# 355/113 and small odd cubes

An important approximation to $$\pi$$ is given by the convergent $$\frac{355}{113}$$.

The numerator and the denominator of this fraction are at the same distance of small consecutive odd cubes.

$$\frac{355}{113} = \frac{7^3+12}{5^3-12}$$

Is this a consequence of some general formula, such as a series or continued fraction?

Attempts, drafts

$$\frac{113}{355} = \frac{1}{10}\left(2+\frac{5^3}{3+\frac{7^3}{4}}\right)-\frac{8}{355} = \frac{1}{5}\left(3-\frac{5^3}{3+\frac{7^3}{4}}\right)$$

Similarly, for $$\frac{22}{7}$$

$$\frac{7}{22} =\frac{1}{2}\left(\frac{3^3}{1+\frac{4^3}{2}}\right) -\frac{1}{11}$$

For the first convergent $$3$$, $$\frac{3^3-6}{1^3+6}=\frac{21}{7}=3$$

The tentative expression $$a(n)=\frac{(4n+3)^3+6(3n-1)}{(4n+1)^3-6(3n-1)}$$

has $$a(0)=3$$ and $$a(1)=\frac{355}{113}$$ but $$\lim_{n \to \infty} a(n) = 1 \neq \pi$$

A similar fraction for the approximation $$\frac{223}{71}$$ (Archimedes' lower bound) is $$\frac{98+5^3}{98-3^3}=\frac{2·7^2+5^3}{2·7^2-3^3}=\frac{223}{71}=3+\frac{1}{7+\frac{1}{10}}$$

• If so would $22/7$ be expressible that way also, being an earlier convergent to the (simple) continued fraction for $\pi$ ? Commented Feb 7, 2021 at 6:49
• Not sure how to reach $\frac{22}{7}$, but $\frac{25-3}{4+3}$ would suggest squares instead of cubes. From cubes, $\frac{22}{7}=\frac{3^3-5}{2^3-1}$, not that nice. Commented Feb 7, 2021 at 7:03
• When a partial quotient is 1, we get convergents that are neighbours in a Farey sequence eg, $22/7, 355/113, 333/106$, with $355 = 22 + 333$ and $113 = 7 + 106$. That can give us linear patterns, but of course it doesn't explain squares & cubes. Commented Feb 7, 2021 at 7:29
• I've found a couple via the convergents of $\sqrt[3]\pi$: $$\frac{517^3+13607}{353^3-13607} = \frac{22}{7}$$ and $$\frac{41^3+10}{28^3-10} = \frac{333}{106}$$ Commented Feb 7, 2021 at 8:58
• I solved for $x$ in $(u^3+x)/(v^3-x) = p/q$, where $p/q$ is a $\pi$ convergent & $u/v$ is a $\sqrt[3]\pi$ convergent. I just did a brute-force search, looking for combinations that result in an integer value for $x$. Commented Feb 7, 2021 at 10:38

For a coprime pair of integers $$1 \le p,q$$, consider the following diophantine equation: $$\frac{x^3+z}{y^3-z}= \frac{p}{q}$$ with $$x,y,z$$ integer unknowns. This is equivalent to $$z= \frac{y^3p-x^3q}{p+q}$$ Thus a solution exists if and only if there exist $$x,y$$ such that $$y^3p-x^3q \equiv 0 \pmod{(p+q)}$$ Now, note that $$-q \equiv p$$, and that it has an inverse $$\mod (p+q)$$. This means that we need $$x^3+y^3 \equiv 0 \pmod{(p+q)}$$
Now, if we pick $$p=355$$ and $$q=113$$, we get $$p+q=468 = 7^3+5^3$$ In other words, the fraction $$355/113$$ has such a special property not because it is a convergent of $$\pi$$, but simply because the sum of numerator and denominator is a sum of two small cubes. We can do the same job with the fraction $$\frac{203}{265} = \frac{5^3+78}{7^3-78}$$ nothing special with $$\pi$$. As for $$22/7$$ we can find $$\frac{22}{7} = \frac{17^3+5^3}{12^3-5^3}$$ which is even more surprising. In my opinion, having three variables gives you a lot of freedom, so that yo can find such nice forms for any rational number.
EDIT: Just for fun, I looked up convergents of a completely different constant, namely $$\gamma$$: Euler-Mascheroni constant. One of its convergents is $$\frac{71}{123}= \frac{3^3+470}{11^3-470}$$ Again, there is nothing special about this rational, simply you can find such a form for any number.
• Great. A graphic for that beautiful $\frac{22}{7}$ would have the same symmetry as the one in the question if $12$ was odd, or the four cubes even. Maybe your answer generalizes the question to how to build a continued fraction for a constant, that includes a particular approximation. Commented Feb 10, 2021 at 5:53
• For $\gamma$, $3$ and $11$ would allow for two matching near-cubes, but $470=2·5·47$ has a factor too big for $3$ and $11$. The symmetry of the $\frac{355}{113}$ picture is due to the small factors in $12=2·2·3$. Commented Feb 10, 2021 at 6:09
• Some p, q combinations generate a lot of solutions, so my script only prints solutions with $|z|<\min(x, y)^2$, sorted (ascending) on $|z|$. Commented Feb 11, 2021 at 18:50
• Very interesting, thank you! I got no solution for $\frac{25}{8}$ (a semiconvergent for $\pi$). Are there rationals that do not accept such a form or some issue in the implementation? Commented Feb 11, 2021 at 22:06