Suppose that $\{a_n\}$ is a sequence such that $a_{n+1}=(1+\frac{k}{n})a_{n}+1$ with $a_{1}=1$.Find all positive integers $k$ such that any $a_n$ be integer.

When the natural numbers are written one after another in an increasing way, you get an infinite succession of digits $123456789101112 ....$ Denote $A_k$ the number formed by the first $k$ digits of this sequence . Prove that for all positive integer $n$ there is a positive integer $m$ which simultaneously verifies the following three conditions: (i) $n$ divides $A_m$, (ii) $n$ divides $m$, (iii) $n$ divides the sum of the digits of $A_m$.

For an integer $n \ge 1$ we consider sequences of $2n$ numbers, each equal to $0, -1$ or $1$. The [i]sum product value[/i] of such a sequence is calculated by first multiplying each pair of numbers from the sequence, and then adding all the results together. For example, if we take $n = 2$ and the sequence $0,1, 1, -1$, then we find the products $0\cdot 1, 0\cdot 1, 0\cdot -1, 1\cdot 1, 1\cdot -1, 1\cdot -1$. Adding these six results gives the sum product value of this sequence: $0+0+0+1+(-1)+(-1) = -1$. The sum product value of this sequence is therefore smaller than the sum product value of the sequence $0, 0, 0, 0$, which equals $0$. Determine for each integer $n \ge 1$ the smallest sum product value that such a sequence of $2n$ numbers could have. [i]Attention: you are required to prove that a smaller sum product value is impossible.[/i]

Define the polymonial sequence $\left \{ f_n\left ( x \right ) \right \}_{n\ge 1}$ with $f_1\left ( x \right )=1$, $$f_{2n}\left ( x \right )=xf_n\left ( x \right ), \; f_{2n+1}\left ( x \right ) = f_n\left ( x \right )+ f_{n+1} \left ( x \right ), \; n\ge 1.$$ Look for all the rational number $a$ which is a root of certain $f_n\left ( x \right ).$

Real numbers $ a_{1}$, $ a_{2}$, $ \ldots$, $ a_{n}$ are given. For each $ i$, $ (1 \leq i \leq n )$, define \[ d_{i} \equal{} \max \{ a_{j}\mid 1 \leq j \leq i \} \minus{} \min \{ a_{j}\mid i \leq j \leq n \} \] and let $ d \equal{} \max \{d_{i}\mid 1 \leq i \leq n \}$. (a) Prove that, for any real numbers $ x_{1}\leq x_{2}\leq \cdots \leq x_{n}$, \[ \max \{ |x_{i} \minus{} a_{i}| \mid 1 \leq i \leq n \}\geq \frac {d}{2}. \quad \quad (*) \] (b) Show that there are real numbers $ x_{1}\leq x_{2}\leq \cdots \leq x_{n}$ such that the equality holds in (*). [i]Author: Michael Albert, New Zealand[/i]

Let $(x_n)_{n\ge2}$ be a sequence of real numbers such that $x_2>0$ and $x_{n+1}=-1+\sqrt[n]{1+nx_n}$ for $n\ge2$. Find (a) $\lim_{n\to\infty}x_n$, (b) $\lim_{n\to\infty}nx_n$.

Let $(a_n)_{n \geq 0}$ and $(b_n)_{n \geq 0}$ be sequences of real numbers such that $ a_0>\frac{1}{2}$ , $a_{n+1} \geq a_n$ and $b_{n+1}=a_n(b_n+b_{n+2})$ for all non-negative integers $n$ . Show that the sequence $(b_n)_{n \geq 0}$ is bounded .

For a positive integer $n$, an [i]$n$-sequence[/i] is a sequence $(a_0,\ldots,a_n)$ of non-negative integers satisfying the following condition: if $i$ and $j$ are non-negative integers with $i+j \leqslant n$, then $a_i+a_j \leqslant n$ and $a_{a_i+a_j}=a_{i+j}$. Let $f(n)$ be the number of $n$-sequences. Prove that there exist positive real numbers $c_1$, $c_2$, and $\lambda$ such that \[c_1\lambda^n<f(n)<c_2\lambda^n\] for all positive integers $n$.

Denote by $l(n)$ the largest prime divisor of $n$. Let $a_{n+1} = a_n + l(a_n)$ be a recursively defined sequence of integers with $a_1 = 2$. Determine all natural numbers $m$ such that there exists some $i \in \mathbb{N}$ with $a_i = m^2$. [i]Proposed by Nikola Velov, North Macedonia[/i]

The sequences $a_n$, $b_n$ and $c_n$ are defined recursively in the following way: $a_0 = 1/6$, $b_0 = 1/2$, $c_0 = 1/3,$ $$a_{n+1}= \frac{(a_n + b_n)(a_n + c_n)}{(a_n - b_n)(a_n - c_n)},\,\, b_{n+1}= \frac{(b_n + a_n)(b_n + c_n)}{(b_n - a_n)(b_n - c_n)},\,\, c_{n+1}= \frac{(c_n + a_n)(c_n + b_n)}{(c_n - a_n)(c_n - b_n)}$$ For each natural number $N$, the following polynomials are defined: $A_n(x) =a_o+a_1 x+ ...+ a_{2N}x^{2N}$ $B_n(x) =b_o+a_1 x+ ...+ a_{2N}x^{2N}$ $C_n(x) =a_o+a_1 x+ ...+ a_{2N}x^{2N}$ Assume the sequences are well defined. Show that there is no real $c$ such that $A_N(c) = B_N(c) = C_N(c) = 0$.

Define a sequence $\{a_n\}$ by\[S_1=1,\ S_{n+1}=\frac{(2+S_n)^2}{ 4+S_n} (n=1,\ 2,\ 3,\ \cdots).\] Where $S_n$ the sum of first $n$ terms of sequence $\{a_n\}$. For any positive integer $n$ ,prove that\[a_{n}\ge \frac{4}{\sqrt{9n+7}}.\]

Sequence $(a_n)_{n\geq 0}$ is defined as $a_{0}=0, a_1=1, a_2=2, a_3=6$, and $ a_{n+4}=2a_{n+3}+a_{n+2}-2a_{n+1}-a_n, n\geq 0$. Prove that $n^2$ divides $a_n$ for infinite $n$. (Romania)

Let $S = \{0, 1, 2, .., 1994\}$. Let $a$ and $b$ be two positive numbers in $S$ which are relatively prime. Prove that the elements of $S$ can be arranged into a sequence $s_1, s_2, s_3,... , s_{1995}$ such that $s_{i+1} - s_i \equiv \pm a$ or $\pm b$ (mod $1995$) for $i = 1, 2, ... , 1994$

A sequence $(u_{n})$ is defined by \[ u_{0}=2 \quad u_{1}=\frac{5}{2}, u_{n+1}=u_{n}(u_{n-1}^{2}-2)-u_{1} \quad \textnormal{for } n=1,\ldots \] Prove that for any positive integer $n$ we have \[ [u_{n}]=2^{\frac{(2^{n}-(-1)^{n})}{3}} \](where $[x]$ denotes the smallest integer $\leq x)$