I was playing with Excel and created a sequence with some interesting features.
In column A, list the sequence $\color{red}{a_n=\lceil \sin (2n) \rceil}$ (using the ceiling function).
In column B, list the sequence of gaps between the terms of the previous sequence.
In column C, list the sequence of gaps between the terms of the previous sequence.
And so on.
Let $b_n$ be the sequence in the top diagonal (highlighted).
Behold the graph of $|b_n/2^n|$ against $n$:
The terms on the approximate horizontal line are very close to $1/44$.
What's going on here?
More specifically:
- Why does dividing the terms of $|b_n|$ by $2^n$ yield such stability (as opposed to making the terms go toward $0$ or infinity)?
- Why are there long strings of values that are very close to $1/44$?
- Why does the graph dip to $0$ at $n=698$ and then go back up?
- How does the graph behave for $n>1000$?
I only have a fuzzy notion that the last three questions may be related to rational approximations for $\pi$.
For comparison purposes only: graphs derived from other starting sequences
If $a_n=\lceil \sin (\color{red}{1}n) \rceil$, here is the graph of $|b_n/2^n|$ against $n$:
If $a_n=\lceil \sin (\color{red}{3}n) \rceil$, here is the graph of $|b_n/2^n|$ against $n$:
If $a_n=\lceil \sin (\color{red}{4}n) \rceil$, here is the graph of $|b_n/2^n|$ against $n$:
If $a_n$ is a random binary sequence with $0$ and $1$ equally likely, here is one possible graph of $|b_n/2^n|$ against $n$:
If $a_n$ is the sequence of prime numbers, here is the graph of $|b_n/2^n|$ against $n$:
(Incidentally, the last two graphs resemble "Mountains of Guilin" functions, i.e. $f_{p_j,n}(x)=|\sin (p_1x)+\sin (p_2x)+\dots+\sin (p_nx)|$ where $p_j$ are linearly independent over $\mathbb{Q}$.)
(The answers to this question might resolve an earlier question of mine: Strange dips in sequence $u_n=\log{|(n-1)^{\text{st}}\text{ difference of first $n$ primes}|}$.)
1. SOME CLAIMS
Fix a positive real number $\alpha$ and consider $$ a_k:=\begin{cases} 1,& \{\alpha k\}<1/2, \\ 0,& \{\alpha k\}\geq 1/2. \end{cases} $$ (Using $\lbrace x\rbrace= x-\lfloor x\rfloor\in [0,1)$, the fractional part of a real number $x$.)
Then, form the sequence $$ b_n=\sum_{k=1}^n \,(-1)^k \,\binom {n-1}{k-1}\, a_k, $$ and also $$ c_n=2^{1-n}|b_n|. $$
I will now make several claims. Hopefully they are true, and hopefully the gaps in the proofs can be filled in.
CLAIM 1. When $\alpha=1/\pi$, this gives rise to the sequence in the post, with the first values of $a_k$ being $$ 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, $$ and the first values of $b_n$ being $$ -1, -1, -1, 0, 3, 9, 18, 27, 27, 0, -81, -243, -486, -728, -715, 105, 2746, 8924, 21537, 45847. $$
This is easy. We have $\lceil\sin(2k)\rceil=$1 or $0$ precisely if $\{k/\pi\}<1/2$ or $>1/2$, respectively, hence $a_k=\lceil\sin(2k)\rceil$ for $\alpha=1/\pi$. Then use induction (or the formula for powers of the difference operator) to prove the formula for $b_n$.
Define, for $q>r\geq 0$ integers, $$ f_{n,r,q}=2^{1-n}\sum_{\substack{k=1,\ldots,n\\ k\equiv r\mod q}}^n(-1)^k \binom{n-1}{k-1}. $$
CLAIM 2. Let $\omega=\exp(2\pi i/q)$. Then $$ f_{n,r,q}=-\frac 1q\sum_{\ell=0}^{q-1}\frac{(1-\omega^\ell)^{n-1}}{2^{n-1}}\omega^{-\ell (r-1)}. $$ As a consequence, as $n\to+\infty$, $$ f_{n,r,q}\to\begin{cases}\frac {(-1)^r}{q},&\mbox{if }q\mbox{ is even},\\ 0,&\mbox{otherwise}.\end{cases} $$
(See these nice notes, Section 6.)
CLAIM 3. We have $\lim_{n\to+\infty}c_n=0$ unless $\alpha=p/q$ is rational with $p,q$ coprime integers and $q$ equal to twice an odd number, in which case $\lim_{n\to+\infty}c_n=1/q$.
