$\sum_{k=1}^{n} \arcsin(\sin(k))= a_n+b_n \pi$, for $a_n,b_n\in\mathbb{Z}$; what can be said about the sequence $(a_n,b_n )$?

This may help: $$\arcsin(\sin(x)) = |((x-\pi/2) \bmod (2\pi))-\pi|-\pi/2$$ or equivalently $$\arcsin(\sin(x)) = \left|x-3\pi/2 -2\pi\left\lfloor\frac{x-\pi/2}{2\pi}\right\rfloor\right|-\pi/2$$

If we write $\arcsin(\sin(k)) = x_k + y_k \pi$ with $x_k, y_k \in \mathbb Z$ we have $$x_k =\operatorname{sgn}(\cos(k))\,k$$ and $$y_k=-\operatorname{sgn}(\cos(k))\left(\frac{3}{2}+2\left\lfloor\frac{k-\pi/2}{2\pi}\right\rfloor\right)-\frac{1}{2}$$

This is very interesting !

For the fun of it, I computed the ratio's $$R_k=-\frac{a_{10^k}}{b_{10^k}}$$ and obtained the following sequence $$\left\{1,\frac{17}{6},3,\frac{22}{7},\frac{22}{7},\frac{931}{296},\frac{10559}{3361},\frac{1093611}{348107},\cdots\right\}$$ For the last one $$\frac{1093611}{348107}-\pi=1.74 \times 10^{-6}$$

I will try to express the conditions needed to answer the following sub-question:

For $S(n) = \sum\limits_{k=1}^n \text{arcsin}(\sin(k)) = a_n+b_n \pi, a_n \in \mathcal{N}\text{ and } b_n \in \mathcal{N}$, for which $n$ is $a_n = b_n = 0$?

I analyzed $S(n)$ in frequency domain by taking the Poisson Sum (omitted due to length, but I can post it if required). After some simplifications, the following was obtained.

Let $m = \left\lfloor \frac{2n+1}{2\pi} + \frac12 \right\rfloor$

$a_n = b_n = 0$ when the following conditions are met

1. $m$ is even

AND

2. $$\sum\limits_{j=0}^{\frac{m}{2}-1} \left( \frac{\log^2(-e^{i (4j+3) \pi^2})-\log^2(-e^{i (4j+1) \pi^2})}{2 \pi^2} +4(2j+1)\pi^2\right) \\= m(2n+1)\pi - n(n+1)$$

Notes:

1. I verified above for $n=24$, $n=44$, $n=68$ and $n=103640$.
2. This is not a fundamental change in the complexity of the problem. Due to irrationality of $\pi$, a simple-looking expression like $\log(-e^{i 5 \pi^2})$ depends on the number of $i 2\pi$ rotations in the $i 5 \pi^2$ to get the correct logarithm branch. Same irrationality hurdle when summing up multiple terms like that. Someone may be able to simplify that.

Edit1

Not that it is any simpler, but I got rid of the logarithm and exponential by writing the condition as follows

$$\sum\limits_{j=0}^{\frac{m}{2}-1} \left( \frac{\left(1+(4j+1)\pi-2\left\lceil \frac{\left\lfloor 1 + (4j+1)\pi \right\rfloor}{2}\right\rceil\right)^2-\left(1+(4j+3)\pi-2\left\lceil \frac{\left\lfloor 1 + (4j+3)\pi \right\rfloor}{2}\right\rceil\right)^2}{2 } +4(2j+1)\pi^2\right) \\= m(2n+1)\pi - n(n+1)$$

Edit2

If the conjecture "$b_n = 0$ if $a_n=0$" holds (I can't prove it currently, will work on it), we can simply equate the "real"(constant) parts and "imaginary" (ones with $d \pi$, $d \in \mathcal{Z}$) on both sides of my equation to derive a simpler looking equation as follows.

$$\sum\limits_{j=0}^{m-1} (-1)^j\left(\left\lceil \frac{\left\lfloor 1 + (2j+1)\pi \right\rfloor}{2}\right\rceil^2 - \left\lceil \frac{\left\lfloor 1 + (2j+1)\pi \right\rfloor}{2}\right\rceil\right) + \frac{n(n+1)}{2} = 0$$

Derivation of the Original Equation

First, let's note that $\text{arcsin}(\sin(x))$ is an odd-symmetric periodic waveform with periodicity $2\pi$ as shown below

When we do the sum $S(n) = \sum\limits_{k=1}^n \text{arcsin}(\sin(k))$ in "time domain", it is taking first $n$ snapshots of this function and summing them up. Poisson summation says that infinte summation of time-domain integer-spaced samples is the same as infinite summation of frequency domain integer-spaced samples. However, the Poisson summation reduces to triviality of $0=0$ for an odd-symmetry waveform. So I converted it to an even-symmetry waveform by flipping the polarity of the waveform on the negative $x$-axis.

We can represent the original waveform as $\frac{\pi}{2}$TriangularWave$\left(\frac{x}{2\pi}\right)$. The even-symmetry waveform can be obtained by multiplying with $\text{sgn}(x)$. Further to limit the summation to just $n$ samples, we can multiply it with a Rect function $\operatorname{rect}\left(\frac{x}{2n+1}\right)$. Note that $2n+1$ (instead of $2n$) was used to fully cover the $n^{th}$ sample, since the behavior of rect function at the discontinuity is poorly defined. Combining it all, we get the truncated even-symmetry time-domain waveform $$E(x) = \frac{\pi}{2}\operatorname{TriangularWave}\left(\frac{x}{2\pi}\right) \operatorname{sgn}(x) \operatorname{rect}\left(\frac{x}{2n+1}\right) $$ Example shown for $n=10$

Now it is ready for Fourier Transformation and subsequent Poisson sum. I couldn't derive a general expression for the Fourier Transform$\mathcal{F_y}(E(x))$. So I brute forced it in Mathematica for a few terms (it was choking for large $n$). My code used variable $i$ for $n$.

