Define the sequence $a_n = \lfloor \alpha n \rfloor$ for a real number $\alpha$. Is there any pair of natural numbers $k, l$ satisfying the following condition?: $$\sum_{n=1}^k a_n = \sum_{n=k+1}^l a_n$$ In other words, can this sequence be divided into two parts with equal sums?
Furthermore, are there infinitely many pairs $k,l$ that satisfy this condition, or does there exist a real number $\alpha$ such that there are either a finite number of solutions or no solutions at all?
My try:
I have searched for solutions for several numbers.
- In a case of $\alpha=1$, the smallest solution is $k=2,l=3$, because $1+2=3$. The next solution is $k=14,l=20$. $k$ is A053141, $l$ is A001652 and the sum is A075528. This pattern holds true for all integer $\alpha$.
- When $\alpha=\pi$, the smallest solution is $k=2,l=3$, $3+6=9$.
- When $\alpha=e$, the smallest solution is $k=19, l=27$, $2+5+8+10+13+16+19+21+24+27+29+32+35+38+40+43+46+48+51=54+57+59+62+65+67+70+73$.
- When $\alpha=e^2$, the smallest solution is $k=12778,l=18071$.
- For fractions with a denominator of $3$, the smallest solution tends to be larger. When $\alpha = 7/3$ it is $k = 60,l = 85$, when $\alpha = 23/3$ it is $k = 102744180, l = 145302213$, and furthermore, when $\alpha = 71/3$ it is $k = 258279187692151687649783580, l = 365261930112947083771975701$.
- Additionally, using the fact that the partial sums of $\{ a_n \}$ can be expressed as a set of quadratic Diophantine equations, I have confirmed that there are infinitely many pairs $k, l$ for rational $\alpha$ where $-10 < \alpha < 10$ and its denominator is up to 20.
Also, I found it to be true for some range of $\alpha$:
- When $0 \leq \alpha < \frac{1}{2}$, $a_n = 0, 0, \ldots$ so the solution is $k=1, l=2$.
- When $\frac{1}{2} \leq \alpha < \frac{2}{3}$, $a_n = 0, 1, 1, \ldots$ so the solution is $k=2, l=3$.
- When $\frac{2}{3} \leq \alpha < \frac{5}{7}$, $a_n = 0, 1, 2, 2, 3, 4, 4$ and since $0+1+2+2+3 = 4+4$, the solution is $k=5, l=7$.
- When $\frac{5}{7} \leq \alpha < \frac{13}{18}$, the solution is $k=17, l=24$.
- Similarly, using a computer, I have confirmed that there is at least one solution in the range $-0.58110\ldots=-1800169841164/3097849628081<\alpha<503914285046/366091395809=1.37647\ldots$.
From a probability perspective, considering that $a_n\simeq\alpha n$, we have $S_n=\sum_{i=1}^n a_i\simeq \frac{\alpha}{2}n^2$. Solving $2S_k=S_l$ in the range of real numbers gives $l\simeq\sqrt{2}k$, and assuming there is no arbitrary distribution for $S_n$, the probability of this being an integer is about $\frac{1}{\sqrt{2}\alpha k}$, and when summing over $k$, it diverges. In other words, it is expected that there are infinitely many $k, l$.
If there is at least one solution for all rational $\alpha$, then there is at least one solution for all real $\alpha$ because we can bound the $\alpha$ between two rational numbers whose ${a_n}$ are the same up to the solution, but this method does not prove that there are infinitely many solutions.
