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A very specific question: What's the sum of the reciprocals of the numbers that can be written as the sum of two positive cubes (Oeis: A003325)?

$$\begin{align} &\sum_{n=1}^\infty \frac{1}{A003325(n)} \\ & =\frac{1}{1^3+1^2}+\frac{1}{1^3+2^3}+\frac{1}{2^3+2^3}+\frac{1}{1^3+3^3}+\frac{1}{2^3+3^3}+\frac{1}{3^3+3^3}+\frac{1}{1^3+4^3}+\dots \\ & = \frac{1}{2}+\frac{1}{9}+\frac{1}{16}+\frac{1}{28}+\frac{1}{35}+\frac{1}{54}+\frac{1}{65}+\frac{1}{72}+\frac{1}{91}+\dots \end{align} $$

Some notation to organize my thinking:

Let $S\subset \mathbb{N}$ and let $1_{S}(x)=\cases{1 \mbox{ when }{x\in S }\\0 \mbox{ when } x\notin S}$

And define $m(S)=\sum_{n=1}^\infty \frac{1_S(n)}{n}$. This would be the sum of the reciprocals of $S$.

$S_{a,b}=\{n\in \mathbb{N}: \exists \vec{x} \in\mathbb{N^{a}}, \sum_{i=1}^a |x_i|^b=n \}$. This is the set of numbers that can be written as sum of $a$ positive numbers raised to the fixed power $b$.

Some immediate results of this notation. And some examples.

$m(S_{1,k})=\zeta(k)$ and $m(S_{4,2})$ is a divergent sum because $S_{4,2}=\mathbb{N}$.

My question above is asking for $m(S_{2,3})$. And somewhat more broadly I am asking: Can we get a handle on the density of this set? Why should we have any chances of answering this?

As we can read here, here, here and here. The first two links are MSE questions and the latter two are publications by Kevin A. Broughan.

$$n\in S_{2,3} \iff \exists m \mid n ,\quad n^{1/3} \leq m \leq 4^{1/3} n^{1/3} \mbox{ s.t. }\\ ( m^{2} - \frac{n}{m})=3l \mbox{ and }(m^{2} - 4l) \mbox{ is a perfect square. }$$

So because we have this nice characterization I was curious if this number could possibly be expressed in relationship to other well-known constants like value of the zeta function.

Note that the related question: What's the sum of the reciprocals of the numbers that can be written as the sum of two non-negative cubes? But we can see that this question is just $\zeta(3)$ away from the title question. That is,
$$m(\{n:(x,y)\in \mathbb{N_0^2}: n=x^3+y^3 \})=m(S_{2,3})+\zeta(3)$$

Here is what I know so far $m(S_{2,3}) \approx 0.9777693455 =\sum_{n=1}^{20000} \frac{1}{A003325(n)}$.

Can anyone provide any more insight into what this number is?

Apologies

Note that in the original version of this I made a typo and wrote $S_{a,b}=\{n\in \mathbb{N}: \exists \vec{x} \in\mathbb{Z^{a}}, \sum_{i=1}^a |x_i|^b=n \}$ this isn't quite what I wanted to type here.

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  • $\begingroup$ Presumably, the number of integers up to $N$ expressible as a sum of two positive cubes is on the order of $N^{2/3}$. $\endgroup$ – Gerry Myerson Sep 19 '19 at 1:52
  • $\begingroup$ @GerryMyerson. Cool! Why is that? $\endgroup$ – Mason Sep 19 '19 at 2:00
  • $\begingroup$ Because the number of cubes up to $N$ is essentially $N^{1/3}$, so the number of pairs of cubes (up to a constant multiplier) is $N^{2/3}$. Some of these pairs have a sum exceeding $N$, but excluding them should just bring in another constant multiplier. $\endgroup$ – Gerry Myerson Sep 19 '19 at 2:27
  • $\begingroup$ And then what would happen for $4$ cubes? It cannot be $N^{4/3}$... I am not quite seeing how that should work. But thanks. It might click into place. I will think on this. $\endgroup$ – Mason Sep 19 '19 at 2:30
  • $\begingroup$ It works so long as it gives an answer significantly less than $N$. $\endgroup$ – Gerry Myerson Sep 19 '19 at 2:37
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I would like to mention that from Parseval theorem and the reflection formula we can achieve (though perhaps not to our full liking) a formulation for $L$ defined below which is related to the OP's query but not quite the same. $$L=\sum_{n,m \ge 1} \frac1{n^3+m^3} = \int_0^\infty f(x)^2dx= \frac1{2\pi} \int_{-\infty}^\infty |F(1/2+it)|^2dt$$ $$=\frac12 \int_{-\infty}^\infty \frac{|\zeta(3/2+3it)|^2}{|\sin(\pi /2+i\pi t)|}dt=\frac12 \int_{-\infty}^\infty \frac{|\zeta(3/2+3it)|^2}{\cosh(\pi t)}dt$$

Unfortunately the residue theorem doesn't apply.

$$f(x)= \sum_{n\ge 1} e^{-n^3x},\qquad F(s)=\Gamma(s)\zeta(3s)= \int_0^\infty x^{s-1}f(x)dx$$

The $L$ defined above should be approximately $2m(S_{2,3})-\zeta(3)$

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  • 2
    $\begingroup$ The first sum in the answer (call it $S$) does not appear to give what OP wants, as for $m \neq n$ it counts twice the reciprocal of $m^2 + n^2$ (one for each ordering of $(m, n)$.) So, OP's desired sum is approximately $\frac{1}{2}(S + \zeta(3))$. But even this adjustment slightly overcounts, as there are some integers that are expressible as the sum of two different pairs of cubes, and so whose reciprocals are counted twice (or more) e.g. $1^3 + 12^3 = 9^3 + 10^3$. $\endgroup$ – Travis Willse Sep 19 '19 at 6:56
  • $\begingroup$ I'd say there infinitely many such pairs of sums of cubes techiedelight.com/numbers-represented-as-sum-of-two-cubes if so it is quite obvious (no generating function...) we can't say anything of the sum over the integers represented as a sum of cubes instead of the sum over the pairs of cubes. The "characterization" of OP's post is trivial and it doesn't help. $\endgroup$ – reuns Sep 19 '19 at 7:13
  • $\begingroup$ I don't disagree. (Well, OP's "characterization" is useful for bounds, but probably we can get tighter ones with the approach in my comment.) I was just pointing out the difference between what OP asked for and what was written here. $\endgroup$ – Travis Willse Sep 19 '19 at 7:22
  • $\begingroup$ And there are certainly infinitely many such pairs of sums---after all, if $a^3 + b^3 = c^3 + d^3$ then $(ka)^3 + (kb)^3 = (kc)^3 + (kd)^3$ for all positive integers $k$. The contributions from any one of these families is controllable, but I would speculate that there are even infinitely many coprime $(a, b, c, d)$ satisfying the condition, i.e., infinitely many such families (and so the problem is still apparently intractable in the sense of your comment). $\endgroup$ – Travis Willse Sep 19 '19 at 7:25
  • $\begingroup$ I proposed a rewrite which includes some of the content of this comment thread. I just got the power to edit users answers without approval. I hate it. Feel free to roll back if you don't approve. $\endgroup$ – Mason Sep 19 '19 at 23:11

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