Note that the first $2$ and last $2$ terms are in every entry (similar, with a pattern except for sign difference), but the number of middle entries increases every once in a while (with alternating sign of the new entry). Initially it looked like for $3$ consecutive $n$, the number of middle terms is the same. BUT, the pattern changes every once in a while to $4$ consecutive $n$ having the same number of middle terms. What could this pattern be? After playing with wolfram and a bit of Gemini AI, I found out that number of middle terms is $m = \left\lfloor \frac{2n+1}{2\pi} + \frac12 \right\rfloor$. $m$ can be odd or even. When it is odd, it is a pain to do the possion summation because of an unpaired term like $-\frac{\cos(3 \pi^2 y)}{\pi^2 y^2}$ (which blows up at $y=0$) in $n=5$ for example.Of course, it can be clubbed with one the first $2$ terms to get a value. Luckily from OP's list, I could figure out that $m$ is even for all the interesting $n$ they found (A rigorous proof is missing in my approach). But having chosen an even $m$, the middle terms come in pairs (canceling the pole at origin) and we can proceed to do poisson summation ($\sum\limits_{y=-\infty}^{\infty} \mathcal{F_y}[E(x)]$ where $y$ increments discretely by $1$) of the first two terms, Poisson summation of middle terms and Poisson summation of last $2$ terms.

The Poisson sum of the middle terms (clubbed into pairs for pole cancelation) is like below: The Poisson summation of the sum of the first $2$ and last $2$ terms is fairly straightforward since they are of the very nice form $\frac{\sin^2(z)}{z^2}$ or $\frac{\sin(z)}{z}$. I have omitted that simple case. But those are the terms that ended up in the RHS of my equation. The middle terms ended up in the LHS of my equation

Note that we can use the identity $\operatorname{Li}_{2}(z) + \operatorname{Li}_{2}\left(\frac1z\right) = \frac{-\pi^2}{6} - \frac{\log^2(-z)}{2}$ to convert PolyLog to $\log^2(e^{i w})$ form. Then I published my original condition where the Possion sum of the middle terms equals the negative of the Poisson sum of the remaining easier terms. Then I edited to convert the logarithm and exponetial functions to reduce them to their principal branches (I may have to verify my Floor/Ceiling once again, but it held up for several examples I tried). Last I made a conjecture that $a_n=0$ implies $b_n=0$ that is experimentally holding up so far and presented a simplified formula

I found this problem very interesting and after failing to prove anything, I decided to run a deeper computational search to test the conjectures (proving them wrong is a proof!) and look for other patterns. I wrote a program to compute $(a_n, b_n)$ for $n$ up to $10^{13}$, a process that took about 48 hours. I used the rug crate for arbitrary precision arithmetic as a fallback when k/ $\pi$ is close to a half-integer, which is crucial for maintaining precision in the calculations. Initially, I thought I disproved OP Conjecture 1, but it was just a precision issue

Simplified formula for $a_n$ and $b_n$

Let's first establish a clear formula for the terms. The sum is $S_n = \sum_{k=1}^n \arcsin(\sin k)$. For any integer $k \ge 1$, let $m_k = \lfloor k/\pi + 1/2 \rfloor$ be the nearest integer to $k/\pi$. The term $\arcsin(\sin k)$ can be expressed as: $$ \arcsin(\sin k) = (-1)^{m_k} (k - m_k \pi) $$ This is because $k-m_k\pi$ is in $[-\pi/2, \pi/2]$, and $\sin k = \sin(k - m_k\pi + m_k\pi) = (-1)^{m_k}\sin(k-m_k\pi)$. The sum can then be written as $S_n = a_n + b_n\pi$, where the integer coefficients are: $$ a_n = \sum_{k=1}^n (-1)^{m_k} k \quad \text{and} \quad b_n = -\sum_{k=1}^n (-1)^{m_k} m_k $$

Finitude of Zeros

The most significant finding from the search up to $10^{13}$ is that the occurrences of $a_n=0$ or $b_n=0$ appear to be finite, with largest being at $n=583,726,823$ where $b_n=0$.

My Conjecture 2: For all $n > 583,726,823$, we have $a_n \neq 0$ and $b_n \neq 0$.

If this conjecture is true, it implies that the following two conjectures, which have been verified up to the search limit, become vacuously true for $n > 583,726,823$.

• OP Conjecture 1: If $b_n=0$ and $n>1$, then $a_n \in \{0, 2\}$.
• @Srini Conjecture: If $a_n=0$, then $b_n=0$.

Proven and Falsified Properties

• Parity of $a_n$ (Proven): The parity of $a_n$ follows the parity of the triangular numbers. We have $a_n = \sum_{k=1}^n (-1)^{m_k} k \equiv \sum_{k=1}^n k = \frac{n(n+1)}{2} \pmod 2$.

• My Conjecture 1 (Falsified): The conjecture that $b_n=0$ implies $n \equiv 0 \text{ or } 1 \pmod 4$ was falsified by a counterexample at $n=1,980,127$. For this $n$, we have $n \equiv 3 \pmod 4$ and $(a_n, b_n) = (0, 0)$. This failure seems related to the fact that $1,980,127\,/\,\pi \approx 630293.9999994$ is exceptionally close to an integer.

OP Conjecture 2: On the limit of $a_n/b_n$

The original observation that the ratio $a_n/b_n$ gets close to $-\pi$ seems correct, thought it's punctuated by rare but significant excursions that do not to prevent full convergence after all.

The following table tracks the range of the ratio over selected dyadic intervals $[2^k, 2^{k+1}-1]$.