The infinite sequence $a_0,a _1, a_2, \dots$ of (not necessarily distinct) integers has the following properties: $0\le a_i \le i$ for all integers $i\ge 0$, and \[\binom{k}{a_0} + \binom{k}{a_1} + \dots + \binom{k}{a_k} = 2^k\] for all integers $k\ge 0$. Prove that all integers $N\ge 0$ occur in the sequence (that is, for all $N\ge 0$, there exists $i\ge 0$ with $a_i=N$).

For a positive integer $k$, define the $k$-[i]pop[/i] of a positive integer $n$ as the infinite sequence of integers $a_1, a_2, ...$ such that $a_1 = n$ and $$a_{i+1}= \left\lfloor \frac{a_i}{k} \right\rfloor , i = 1, 2, ..$$ where $ \lfloor x\rfloor $ denotes the greatest integer less than or equal to $x$. Furthermore, define a positive integer $m$ to be $k$-[i]pop avoiding[/i] if $k$ does not divide any nonzero term in the $k$-pop of $m$. For example, $14$ is 3-pop avoiding because $3$ does not divide any nonzero term in the $3$-pop of $14$, which is $14, 4, 1, 0, 0, ....$ Suppose that the number of positive integers less than $13^{2018}$ which are $13$-pop avoiding is equal to N. What is the remainder when $N$ is divided by $1000$?

The Fibonacci numbers are a sequence of numbers defined recursively as follows: $F_1=1$, $F_2=1$, and $F_n=F_{n-1}+F_{n-2}$. Using this definition, compute the sum $$\sum_{k=1}^{10}\frac{F_k}{F_{k+1}F_{k+2}}.$$

The sequence $(a_n)$ is defined recursively by $a_0=1$, $a_1=\sqrt[19]{2}$, and $a_n=a_{n-1}a_{n-2}^2$ for $n \ge 2$. What is the smallest positive integer $k$ such that the product $a_1a_2 \cdots a_k$ is an integer? $\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 19 \qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ 21$

Consider an infinite sequence consisting of distinct positive integers such that each term (except the rst one) is either an arithmetic mean or a geometric mean of two neighboring terms. Does it necessarily imply that starting at some point the sequence becomes either arithmetic progression or a geometric progression?

Consider increasing integer sequences with elements from $1,\ldots,10^6$. Such a sequence is [i]Adriatic[/i] if its first element equals 1 and if every element is at least twice the preceding element. A sequence is [i]Tyrrhenian[/i] if its final element equals $10^6$ and if every element is strictly greater than the sum of all preceding elements. Decide whether the number of Adriatic sequences is smaller than, equal to, or greater than the number of Tyrrhenian sequences. (Proposed by Gerhard Woeginger, Austria)

A sequence of real numbers $ a_{0},\ a_{1},\ a_{2},\dots$ is defined by the formula \[ a_{i \plus{} 1} \equal{} \left\lfloor a_{i}\right\rfloor\cdot \left\langle a_{i}\right\rangle\qquad\text{for}\quad i\geq 0; \]here $a_0$ is an arbitrary real number, $\lfloor a_i\rfloor$ denotes the greatest integer not exceeding $a_i$, and $\left\langle a_i\right\rangle=a_i-\lfloor a_i\rfloor$. Prove that $a_i=a_{i+2}$ for $i$ sufficiently large. [i]Proposed by Harmel Nestra, Estionia[/i]

A strategical video game consists of a map of finitely many towns. In each town there are $k$ directions, labelled from $1$ through $k$. One of the towns is designated as initial, and one – as terminal. Starting from the initial town the hero of the game makes a finite sequence of moves. At each move the hero selects a direction from the current town. This determines the next town he visits and a certain positive amount of points he receives. Two strategical video games are equivalent if for every sequence of directions the hero can reach the terminal town from the initial in one game, he can do so in the other game, and, in addition, he accumulates the same amount of points in both games. For his birthday John receives two strategical video games – one with $N$ towns and one with $M$ towns. He claims they are equivalent. Marry is convinced they are not. Marry is right. Prove that she can provide a sequence of at most $N +M$ directions that shows the two games are indeed not equivalent. [i]Stefan Gerdjikov, Bulgaria[/i]

