Found problems: 1239
The sequence $x_{n}$ is de fined by:
$x_{0}=1, x_{1}=0, x_{2}=1,x_{3}=1, x_{n+3}=\frac{(n^2+n+1)(n+1)}{n}x_{n+2}+(n^2+n+1)x_{n+1}-\frac{n+1}{n}x_{n} (n=1,2,3..)$
Prove that all members of the sequence are perfect squares.
Let $(b_n)_{n \ge 0}$ be a sequence of positive integers satisfying $b_n=d\left(\sum_{i=0}^{n-1} b_k\right)$ for all $n \ge 1$. (By $d(m)$ we denote the number of positive divisors of $m$.)
a) Prove that $(b_n)_{n \ge 0}$ is unbounded.
b) Prove that there are infinitely many $n$ such that $b_n>b_{n+1}$.
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
A sequence $a_1, a_2, a_3, \ldots $ is defined as follows: $a_1$ is a positive integer and
$$a_{n+1} = \left\lfloor \frac{3}{2} a_n \right\rfloor +1$$
for all $n \in \mathbb{N}$. Can $a_1$ be chosen in such a way that the first $100000$ terms of the sequence are even, but the $100001$-th term is odd?
Let $ n > 1$ be an integer. Find all sequences $ a_1, a_2, \ldots a_{n^2 \plus{} n}$ satisfying the following conditions:
\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 \plus{} n;
\]
\[ \text{ (b) } a_{i \plus{} 1} \plus{} a_{i \plus{} 2} \plus{} \ldots \plus{} a_{i \plus{} n} < a_{i \plus{} n \plus{} 1} \plus{} a_{i \plus{} n \plus{} 2} \plus{} \ldots \plus{} a_{i \plus{} 2n} \text{ for all } 0 \leq i \leq n^2 \minus{} n.
\]
[i]Author: Dusan Dukic, Serbia[/i]
Let be a sequence $ \left( x_n \right)_{n\ge 1} $ having the property that
$$ \lim_{n\to\infty } \left( 14(n+2)x_{n+2} -15(n+1)x_{n+1} +nx_n \right) =13. $$
Show that $ \left( x_n \right)_{n\ge 1} $ is convergent and calculate its limit.
[i]Cosmin Nițu[/i]
A sequence of integers $\{f(n)\}$ for $n=0,1,2,\ldots$ is defined as follows: $f(0)=0$ and for $n>0$,
$$\begin{matrix}f(n)=&f(n-1)+3,&\text{if }n=0\text{ or }1\pmod6,\\&f(n-1)+1,&\text{if }n=2\text{ or }5\pmod6,\\&f(n-1)+2,&\text{if }n=3\text{ or }4\pmod6.\end{matrix}$$Derive an explicit formula for $f(n)$ when $n\equiv0\pmod6$, showing all necessary details in your derivation.
A sequence of integers is constructed as follows: $a_1$ is an arbitrary three-digit number, $a_2$ is the sum of squares of the digits of $a_1, a_3$ is the sum of squares of the digits of $a_2$, etc. Prove that either $1$ or $4$ must occur in the sequence $a_1, a_2, a_3, ....$
Find the formula for the general term of the sequence an, for which $a_1 = 1$, $a_2 = 3$, $a_{n+1} = 3a_n-2a_{n-1}$ (you need to express an in terms of $n$).
Let $a$ be a positive integer and $(a_n)_{n\geqslant 1}$ be a sequence of positive integers satisfying $a_n<a_{n+1}\leqslant a_n+a$ for all $n\geqslant 1$. Prove that there are infinitely many primes which divide at least one term of the sequence.
[i]Moldavia Olympiad, 1994[/i]
Let $(a_n)_{n\ge1}$ be a sequence given by
\begin{align*}
a_1 &= 1, \\
a_{2^k+j} &= -a_j\text{ for any } k\ge0,1\le j\le 2^k.
\end{align*}
Show that the sequence is not periodic.
The sequences $(a_n)$ and $(b_n)$ are defined by $a_1 = b_1 = 1$ and $a_{n+1} = a_n +b_n, b_{n+1} = a_nb_n$ for $n = 1,2,...$ Show that every two distinct terms of the sequence $(a_n)$ are coprime
Let b be a given real number. The sequence of integers $a_1, a_2,a_3, ...$ is such that $a_1 =(b]$ and $a_{n+1}=(a_n+b]$ for all $n\ge 1$ Prove that the sum $a_1+\frac{a_2}{2}+\frac{a_3}{3}+...+\frac{a_n}{n}$ is an integer number for any natural $n$ .
(In the condition of the problem, $(x]$ denotes the smallest integer that is greater than or equal to $x$)
Let $a_n$ be a sequence de fined by some $a_0$ and the recursion $a_{n+1} = a_n + 2 \cdot 3^n$ for $n \ge 0$.
