Every term of an infinite geometric progression is also a term of a given infinite arithmetic progression. Prove that the common ratio of the geometric progression is an integer. (4)

Let $(a_n)_{n \ge 1}$ be a sequence of real numbers satisfying the properties: [list=1] [*] the sequence $x_n=\sum\limits_{k=1}^n a_k^2$ is convergent; [*] the sequence $y_n=\sum\limits_{k=1}^n a_k$ is unbounded. [/list] Prove that the sequence $(b_n)_{n \ge 1}$ given by $b_n=\{y_n\}$ is divergent. Note: $\{ x \}$ denotes the fractional part of $x.$

Find all sequences $(a_n)$ of positive integers satisfying the equality $a_n=a_{a_{n-1}}+a_{a_{n+1}}$ a) for all $n\ge 2$ b) for all $n \ge 3$ (I. Gorodnin)

The sequence $a_i$ is defined as $a_1 = 2, a_2 = 3$, and $a_{n+1} = 2a_{n-1}$ or $a_{n+1} = 3a_n - 2a_{n-1}$ for all integers $n \ge 2$. Prove that no term in $a_i$ is in the range $[1612, 2012]$.

$a_1,a_2,\ldots,a_n$ is a sequence of positive integers that has at least $\frac {2n}{3}+1$ distinct numbers and each positive integer has occurred at most three times in it. Prove that there exists a permutation  $b_1,b_2,\ldots,b_n$ of $a_i $'s such that all the $n$ sums $b_i+b_{i+1}$ are distinct ($1\le i\le n $ , $b_{n+1}\equiv b_1 $) [i]Proposed by Mohsen Jamali[/i]

The sequence of numbers $a_1, a_2, a_3, ...$ is defined recursively by $a_1 = 1, a_{n + 1} = \lfloor \sqrt{a_1+a_2+...+a_n} \rfloor $ for $n \ge 1$. Find all numbers that appear more than twice at this sequence.

Let $(x_n)_{n\geq 1}$ be the sequence defined recursively as such: \[x_1=1, \ x_{n+1}=\frac{x_1}{n+1}+\frac{x_2}{n+2}+\cdots+\frac{x_n}{2n} \ \forall n\geq 1.\]Consider the sequence $(y_n)_{n\geq 1}$ such that $y_n=(x_1^2+x_2^2+\cdots x_n^2)/n$ for all $n\geq 1.$ Prove that [list=a] [*]$x_{n+1}^2<y_n/2$ and $y_{n+1}<(2n+1)/(2n+2)\cdot y_n$ for all $n\geq 1;$ [*]$\lim_{n\to\infty}x_n=0.$ [/list]

There are $30$ children, $a_1,a_2,...,a_{30}$ seated clockwise in a circle on the floor. The teacher walks behind the children in the clockwise direction with a box of $1000$ candies. She drops a candy behind the first child $a_1$. She then skips one child and drops a candy behind the third child, $a_3$. Now she skips two children and drops a candy behind the next child, $a_6$. She continues this way, at each stage skipping one child more than at the preceding stage before dropping a candy behind the next child. How many children will never receive a candy? Justify your answer.

Let $\{a_n\}_{n \in N}$ be a sequence of real numbers with $a_1 = 2$ and $a_n =\frac{n^2 + 1}{\sqrt{n^3 - 2n^2 + n}}$ for all positive integers $n \ge 2$. Let $s_n = a_1 + a_2 + ...+ a_n$ for all positive integers $n$. Prove that $$\frac{1}{s_1s_2}+\frac{1}{s_2s_3}+ ...+\frac{1}{s_ns_{n+1}}<\frac15$$ for all positive integers $n$.

Consider pairs of the sequences of positive real numbers \[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\] and the sums \[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\] For any pair define $c_n = \min\{a_i,b_i\}$ and $C_n = c_1 + \cdots + c_n$, $n=1,2,\ldots$. (1) Does there exist a pair $(a_i)_{i\geq 1}$, $(b_i)_{i\geq 1}$ such that the sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are unbounded while the sequence $(C_n)_{n\geq 1}$ is bounded? (2) Does the answer to question (1) change by assuming additionally that $b_i = 1/i$, $i=1,2,\ldots$? Justify your answer.

Find the number of sequences with $2022$ natural numbers $n_1, n_2, n_3, \ldots, n_{2022}$, such that in every sequence: $\bullet$ $n_{i+1}\geq n_i$ $\bullet$ There is at least one number $i$, such that $n_i=2022$ $\bullet$ For every $(i, j)$ $n_1+n_2+\ldots+n_{2022}-n_i-n_j$ is divisible to both $n_i$ and $n_j$

The sequence ($a_n$) of natural numbers is defined by $a_1 = 2$ and $a_{n+1}$ equals the sum of tenth powers of the decimal digits of $a_n$ for all $n \ge 1$. Are there numbers which appear twice in the sequence ($a_n$)?

