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The taxicab numbers are those integers of the form

$$N=a^3+b^3=c^3+d^3$$

where $a, b, c, d \in \mathbb{N}$ and $\{ a, b \} \neq \{c, d \}$. Famously, the first taxicab number is $1729 = 1^3+12^3 = 9^3 + 10^3$.

Let us consider the list of taxicab numbers in ascending order and keep track of what proportion of them are even. After the first $30,000$ terms, it appears that a bias towards evenness is present:

Graphs showing the proportion of even taxicab numbers amongst taxicab numbers, which clearly shows a bias towards even taxicabs.

This seems surprising to me. After $30,000$ terms, we have about $\sim 63\%$ of taxicab numbers being even, quite substantially more than half. Why does this appear to be the case?

This apparent bias seems to come from the fact that the numbers are sums of cubes in two different ways. Indeed, the weaker property of a number merely being the sum of two cubes results an even split between even and odd numbers - intuitively this is to be expected, with formal argument can be made along the lines of observing that the sums of cubes modulo $2$ are $0^3+0^3, 0^3+1^3,1^3+0^3$ and $1^3+1^3$, two of which are $0$ and two of which are $1 \mod 2$.

It's possible that this trend does not continue and after some extensive enumeration of the taxicabs, perhaps the proportion drops back down again (similar surprising things can happen). But the consistency of the trend is so consistent$^1$ and remarkable that I thought a question about it is of note regardless, even if it is merely a coincidence.

$^1$ indeed, the only times that we ever have more odd than even taxicabs enumerated in my data is after enumerating the first $1, 35$ and $37$ taxicab numbers.

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    $\begingroup$ My guess, if someone wants to explore it: some of the taxicab numbers are “imprimitive”—just multiples of earlier taxicab numbers, like $1729d^3$ for any $d\ge2$. I suspect that the primitive taxicab numbers are equally likely to be even or odd, but including the imprimitive ones introduces the bias towards evenness (and towards multiples of any fixed integer). $\endgroup$ Commented 2 days ago
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    $\begingroup$ May be it can help to understand the phenomen if you exclude all imprimitives taxicab numbers and calculate proportions with this new rule. $\endgroup$ Commented 2 days ago
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    $\begingroup$ @GregMartin To test this, the primitive taxicab numbers are also enumerated on OEIS as A018850. This includes a table going out to the 9859th such term. After importing to Mathematica the split seems to be 5904 odd to 3955 even, i.e., roughly 60:40 in favor of odds over evens. So the primitive taxicab numbers also seem to be biased, but in the opposite direction! $\endgroup$ Commented yesterday

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Here's a heuristic argument. Let's look at pairs $(a, b)$ and the resulting parity of $a^3 + b^3$ $\pmod 8$

Even cubes, $(2k)^3 \equiv 0$

Odd cubes, $(2k+1)^3 \equiv 2k+1$. So they're evenly distributed between ${1,3,5,7}$

An odd sum of cubes comes from the sum of an even and odd cube, so they're evenly distributed among $1,3,5,7$. (representing $1/2$ of integers).

An even sum of cubes comes from either two evens or two odds. The (even + even) pairs are equivalent to $0$ ($1/8$ of integers). The (odd + odd) pairs are evenly distributed among the remainders $0, 2, 4$, and $6$ (representing $1/2$ of the integers).

Now we can calculate the expected proportions of even and odd taxicab numbers.

  1. Odd Taxicabs (Sums $\equiv 1, 3, 5, 7 \pmod 8$). $1/2$ of all pairs land here, distributed evenly across $1/2$ of the integers.$$\text{Odd Taxicabs} \propto \frac{(1/2)^2}{1/2} =\frac{1}{2}$$

  2. Even Taxicabs (Sums $\equiv 0 \pmod 8$). This specific remainder makes up $1/8$ of the integers. The pairs that land here are the (even + even) pairs ($1/4$ of total pairs), plus $1/4$ of the (odd + odd) pairs. $$(\text{Even Taxicabs}\equiv 0) \propto \frac{(1/4+(1/4)(1/4))^2}{1/8} = \frac{25}{32}$$

  3. Even Taxicabs (Sums $\equiv 2, 4, 6 \pmod 8$). Together, these three remainders make up $3/8$ of the integers. The pairs that land here are the remaining (odd + odd) pairs. $$(\text{Even Taxicabs}\equiv 2, 4, 6) \propto \frac{((3/4)(1/4))^2}{3/8} = \frac{3}{32}$$

So, the expected proportion of Even Taxicabs is:$$\frac{\text{Even}}{\text{Even} + \text{Odd}} = \frac{25/32+3/32}{25/32+3/32 + 16/32} = \frac{7}{11}\approx63.6\%$$

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  • $\begingroup$ A more direct way, compute table $(i^3+j^3)_{i,j}$ in $\mathbb{Z}/8$, count occurences of each number $N(k)$, then calculate $(\sum_{\text{even}\;k}N(k)^2)/(\sum_{\text{all}\;k}N(k)^2)$. But I still wonder why $8$ is chosen? Choosing other numbers gives different results (however, bias remains). $\endgroup$ Commented 2 hours ago

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