# A simple irrational number with the same first 11,667,755 digits as $\frac{2}{3}$

Let $$f(x) = \frac{1}{2}+\sum_{k=1}^\infty \frac{\mbox{sgn}(\sin kx)}{2^{k+1}} .$$ Here $$\mbox{sgn}$$ represents the sign function. Many simple integer and rational values of $$x$$ result in $$f(x)$$ very closely approximating some simple rational numbers, and you don't have to spend much time to identify plenty of them. Yet it seems obvious that if $$x$$ is rational, then $$f(x)$$ is irrational. One number that stands out is $$x = 10^5 + \frac{1}{10}\cdot\Big(\frac{3}{5}\Big)^2.$$

Surprisingly, $$f(x)$$ is almost equal to $$\frac{2}{3}$$, as the first $$12,897$$ binary digits of both numbers agree. Just after that, they disagree. You don't need a sophisticated algorithm to check this. Just compute $$\mbox{sgn}(\sin kx)$$ for $$k=1, 2, \cdots, 12,897$$. These signs alternate perfectly depending on whether $$k$$ is odd or even, just like the binary digits of $$\frac{2}{3}$$.

Question

I started to have some doubts about the fact that if $$x$$ is rational, then the sequence $$z_k = \mbox{sgn}(\sin kx)$$ can not be periodic. Can someone prove that I am right, and that this weird number $$x$$ is just a coincidence, not leading to periodicity anyway. Do you have some explanation for these coincidences for so many different $$f(x)$$ values: very often, 20 or 30 binary digits match those of a simple rational, sometimes 40 and even 87 digits for the number $$x=10^5$$ itself -- but no pattern for $$x=10^5-1, x=10^5-\frac{1}{10}, \mbox{ or } x = 10^5+1$$. A pattern again for $$x=2\cdot 10^5$$ and for so many other numbers, starting with $$x=1$$ resulting in $$f(x)=0.11111113\cdots$$ (in base $$10$$).

Update

Another number leading to almost periodicity is $$x=\log_2 3$$ resulting in $$f(x) = 2/5$$ (almost). But $$x=\sqrt{2}/2$$ does not yield the same spectacular result. It is a hit and miss.

Finally, try $$x=\frac{355}{113}$$. The first $$11776655$$ binary digits of $$f(x)$$ are identical to those of $$\frac{2}{3}$$. Not only $$11776655$$ is large, but even more surprising, look at the base-$$10$$ digits of $$11776655$$: two $$1$$, two $$7$$, two $$6$$, two $$5$$. Note that if you concatenate the base-$$10$$ digits of $$355$$ with those of $$113$$, you get $$113355$$.

The reason for the near-periodic behavior of the choice $$x = 10^5 + \frac{9}{250}$$ has to do with the fact that $$\frac{x}{\pi} \approx 31831.00007753496977,$$ which is almost an integer with error $$\epsilon \approx 0.000077 < 10^{-4}$$. Moreover, $$\frac{1}{\epsilon} \approx 12897.4062021729,$$ and now you can see why this many terms are needed.
The above also suggests that if you can find some choice of $$x$$ such that $$\frac{x}{\pi} - \left\lfloor \frac{x}{\pi} \right\rfloor$$ is extremely tiny, you can make this phenomenon extend to as large of a value of $$k$$ as you please. It just so happens that the particular choice $$10^5 + \frac{9}{250}$$ is also close to a round number in base 10.
• Based on your answer, I updated my post, now with $x=\frac{355}{113}$ and the first $11776655$ binary digits of $f(x)$ equal to those of $\frac{2}{3}$. – Vincent Granville Sep 2 '19 at 15:31