The sequence $a_0, a_1, a_2, ..$ is defined by $a_0 = a, a_{n+1} = a_n + L(a_n)$, where $L(m)$ is the last digit of $m$ (eg $L(14) = 4$). Suppose that the sequence is strictly increasing. Show that infinitely many terms must be divisible by $d = 3$. For what other d is this true?

Let $f(0), f(1), f(2), \dots$ be a sequence satisfying \[ f(0) = 0 \quad \text{and} \quad f(n) = n - f(f(n-1)) \] for $n=1,2,3,\dots$. Give a formula for $f(n)$ such that its value can be immediately computed using $n$ without having to compute the previous terms.

From the sequence $1,\frac12, \frac13, ...$ can one choose (a) a subsequence of $100$ different numbers, (b) an infinite subsequence such that each number (beginning from the third) is equal to the difference between the two preceding numbers ($a_k=a_{k-2}-a_{k-1}$)? (SI Tokarev)

[b](a)[/b] Find the rearrangement $\{a_1, \dots , a_n\}$ of $\{1, 2, \dots, n\}$ that maximizes \[a_1a_2 + a_2a_3 + \cdots + a_na_1 = Q.\] [b](b)[/b] Find the rearrangement that minimizes $Q.$

The sum $$\frac{1^2-2}{1!} + \frac{2^2-2}{2!} + \frac{3^2-2}{3!} + \cdots + \frac{2021^2 - 2}{2021!}$$ $ $ \\ can be expressed as a rational number $N$. Find the last 3 digits of $2021! \cdot N$.

Some positive integers, sum of which is $23$, are written in sequential form. Neither one of the terms nor the sum of some consecutive terms in the sequence is equal to $3$. [b]a) [/b]Is it possible that the sequence contains exactly $11$ terms? [b]b)[/b]Is it possible that the sequence contains exactly $12$ terms?

Let $ a_0$, $ a_1$, $ a_2$, $ \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $ \gcd (a_i, a_{i \plus{} 1}) > a_{i \minus{} 1}$. Prove that $ a_n\ge 2^n$ for all $ n\ge 0$. [i]Proposed by Morteza Saghafian, Iran[/i]

Let $a$ and $b$ be fixed positive integers. Show that the set of primes that divide at least one of the terms of the sequence $a_n = a \cdot 2017^n + b \cdot 2016^n$ is infinite.

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]

Let \( a_1 \) be an integer greater than or equal to 2. Consider the sequence such that its first term is \( a_1 \), and for \( a_n \), the \( n \)-th term of the sequence, we have \[ a_{n+1} = \frac{a_n}{p_k^{e_k - 1}} + 1, \] where \( p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \) is the prime factorization of \( a_n \), with \( 1 < p_1 < p_2 < \cdots < p_k \), and \( e_1, e_2, \dots, e_k \) positive integers. For example, if \( a_1 = 2024 = 2^3 \cdot 11 \cdot 23 \), the next two terms of the sequence are \[ a_2 = \frac{a_1}{23^{1-1}} + 1 = \frac{2024}{1} + 1 = 2025 = 3^4 \cdot 5^2; \] \[ a_3 = \frac{a_2}{5^{2-1}} + 1 = \frac{2025}{5} + 1 = 406. \] Determine for which values of \( a_1 \) the sequence is eventually periodic and what all the possible periods are. [b]Note:[/b] Let \( p \) be a positive integer. A sequence \( x_1, x_2, \dots \) is eventually periodic with period \( p \) if \( p \) is the smallest positive integer such that there exists an \( N \geq 0 \) satisfying \( x_{n+p} = x_n \) for all \( n > N \).

Let $m$ be a fixed integer greater than $1$. The sequence $x_0$, $x_1$, $x_2$, $\ldots$ is defined as follows: \[x_i = \begin{cases}2^i&\text{if }0\leq i \leq m - 1;\\\sum_{j=1}^mx_{i-j}&\text{if }i\geq m.\end{cases}\] Find the greatest $k$ for which the sequence contains $k$ consecutive terms divisible by $m$ . [i]Proposed by Marcin Kuczma, Poland[/i]

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.

Let $n$ be a positive integer and let $a_1, \ldots, a_{n-1} $ be arbitrary real numbers. Define the sequences $u_0, \ldots, u_n $ and $v_0, \ldots, v_n $ inductively by $u_0 = u_1 = v_0 = v_1 = 1$, and $u_{k+1} = u_k + a_k u_{k-1}$, $v_{k+1} = v_k + a_{n-k} v_{k-1}$ for $k=1, \ldots, n-1.$ Prove that $u_n = v_n.$