Found problems: 963
1975 IMO Shortlist, 11
Let $a_{1}, \ldots, a_{n}$ be an infinite sequence of strictly positive integers, so that $a_{k} < a_{k+1}$ for any $k.$ Prove that there exists an infinity of terms $ a_{m},$ which can be written like $a_m = x \cdot a_p + y \cdot a_q$ with $x,y$ strictly positive integers and $p \neq q.$
2006 IMO Shortlist, 2
The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$.
[i]Proposed by Mariusz Skalba, Poland[/i]
1980 IMO Longlists, 18
Given a sequence $\{a_n\}$ of real numbers such that $|a_{k+m} - a_k - a_m| \leq 1$ for all positive integers $k$ and $m$, prove that, for all positive integers $p$ and $q$, \[|\frac{a_p}{p} - \frac{a_q}{q}| < \frac{1}{p} + \frac{1}{q}.\]
2005 Czech And Slovak Olympiad III A, 6
Decide whether for every arrangement of the numbers $1,2,3, . . . ,15$ in a sequence one can color these numbers with at most four different colors in such a way that the numbers of each color form a monotone subsequence.
2022 Thailand TSTST, 3
Let $n > 1$ be a given integer. Prove that infinitely many terms of the sequence $(a_k )_{k\ge 1}$, defined by \[a_k=\left\lfloor\frac{n^k}{k}\right\rfloor,\] are odd. (For a real number $x$, $\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.)
[i]Proposed by Hong Kong[/i]
2008 Thailand Mathematical Olympiad, 3
For each positive integer $n$, define $a_n = n(n + 1)$. Prove that
$$n^{1/a_1} + n^{1/a_3} + n^{1/a_5} + ...+ n^{1/a_{2n-1}} \ge n^{a_{3n+2}/a_{3n+1}}$$
.
2013 Korea Junior Math Olympiad, 3
$\{a_n\}$ is a positive integer sequence such that $a_{i+2} = a_{i+1} +a_i$ (for all $i \ge 1$).
For positive integer $n$, de fine as $$b_n=\frac{1}{a_{2n+1}}\Sigma_{i=1}^{4n-2}a_i$$
Prove that $b_n$ is positive integer.
2007 Mathematics for Its Sake, 1
Prove that the parity of each term of the sequence $ \left( \left\lfloor \left( \lfloor \sqrt q \rfloor +\sqrt{q} \right)^n \right\rfloor \right)_{n\ge 1} $ is opposite to the parity of its index, where $ q $ is a squarefree natural number.
1967 IMO Longlists, 12
Given a segment $AB$ of the length 1, define the set $M$ of points in the
following way: it contains two points $A,B,$ and also all points obtained from $A,B$ by iterating the following rule: With every pair of points $X,Y$ the set $M$ contains also the point $Z$ of the segment $XY$ for which $YZ = 3XZ.$
2013 BAMO, 5
Let $F_1,F_2,F_3,...$ be the [i]Fibonacci sequence[/i], the sequence of positive integers with $F_1 =F_2 =1$ and $F_{n+2}=F_{n+1}+F_n$ for all $n \ge 1$. A [i]Fibonacci number[/i] is by definition a number appearing in this sequence.
Let $P_1,P_2,P_3,...$ be the sequence consisting of all the integers that are products of two Fibonacci numbers (not
necessarily distinct) in increasing order. The first few terms are $1,2,3,4,5,6,8,9,10,13,...$ since, for example $3 = 1 \cdot 3, 4 = 2 \cdot 2$, and $10 = 2 \cdot 5$.
Consider the sequence $D_n$ of [i]successive [/i] differences of the $P_n$ sequence, where $D_n = P_{n+1}-P_n$ for $n \ge 1$. The first few terms of D_n are $1,1,1,1,1,2,1,1,3, ...$ .
Prove that every number in $D_n$ is a [i]Fibonacci number[/i].
1998 Belarusian National Olympiad, 5
Is there an infinite sequence of positive real numbers $x_1,x_2,...,x_n$ satisfying for all $n\ge 1$ the relation $x_{n+2}= \sqrt{x_{n+1}}-\sqrt{x_n}$?
2013 China Northern MO, 7
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.
2017 Azerbaijan Team Selection Test, 1
Consider the sequence of rational numbers defined by $x_1=\frac{4}{3}$, and $x_{n+1}=\frac{x_n^2}{x_n^2-x_n+1}$. Show that the nu,erator of the lowest term expression of each sum $x_1+x_2+...+x_k$ is a perfect square.
