Let $k$ be a fixed positive integer. Consider the sequence definited by \[a_{0}=1 \;\; , a_{n+1}=a_{n}+\left\lfloor\root k \of{a_{n}}\right\rfloor \;\; , n=0,1,\cdots\] where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. For each $k$ find the set $A_{k}$ containing all integer values of the sequence $(\sqrt[k]{a_{n}})_{n\geq 0}$.

$(x_n)$ be a sequence with $x_1=0$, $$x_{n+1}=5x_n + \sqrt{24x_n^2+1}$$. Prove that for $k \geq 2$ $x_k$ is a natural number.

We define the sequence $x_1=n,y_1=1,x_{i+1}=[\frac{x_i+y_i}{2}],y_{i+1}=[\frac{n}{x_{i+1}} ]$. Prove that $min\{ x_1, x_2, ..., x_n\}=[\sqrt{n}]$ .

Let $r$ be a positive integer, and let $a_0 , a_1 , \cdots $ be an infinite sequence of real numbers. Assume that for all nonnegative integers $m$ and $s$ there exists a positive integer $n \in [m+1, m+r]$ such that \[ a_m + a_{m+1} +\cdots +a_{m+s} = a_n + a_{n+1} +\cdots +a_{n+s} \] Prove that the sequence is periodic, i.e. there exists some $p \ge 1 $ such that $a_{n+p} =a_n $ for all $n \ge 0$.

Consider $k$ positive integers $a_1,a_2,\ldots,a_k$ satisfying $1\le a_1<a_2<\ldots<a_k\le n$ and $\operatorname{lcm}(a_i,a_j)\le n$ for any $i,j$. Prove that $$k\le2\lfloor\sqrt n\rfloor.$$

Given any $n$ positive integers, and a sequence of $2^n$ integers (with terms among them), prove there exists a subsequence made of consecutive terms, such that the product of its terms is a perfect square. Also show that we cannot replace $2^n$ with any lower value (therefore $2^n$ is the threshold value for this property).

(a) Prove that $(2^{n+1}-1)!$ is divisible by $ \prod_{i=0}^n (2^{n+1-i}-1)^{2^i }$, for every natural number n (b) Define the sequence ($c_n$) by $c_1=1$ and $c_{n}=\frac{4n-6}{n}c_{n-1}$ for $n\ge 2$. Show that each $c_n$ is an integer.

[b]1)[/b] Prove that there are finite sequences, of any length, of nonegative integers having the property that the arithmetic mean of any choice of its elements is natural. [b]2)[/b] Study if there is an increasing infinite sequence of nonegative integers having the property that the arithmetic mean of any finite choice of its elements is natural.

A student firstly wrote $x=3$ on the board. For each procces, the stutent deletes the number x and replaces it with either $(2x+4)$ or $(3x+8)$ or $(x^2+5x)$. Is this possible to make the number $(20^{17}+2016)$ on the board? \\ (Explain your answer) \\ [hide=Note]This type of the question is well known but I am going to make a collection so, :blush: [/hide]

Given infinite sequence $a_n$. It is known that the limit of $$b_n=a_{n+1}-a_n/2$$ equals zero. Prove that the limit of $a_n$ equals zero.

A sequence $a_n$ is defined by $$a_0=0,\qquad a_1=3;$$$$a_n=8a_{n-1}+9a_{n-2}+16\text{ for }n\ge2.$$Find the least positive integer $h$ such that $a_{n+h}-a_n$ is divisible by $1999$ for all $n\ge0$.

Given an infinite sequence of numbers $a_1, a_2, a_3, ...$ such that for each positive integer $k$, there exists positive integer $t$ for which $a_k = a_{k+t} = a_{k+2t} = ....$ Does this sequences must be periodic?

Let \( p_1, p_2, \cdots, p_{2025} \) be real numbers. For \( 1 \leq i \leq 2025 \), let \[\{a_n^{(i)}\}_{n \geq 0}\] be an infinite real sequence satisfying \[a_0^{(i)} = 0.\] It is known that: (1) \[a_1^{(1)}, a_1^{(2)}, \cdots, a_1^{(2025)}\] are not all zero. (2) For any integer \( n \geq 0 \) and any \( 1 \leq i \leq 2025 \), the following holds: \[p_i \cdot a_n^{(i+1)} = a_{n-1}^{(i)} + a_n^{(i)} + a_{n+1}^{(i)},\] where the sequence \[\{a_n^{(2026)}\}\] satisfies \[a_n^{(2026)} = a_n^{(1)}, \, n = 0, 1, 2, \cdots.\] Prove that there exists a positive real number \( r \) such that for infinitely many positive integers \( n \), \[\max \left\{ |a_n^{(1)}|, |a_n^{(2)}|, \cdots, |a_n^{(2025)}|\right\} \geq r.\]

Suppose that $u_0 , u_1 ,\ldots$ is a sequence of real numbers such that $$u_n = \sum_{k=1}^{\infty} u_{n+k}^{2}\;\;\; \text{for} \; n=0,1,2,\ldots$$ Prove that if $\sum u_n$ converges, then $u_k=0$ for all $k$.

Let $p$ be a prime number. Determine all positive integers $a$ such that the sequence $(a_n)_{n\geq 0}$ defined by $a_0=a$ and $a_{n+1}=pa_n-(p-1)\lfloor \sqrt[p]{a_ n} \rfloor^p$, for every $n\geq0$, is eventually constant.