A sequence of primes $a_n$ is defined as follows: $a_1 = 2$, and, for all $n \geq 2$,$ a_n$ is the largest prime divisor of $a_1a_2...a_{n-1} + 1$. Prove that $a_n \neq 5$ for all n. I'm presuming it must involve proving it's never equal to 0 mod 5, but I don't know what to do. Thanks

$(GDR 3)$ Find the number of permutations $a_1, \cdots, a_n$ of the set $\{1, 2, . . ., n\}$ such that $|a_i - a_{i+1}| \neq 1$ for all $i = 1, 2, . . ., n - 1.$ Find a recurrence formula and evaluate the number of such permutations for $n \le 6.$

Sequence $ \{ f_n(a) \}$ satisfies $ \displaystyle f_{n\plus{}1}(a) \equal{} 2 \minus{} \frac{a}{f_n(a)}$, $ f_1(a) \equal{} 2$, $ n\equal{}1,2, \cdots$. If there exists a natural number $ n$, such that $ f_{n\plus{}k}(a) \equal{} f_{k}(a), k\equal{}1,2, \cdots$, then we call the non-zero real $ a$ a $ \textbf{periodic point}$ of $ f_n(a)$. Prove that the sufficient and necessary condition for $ a$ being a $ \textbf{periodic point}$ of $ f_n(a)$ is $ p_n(a\minus{}1)\equal{}0$, where $ \displaystyle p_n(x)\equal{}\sum_{k\equal{}0}^{\left[ \frac{n\minus{}1}{2} \right]} (\minus{}1)^k C_n^{2k\plus{}1}x^k$, here we define $ \displaystyle \frac{a}{0}\equal{} \infty$ and $ \displaystyle \frac{a}{\infty} \equal{} 0$.

Let $ f:[0,1]\longrightarrow\mathbb{R} $ be a continuous and nondecreasing function. [b]a)[/b] Show that the sequence $ \left( \frac{1}{2^n}\sum_{i=1}^{2^n} f\left(\frac{i}{2^n}\right) \right)_{n\ge 1} $ is nonincreasing. [b]b)[/b] Prove that, if there exists some natural index at which the sequence above is equal to $ \int_0^1 f(x)dx, $ then $ f $ is constant.

For positive integers $m$ and $n$, let $d(m, n)$ be the number of distinct primes that divide both $m$ and $n$. For instance, $d(60, 126) = d(2^2 \cdot 3 \cdot 5, 2 \cdot 3^2 \cdot 7) = 2.$ Does there exist a sequence $(a_n)$ of positive integers such that: [list] [*] $a_1 \geq 2018^{2018};$ [*] $a_m \leq a_n$ whenever $m \leq n$; [*] $d(m, n) = d(a_m, a_n)$ for all positive integers $m\neq n$? [/list] [i](Dominic Yeo, United Kingdom)[/i]

The sequence $f_1, f_2, \cdots, f_n, \cdots $ of functions is defined for $x > 0$ recursively by \[f_1(x)=x , \quad f_{n+1}(x) = f_n(x) \left(f_n(x) + \frac 1n \right)\] Prove that there exists one and only one positive number $a$ such that $0 < f_n(a) < f_{n+1}(a) < 1$ for all integers $n \geq 1.$

Let $T_k = k - 1$ for $k = 1, 2, 3,4$ and \[T_{2k-1} = T_{2k-2} + 2^{k-2}, T_{2k} = T_{2k-5} + 2^k \qquad (k \geq 3).\] Show that for all $k$, \[1 + T_{2n-1} = \left[ \frac{12}{7}2^{n-1} \right] \quad \text{and} \quad 1 + T_{2n} = \left[ \frac{17}{7}2^{n-1} \right],\] where $[x]$ denotes the greatest integer not exceeding $x.$