Found problems: 307
1985 IMO Longlists, 33
A sequence of polynomials $P_m(x, y, z), m = 0, 1, 2, \cdots$, in $x, y$, and $z$ is defined by $P_0(x, y, z) = 1$ and by
\[P_m(x, y, z) = (x + z)(y + z)P_{m-1}(x, y, z + 1) - z^2P_{m-1}(x, y, z)\]
for $m > 0$. Prove that each $P_m(x, y, z)$ is symmetric, in other words, is unaltered by any permutation of $x, y, z.$
2021 Dutch IMO TST, 1
The sequence of positive integers $a_0, a_1, a_2, . . .$ is defined by $a_0 = 3$ and $$a_{n+1} - a_n = n(a_n - 1)$$ for all $n \ge 0$. Determine all integers $m \ge 2$ for which $gcd (m, a_n) = 1$ for all $n \ge 0$.
2017 Grand Duchy of Lithuania, 1
The infinite sequence $a_0, a_1, a_2, a_3,... $ is defined by $a_0 = 2$ and
$$a_n =\frac{2a_{n-1} + 1}{a_{n-1} + 2}$$ , $n = 1, 2, 3, ...$ Prove that $1 < a_n < 1 + \frac{1}{3^n}$ for all $n = 1, 2, 3, . .$
2005 VJIMC, Problem 4
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) $\lim_{n\to\infty}nx_n$.
2014 Costa Rica - Final Round, 6
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 $A_N(c) = B_N(c) = C_N(c) = 0$.
2015 Balkan MO Shortlist, N2
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)
1976 IMO, 3
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 $\leq x)$