2023 Brazil Team Selection Test, 2
Let $a > 1$ be a positive integer and $d > 1$ be a positive integer coprime to $a$. Let $x_1=1$, and for $k\geq 1$, define
$$x_{k+1} = \begin{cases}
x_k + d &\text{if } a \text{ does not divide } x_k \\
x_k/a & \text{if } a \text{ divides } x_k
\end{cases}$$
Find, in terms of $a$ and $d$, the greatest positive integer $n$ for which there exists an index $k$ such that $x_k$ is divisible by $a^n$.
2014 Regional Competition For Advanced Students, 3
The sequence $(a_n)$ is defined with the recursion $a_{n + 1} = 5a^6_n + 3a^3_{n-1} + a^2_{n-2}$ for $n\ge 2$ and the set of initial values $\{a_0, a_1, a_2\} = \{2013, 2014, 2015\}$. (That is, the initial values are these three numbers in any order.)
Show that the sequence contains no sixth power of a natural number.
2022 Macedonian Mathematical Olympiad, Problem 3
The sequence $(a_n)_{n \ge 1}^\infty$ is given by: $a_1=2$ and $a_{n+1}=a_n^2+a_n$ for all $n \ge 1$.
For an integer $m \ge 2$, $L(m)$ denotes the greatest prime divisor of $m$. Prove that there exists some $k$, for which $L(a_k) > 1000^{1000}$.
[i]Proposed by Nikola Velov[/i]
2020 IMO Shortlist, N4
For any odd prime $p$ and any integer $n,$ let $d_p (n) \in \{ 0,1, \dots, p-1 \}$ denote the remainder when $n$ is divided by $p.$ We say that $(a_0, a_1, a_2, \dots)$ is a [i]p-sequence[/i], if $a_0$ is a positive integer coprime to $p,$ and $a_{n+1} =a_n + d_p (a_n)$ for $n \geqslant 0.$
(a) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $(a_0, a_1, a_2, \dots)$ and $(b_0, b_1, b_2, \dots)$ such that $a_n >b_n$ for infinitely many $n,$ and $b_n > a_n$ for infinitely many $n?$
(b) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $(a_0, a_1, a_2, \dots)$ and $(b_0, b_1, b_2, \dots)$ such that $a_0 <b_0,$ but $a_n >b_n$ for all $n \geqslant 1?$
[I]United Kingdom[/i]
2014 Contests, 1
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
[i]Proposed by Gerhard Wöginger, Austria.[/i]
2023 China MO, 1
Define the sequences $(a_n),(b_n)$ by
\begin{align*}
& a_n, b_n > 0, \forall n\in\mathbb{N_+} \\
& a_{n+1} = a_n - \frac{1}{1+\sum_{i=1}^n\frac{1}{a_i}} \\
& b_{n+1} = b_n + \frac{1}{1+\sum_{i=1}^n\frac{1}{b_i}}
\end{align*}
1) If $a_{100}b_{100} = a_{101}b_{101}$, find the value of $a_1-b_1$;
2) If $a_{100} = b_{99}$, determine which is larger between $a_{100}+b_{100}$ and $a_{101}+b_{101}$.
1991 Romania Team Selection Test, 4
A sequence $(a_n)$ of positive integers satisfies$(a_m,a_n) = a_{(m,n)}$ for all $m,n$.
Prove that there is a unique sequence $(b_n)$ of positive integers such that $a_n = \prod_{d|n} b_d$
2007 Germany Team Selection Test, 1
The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$.
[i]Proposed by Mariusz Skalba, Poland[/i]
1998 Bundeswettbewerb Mathematik, 2
Prove that there exists an infinite sequence of perfect squares with the following properties:
(i) The arithmetic mean of any two consecutive terms is a perfect square,
(ii) Every two consecutive terms are coprime,
(iii) The sequence is strictly increasing.
2011 IMO Shortlist, 2
Determine all sequences $(x_1,x_2,\ldots,x_{2011})$ of positive integers, such that for every positive integer $n$ there exists an integer $a$ with \[\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1\]
[i]Proposed by Warut Suksompong, Thailand[/i]
1980 VTRMC, 3
Let $$a_n = \frac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}.$$
(a) Prove that $\lim_{n\to \infty}a_n$ exists.
(b) Show that $$a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}.$$
(c) Find $\lim_{n\to\infty}a_n$ and justify your answer
2023 India EGMO TST, P5
Let $k$ be a positive integer. A sequence of integers $a_1, a_2, \cdots$ is called $k$-pop if the following holds: for every $n \in \mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\{a_1, \cdots , a_{n+k} \}$. Determine, as a function of $k$, how many $k$-pop sequences there are.
[i]Proposed by Sutanay Bhattacharya[/i]