Ideas for a proof: If $\alpha=p/q$ is rational ($p,q$ coprime integers) then $$ 2^{1-n}b_n=\sum_{\substack{r=0,\dots,q-1 \\ \lbrace rp/q\rbrace<1/2}}f_{n,r,q} $$ and so, by the previous point, we get $$ \lim_{n\to+\infty}c_n=\biggl|\lim_{n\to+\infty}2^{1-n}b_n\biggr| = \frac 1q\biggl|\sum_{\substack{r=0,\dots,q-1 \\ \lbrace rp/q\rbrace<1/2}}(-1)^r\biggr|. $$ It should not be too hard to prove that $$ \sum_{\substack{r=0,\dots,q-1 \\ \lbrace rp/q\rbrace<1/2}}(-1)^r =\begin{cases} \pm1,& \mbox{if }q=2\times\mbox{ odd integer},\\ 0,&\mbox{otherwise}. \end{cases} $$
If instead $\alpha\not\in\mathbb Q$, we can pick (Dirichlet approximation) a fraction $p/q$ ($p,q$ coprime) such that $|\alpha-p/q|<1/q^2$ with $q$ arbitrarily large. Then $a_k(\alpha)=a_k(p/q)$ for all $k\leq q$ (or, at least, for most of the values, in a way to be made precise!), such that $c_n(\alpha)$ is essentially $c_n(p/q)$ for all $n\leq q$ and one should bound $c_n(p/q)$ in a smart way, using the explicit expression for $f_{n,r,q}$. I may try to sort out the details in the future.
2. NUMEROLOGY IN THE PLOT AND THE CONTINUED FRACTION EXPANSION OF $\pi$.
The above facts being established (rather, claimed), let us look at the following plots of $c_n$ for $\alpha=1/\pi$ (i.e., the original quantity of interest in the question, up to a factor of $2$): first for $n=1,\ldots,10000$
and then for $n=100,\ldots,50000$ (plotting for values multiple of $50$ only)
(The last plot confirms that eventually, in the long run, $c_n\to 0$ for $\alpha=1/\pi$).
It is natural to guess that the following two facts observed in the question might have an interesting explanation in terms of the first few convergents of $\pi$ $$ 3,\frac{\color{red}{22}}{7},\frac{333}{106},\frac{\color{red}{355}}{113},\frac{103993}{33102},\ldots $$ ($355/113$ being amazingly good).
The values of $c_n$ at the plateaux are very close to $1/\color{red}{22}$ for the first several plateaux.
The separation between plateaux occurs at values which are very close to even multiples of $\color{red}{355}$.
In fact, an explanation for 1. is that the values of $a_k$ for $\alpha=1/\pi$ and $\alpha=7/22$ coincide for all $k<355$. Since $22$ is twice an odd integer, the graph for $\alpha=7/22$ stabilizes to $1/22$. Since the graph for $\alpha=1/\pi$ must coincide with that for some time, this essentially explains the plateaux and its value $1/22$.
Similarly, the $a_k$'s for $\alpha=1/\pi$ and $113/355$ coincide for all $k<52174$. So, even the larger plot reported above coincides with the one for $113/355$, whence it should not be surprising the all the properties of the plot so far are determined by these first convergents.
As a similar example we may consider the convergents of $3/55$, which are $\frac 1{18}, \frac 3{55}$. Here are the plots of $c_n$ for these two values: they coincide for a while, then one stabilizes to $\frac 1{18}$ and the other one eventually goes to zero exhibiting a periodicity $2\times 55$. Again, the initial plateaux of the plot for $3/55$ are explained by the first convergent $1/18$.
3. FINAL COMMENTS
In conclusion, comments about the 4 questions in the original post.
Dividing by $2^n$ (rather, by $2^{n-1}$, as it seemed more natural to me) "stabilizes" the sequence $b_n$ probably because $b_n$ is a sum of terms $-a_1,\dots,(-1)^na_n$ (more or less half of which are $\pm 1$s and half $0$s) against binomial coefficients $\binom{n-1}{k-1}$. Therefore, $$ \sum_{k=1}^n\binom{n-1}{k-1}=2^{n-1} $$ is the appropriate normalization for such a quantity (which can now be interpreted as an average of terms equal to $\pm 1$ and $0$).
In view of this, plateaux and dips in the plot probably correspond to a different balances of $+1$ and $-1$ in these sums weighted by binomials (constructive or destructive interference of terms of different signs).
The plot eventually goes to zero.