Problemset paper (in Vietnamese): https://www.facebook.com/share/p/PpV3EQVE4chTZnZC/?mibextid=TrneLp
I am looking at the problem set of the VMO2024, but I can't solve any of them, nor have an idea of what to do =(. However, they look really interesting, and here is the first question:
For all real numbers $x$, define $\lfloor x \rfloor$ as the largest integer not greater than x. Example $\lfloor \frac{9}{2} \rfloor = 4, \lfloor -2\rfloor = -2, \lfloor -341.4 \rfloor = -342 $.
For integer $n \geq 1$, define two sequences $$\{ a_n \}_{n=1}^\infty = \frac{1}{4^{\lfloor -log_4n \rfloor}} \\ \{b_n\}_{n = 1}^\infty = \frac{1}{n^2} \left( \sum_{k = 1}^n a_k- \frac{1}{a_1 + a_2} \right)$$
$a)$ Find a polynomial $P(x)$ with real coefficients such that $\forall n \geq 1: b_n = P\left( \frac{a_n}{n} \right)$
b) Proof that there exists an (increasing) sequence $\{n_k\}_{k=1}^\infty$ such that $\lim_{k \rightarrow \infty} b_{n_k} = \frac{2024}{2025}$
My initial idea is to evaluate what some of the first $a_n, b_n$ are. Using the properties of the floor function: $$a_n = \frac{1}{4^{\lfloor -log_4 n \rfloor}} = \frac{1}{4^{-\lceil log_4 n \rceil}} = 4^{\lceil log_4 n \rceil}$$ or in other words, $$a_n = \text{Smallest power of 4 that is larger to equal to n}$$
If it helps, $\frac{1}{a_1 + a_2} = \frac{1}{5}$.
I wrote a Python program to numerically evaluate a few first terms. If you want to see the values in a markdown-table style, change the value st in line $45$ from $0$ to $1$. Even with that, my head was still blank.
What do you think?
First, notice that
$$ [-\log_4 n] = -k \quad (k \geq 0) \iff n \in [ 4^{k-1}+1, 4^k]:= \{ 4^{k-1}+1,..., 4^k\}, $$
then suppose $n \in [ 4^{m-1}+1, 4^m], m \geq 1 $
$$ \sum_{k=1}^n a_k = a_1 + \sum_{k =1}^{m-2} \sum_{j \in {[4^{k-1}+1,4^k]}}a_j + \sum_{j \in {[4^{m-1}+1,n]}}a_j = 1 + \frac{4}{5}(4^{2m-2} -1) + (n-4^{m-1})4^m. $$
Hence,
$$ b_n = \frac{4^m}{n} - \frac{4^{2m}}{5n^2} = \frac{a_n}{n} - \frac{1}{5}\frac{a^2_n}{n^2}. $$
So the polynomial you are looking for is $P(x) = x - x^2/5$.
For b) the hint is the dyadic numbers of the form $m/4^k, m \in \{0, ..., 4^k \}$ is dense in $[0,1]$.