Determine all rational values of $a_0$ such that $a^j_k / a^k_j$ is an integer for all integers $j$ and $k$ with $0 < j < k$.
Let a sequence $(a_i)_{i=10}^{\infty}$ be defined as follows:
[list=a]
[*] $a_{10}$ is some positive integer, which can of course be written in base 10.
[*] For $i \geq 10$ if $a_i > 0$, let $b_i$ be the positive integer whose base-$(i + 1)$ representation is the same as $a_i$'s base-$i$ representation. Then let $a_{i + 1} = b_i - 1$. If $a_i = 0$, $a_{i + 1} = 0$.
[/list]
For example, if $a_{10} = 11$, then $b_{10} = 11_{11} (= 12_{10})$; $a_{11} = 11_{11} - 1 = 10_{11} (= 11_{10})$; $b_{11} = 10_{12} (= 12_{10})$; $a_{12} = 11$.
Does there exist $a_{10}$ such that $a_i$ is strictly positive for all $i \geq 10$?
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]
Consider a sequence of polynomials such that $P_0(x)=2,P_1(x)=x$ and for all $n\ge1$ \[P_{n+1}(x)+P_{n-1}(x)=xP_n(x).\]
a) Determine the polynomial \[Q_n(x)=P^2_n(x)-xP_n(x)P_{n-1}(x)+P^2_{n-1}(x)\] for $n=1972.$
b) Express the polynomial \[\bigl(P_{n+1}(x)-P_{n-1}(x)\bigr)^2\] in terms of $P_n(x),Q_n(x).$
Define a sequence $(a_n)$ by $a_1 =1, a_2 =2,$ and $a_{k+2}=2a_{k+1}+a_k$ for all positive integers $k$. Determine all real numbers $\beta >0$ which satisfy the following conditions:
(A) There are infinitely pairs of positive integers $(p,q)$ such that $\left| \frac{p}{q}- \sqrt{2} \, \right| < \frac{\beta}{q^2 }.$
(B) There are only finitely many pairs of positive integers $(p,q)$ with $\left| \frac{p}{q}- \sqrt{2} \,\right| < \frac{\beta}{q^2 }$ for which there is no index $k$ with $q=a_k.$
Let $N$ be a positive integer. A non-decreasing sequence $a_1 \le a_2 \le \dots$ of positive integers is said to be $N$-rioplatense if there exists an index $i$ such that $N = \frac{i}{a_i}$. Show that every sequence $2024$-rioplatense is $k$-rioplatense for $k=1, 2, 3, \dots, 2023$.
Define the numbers $a_0, a_1, \ldots, a_n$ in the following way:
\[ a_0 = \frac{1}{2}, \quad a_{k+1} = a_k + \frac{a^2_k}{n} \quad (n > 1, k = 0,1, \ldots, n-1). \]
Prove that \[ 1 - \frac{1}{n} < a_n < 1.\]
For two rational numbers $r,s$ we say:$$r\mid s$$whenever there exists $k\in\mathbb{Z}$ such that:$$s=kr$$
${(a_n)}_{n\in\mathbb{N}}$ is an increasing sequence of pairwise coprime natural numbers and ${(b_n)}_{n\in\mathbb{N}}$ is a sequence of distinct natural numbers. Assume that for all $n\in\mathbb{N}$ we have:
$$\sum_{i=1}^{n}\frac{1}{a_i}\mid\sum_{i=1}^{n}\frac{1}{b_i}$$
Prove that [b]for all[/b] $n\in\mathbb{N}$ we have: $a_n=b_n$.
Let $k\geqslant 2$ be an integer. Consider the sequence $(x_n)_{n\geqslant 1}$ defined by $x_1=a>0$ and $x_{n+1}=x_n+\lfloor k/x_n\rfloor$ for $n\geqslant 1.$ Prove that the sequence is convergent and determine its limit.
Given is a number $a$ with 0 $\le \alpha \le \pi$. A sequence $c_0,c_1, c_2,...$ is defined as
$$c_0=\cos \alpha$$
$$C_{n+1}=\sqrt{\frac{1+c_n}{2}} \,\, for \,\,\, n=0,1,2,...$$
Calculate $\lim_{n\to \infty}2^{2n+1}(1-c_n)$
A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are 57, 60, and 91. What is the fourth term of this sequence?
$\textbf{(A) }190\qquad\textbf{(B) }194\qquad\textbf{(C) }198\qquad\textbf{(D) }202\qquad\textbf{(E) }206$
The sequence $a_1, a_2,...$ is de fined by the equalities $a_1 = 2, a_2 = 12$ and $a_{n+1} = 6a_n-a_{n-1}$ for every positive integer $n \ge 2$. Prove that no member of this sequence is equal to a perfect power (greater than one) of a positive integer.