorted, this is OEIS sequence A088915. What is its asymptotic behavior? It seems approximately $a(n)=O(n^{1.5})$, but not quite.

I'm actually even more interested in the asymptotic behavior of the sequence given by $AA$, the set of products of two elements of $A$.

Any ideas on how to approach these questions?

For background, there are monic fully reducible cubic polynomials $P,P+a \in \mathbb{Z}[X]$ if and only if $a\in AA$. Fully reducible means that $P$ and $P+a$ are both products of 3 linear terms.

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    $\begingroup$ I'd guess that the multiplicity with which an integer can be written as $mn(m+n)$ is usually small (maybe just $2$ typically). In any case the multiplicity is on average no more than powers of $\log$. Counting with multiplicity it would be easy to get an asymptotic for the number of $m$ and $n$ with $mn(m+n) \le x$: this is readily seen to be a constant times $x^{2/3}$. This is in keeping with what you notice on $a(n)$. Same comments apply to $AA$. $\endgroup$ – Lucia Mar 16 at 21:48
  • $\begingroup$ I notice that $30$ is only listed once on spite of 2,3 and 1,5. So are you taking multiplicity into account or not? $\endgroup$ – Aaron Meyerowitz Mar 17 at 1:54
  • $\begingroup$ @AaronMeyerowitz: a sorted set, so no multiplicies. $\endgroup$ – Yaakov Baruch Mar 17 at 8:06
  • $\begingroup$ Expanding on Lucia's insightful comment, $mn(m+n)\le N$ implies both $n\lessapprox\sqrt{N}$ and $m\le \frac{1}{2}\big(-n+\sqrt{n^2+\frac{4N}{n}}\big)$, therefore if $S$ is the count with multiplicities of the number of terms in $a$ up to $N$, then $S\approx \frac{1}{2}\int_1^{\sqrt{N}}\big(\sqrt{n^2+\frac{4N}{n}}-n\big)dn$... (cont) $\endgroup$ – Yaakov Baruch Mar 17 at 12:56
  • $\begingroup$ Substituting $N^{1/3}x$ for $n$: $S\approx \frac{N^{2/3}}{2}\int_{N^{-1/3}}^{N^{1/6}}\big(\sqrt{x^2+\frac{4}{x}}-x\big)dx\approx \frac{N^{2/3}}{2}\int_0^\infty\big(\sqrt{x^2+\frac{4}{x}}-x\big)dx\approx 2.6499581...\times N^{2/3}$. Empirically, if $T$ is the count without multiplicities, $T/(2.6499581...\times N^{2/3})$ seems to be increasing to a limit $\le1$, so that likely $T=O(N^{2/3})$. The definite integral above seems to match OEIS sequence A197374. $\endgroup$ – Yaakov Baruch Mar 17 at 12:56

It can be shown that the cardinality of

$$R(X) = \{n \in \mathbb{N} : \exists u,v \in \mathbb{Z} \text{ s.t. } n = uv(u+v), n \leq X\}$$


$$\displaystyle R(X) = C X^{2/3}(1 + o(1))$$

for some explicit constant $C$. Indeed, by the main result of this paper and the fact that $R(X)$ above is equal to half the size of the counting function in that paper, one gets that

$$\displaystyle R(X) = \frac{\Gamma^2(1/3)}{4\Gamma(2/3)} X^{2/3} + O(X^{1/2}).$$

We do not need the full strength of my paper with Cam Stewart in this case, since things are easier when the binary form is totally reducible. The totally reducible case has already been mostly done by Christopher Hooley in an earlier paper.

The counting function $R(X)$ I used is not exactly the one that you asked... since you insist that $m,n$ are positive. I suspect that a little bit more work would allow you to tweak the constant. The error term will not be problematic.

In general, it is not known whether there exist an infinite increasing sequence $\{k_\ell\}$ of natural numbers such that the sets $S_\ell = \{(m,n) \in \mathbb{Z}^2 : \gcd(m,n) = 1, mn(m+n) = k_\ell\}$ satisfy $|S_{\ell + 1}| > |S_{\ell}|$ for all $\ell \geq 1$. If such a sequence exists, then there exists a sequence of elliptic curves whose Mordell-Weil rank tends to infinity.

However, one can show (as is done in full generality in the above linked paper) that such occurrences are rare: that is, for 100% of integers representable by a non-singular integral binary form of degree $d \geq 3$, the only repetitions are accounted for by $\operatorname{GL}_2(\mathbb{Q})$-automorphisms.

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    $\begingroup$ Notice that $\frac{\Gamma^2(1/3)}{4\Gamma(2/3)}$ is exactly half of my estimate with multiplicities, which is $2.6499581...×C^{2/3}$. This means that the multiplicity of a number is on average $1$ (aside from switching $m$ and $n$). $\endgroup$ – Yaakov Baruch Mar 19 at 21:10

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