Let $N,a,b,x,y$ be distinct positive integers such that
$$N = a^3 + b^3 = x^3 + y^3$$
Also known as Taxicab numbers or Taxi-cab numbers.
see also : https://oeis.org/A001235
let $t(n)$ be the $n$ th Taxi-cab number. So the first one is $t(1) = 1729$ because $1729 = 12^3 + 1^3 = 10^3 + 9^3$.
Let $\pi(n)$ be the counting function of the primes up to $n$. And $\zeta(s)$ is the regular Riemann zeta function.
Then it appears that for all positive $n$
$$ \lfloor(\frac{\pi(n)}{\zeta(3)})^{T}\rfloor< t(n)$$
with
$$T = 1+\dfrac{2+\dfrac{3+\dfrac{4+\cdots}{5+\cdots}}{4+\dfrac{5+\cdots}{6+\cdots}}} {3+\dfrac{4+\dfrac{5+\cdots}{6+\cdots}}{5+\dfrac{6+\cdots}{7+\cdots}}} $$
(The pattern is $+1$ when going up and $+2$ when going down).
Or even the slightly stronger version :
$$ \lfloor(\frac{\pi(n)}{\zeta(3)})^{\sqrt 3}\rfloor< t(n)$$
Notice $T = 1.728..$ (not $1.729$ lol) and $\sqrt 3 = 1.732..$
(No closed form is know for $T$ as far as I know. I incorrectly assumed it was $\sqrt 3$ here : Simplifying a complicated continued fraction expression. )
Let $w(n)$ be the counting function of the Taxi-cab numbers. So $(w(t(n)) = t(w(n)) = n$.
Then by looking at families of parametric solutions to
$$N = a^3 + b^3 = x^3 + y^3$$
it seems manageable to prove that for large $n$ we have
$$w(n) < \sqrt n + \sqrt{\sqrt n} + \sqrt{\sqrt{\sqrt n}} $$ but this estimates $t(n)$ around $n^2$ what appears too high for large $n$.
I tried some statistical ideas and sieve ideas but I did not get beyond this high estimate $t(n)$ around $n^2$.
I tried some analytic number theory with a generalized theta function $f(s)^k = (\sum_{n>0} s^{n^3})^k$ but also without succes.
How to prove one of these conjectures or get closer ? Or any good arguments pro or contra ?
edit :
Maybe this helps :
The complete solution in positive integers to,
$$x_1^3+x_2^3 = x_3^3+x_4^3$$
was given by Choudhry's On Equal Sums of Cubes (1998). For positive integers $a,b,c$,
$$\begin{aligned} d\,x_1 &= (a^2 + a b + b^2)^2 + (2a + b)c^3\\ d\,x_2 &= (-a^3 + b^3 + c^3)c\\ d\,x_3 &= (a^2 + a b + b^2)^2 - (a - b)c^3\\ d\,x_4 &= (a^3 + (a + b)^3 + c^3)c\end{aligned}$$
where, $$(a^3-b^3)^{1/3}<\,c\,<\frac{(a^3-b^3)^{2/3}}{a-b}$$
and $d=1$, or chosen such that $\text{GCD}(a,b,c)=1$.
P.S. For Choudhry's complete solution in positive integers to $x_1^3+x_2^3+ x_3^3=x_4^3$, see this post.
edit 2
A simple heuristic to get a power estimate is this :
$$a^3 + b^3$$
has the density $\sqrt n^3 = n^{\frac{3}{2}}$.
so
$$N = a^3 + b^3 = x^3 + y^3$$
has the density $ n^{(\frac{3}{2})^2} = n^{\frac{9}{4}}$.
or $n^{2.25}$
But that is much larger than $n^{\sqrt 3}$ and it also much larger than what the data suggests under some logical assumptions and extrapolation.
Say we estimate $t(n)$ as $C n^a $ or more precisely $1729 * n^a$.
Then we have $t(30000) = 520890296211 = 1729 * 30000^{1.89..}$ I ofcourse assumed the constant $C = 1729$ is not estimated to high.
see https://oeis.org/A001235/b001235.txt
If you estimate the constant $C$ much smaller such as $43$ we get $$ 43 * 30000^{2.25} <t(30000) < 44 * 30000^{2.25}$$
So the final words have not been said.
But from the data and arguments it seems the exponent is going down for large $n$.
We can also say $t(30000)$ is close to $9165 * 30000^{\sqrt 3}$ where $9165$ is going down as $n$ grows.
