Found problems: 1239
The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n+2} = F_{n+1}+F_n$ for every integer $n$. A sequence $(a_n)$ of integers is said to be $\textit{phirme}$ if there is a fixed integer $k$ such that $a_n + a_{n+1} = F_{n+k}$ for all $n \geq 1$. Show that if $(a_n)$ is a $\textit{phirme}$ sequence, then there exists an integer $c$ such that $$a_n = F_{n+k-2} + (-1)^nc.$$
Let $n\ge 2$ be a natural number. Let $a_1\le a_2\le a_3\le \cdots \le a_n$ be real numbers such that $a_1+a_2+\cdots +a_n>0$ and $n(a_1^2+a_2^2+\cdots +a_n^2)=2(a_1+a_2+\cdots +a_n)^2.$ If $m=\lfloor n/2\rfloor+1$, the smallest integer larger than $n/2$, then show that $a_m>0.$
Let $a_1,a_2,a_3,\dots$ be an infinite sequence of non-zero reals satisfying \[a_{i} = \frac{a_{i-1}a_{i-2}-2}{a_{i-3}}\]for all $i\geq 4$. Determine all positive integers $n$ such that if $a_1,a_2,\dots,a_n$ are integers, then all elements of the sequence are integers.
Let the sequence ($a_n$) be defined by $a_n = n^6 +5n^4 -12n^2 -36, n \ge 2$.
(a) Prove that any prime number divides some term in this sequence.
(b) Prove that there is a positive integer not dividing any term in the sequence.
(c) Determine the least $n \ge 2$ for which $1989 | a_n$.
For each $x>e^e$ define a sequence $S_x=u_0,u_1,\ldots$ recursively as follows: $u_0=e$, and for $n\ge0$, $u_{n+1}=\log_{u_n}x$. Prove that $S_x$ converges to a number $g(x)$ and that the function $g$ defined in this way is continuous for $x>e^e$.
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]
$a_1,a_2,\ldots,a_n$ 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 $<a_n>$ is defined as follows, $a_1=a_2=1$, $a_3=2$,
$$a_{n+3}=\frac{a_{n+2}a_{n+1}+n!}{a_n},$$
$n \ge 1$.
Prove that all the terms in the sequence are integers.
How many $0$'s are there in the sequence $x_1, x_2,..., x_{2014}$ where $x_n =\big[ \frac{n + 1}{\sqrt{2015}}\big] -\big[ \frac{n }{\sqrt{2015}}\big]$ , $n = 1, 2,...,2014$ ?
(A): $1128$, (B): $1129$, (C): $1130$, (D): $1131$, (E) None of the above.
[b]Problem.[/b] Let $f:\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$ be a function defined as $$f(n)=n+\lfloor\sqrt{n}\rfloor~\forall~n\in\mathbb{R}_0^+.$$ Prove that for any positive integer $m,$ the sequence $$m,f(m),f(f(m)),f(f(f(m))),\ldots$$ contains a perfect square.
For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as
$$a_{n+1} =
\begin{cases}
\sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\
a_n + 3 & \text{otherwise.}
\end{cases}
$$
Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.
[i]Proposed by Stephan Wagner, South Africa[/i]
Let $(a_n)$ be defined by:
$$ a_1 = 2, \qquad a_{n+1} = a_n^3 - a_n + 1 $$
Consider positive integers $n,p$, where $p$ is an odd prime. Prove that if $p | a_n$, then $p > n$.
The sequence $\{a_n\}$ is defined as follows: $a_1$ is a positive rational number, $a_n= \frac{p_n}{q_n}$, ($n= 1,2,…$) is a positive integer, where $p_n$ and $q_n$ are positive integers that are relatively prime, then $a_{n+1} = \frac{p_n^2+2015}{p_nq_n}$ Is there a$_1>2015$, making the sequence $\{a_n\}$ a bounded sequence? Justify your conclusion.
Find all possible values of $C\in \mathbb R$ such that there exists a real sequence $\{a_n\}_{n=1}^\infty$ such that
$$a_na_{n+1}^2\ge a_{n+2}^4 +C$$
for all $n\ge 1$.
Determine all functions $f: \mathbb{Z}_{\geq 2025} \to \mathbb{Z}_{>0}$ such that $mn+1$ divides $f(m)f(n) + 1$ for any integers $m,n \geq 2025$ and there exists a polynomial $P$ with integer coefficients, such that $f(n) \leq P(n)$ for all $n\geq 2025$.
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.
Let $n$ be a positive integer. Given a sequence $\varepsilon_1$, $\dots$, $\varepsilon_{n - 1}$ with $\varepsilon_i = 0$ or $\varepsilon_i = 1$ for each $i = 1$, $\dots$, $n - 1$, the sequences $a_0$, $\dots$, $a_n$ and $b_0$, $\dots$, $b_n$ are constructed by the following rules: \[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\] \[\begin{array}{lll}
a_{i+1} =
\begin{cases}
2a_{i-1} + 3a_i, \\
3a_{i-1} + a_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_i = 0, \\
\text{if } \varepsilon_i = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1, \\[15pt]
b_{i+1}=
\begin{cases}
2b_{i-1} + 3b_i, \\
3b_{i-1} + b_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_{n-i} = 0, \\
\text{if } \varepsilon_{n-i} = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1.
\end{array}\] Prove that $a_n = b_n$.
[i]Proposed by Ilya Bogdanov, Russia[/i]
Let be a $ \beta >1. $ Calculate $ \lim_{n\to\infty} \frac{k(n)}{n} ,$ where $ k(n) $ is the smallest natural number that satisfies the inequality $ (1+n)^k\ge n^k\beta . $
[i]Neculai Hârţan[/i]
The Fibonacci sequence is defined by $F_0=F_1=1$ and $F_{n+2}=F_{n+1}+F_n$ for $n\ge0$. Prove that the sum of $2000$ consecutive terms of the Fibonacci sequence is never a term of the sequence.
Let $ a_1$, $ a_2$, $ \ldots$, $ a_n$ be distinct positive integers, $ n\ge 3$. Prove that there exist distinct indices $ i$ and $ j$ such that $ a_i \plus{} a_j$ does not divide any of the numbers $ 3a_1$, $ 3a_2$, $ \ldots$, $ 3a_n$.
[i]Proposed by Mohsen Jamaali, Iran[/i]
Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0=37$, $a_1=72$, $a_m=0$, and \[a_{k+1}=a_{k-1}-\frac{3}{a_k}\] for $k=1,2, \dots, m-1$. Find $m$.
Find the limit of the sequence $x_n$ defined by recurrence relation $$x_{n+2}=\frac{1}{12}x_{n+1}+\frac{1}{2}x_{n}+1$$ where $n=0,1,2,...$ for any initial values $x_2,x_1$.
Suppose $a_1 =\frac16$ and $a_n = a_{n-1} - \frac{1}{n}+ \frac{2}{n + 1} - \frac{1}{n + 2}$ for $n > 1$. Find $a_{100}$.
The sequence of real numbers $a_1, a_2, a_3, ...$ is defined as follows: $a_1 = 2019$, $a_2 = 2020$, $a_3 = 2021$ and for all $n \ge 1$
$$a_{n+3} = 5a^6_{n+2} + 3a^3_{n+1} + a^2_n.$$
Show that this sequence does not contain numbers of the form $m^6$ where $m$ is a positive integer.
Let \(a_0, a_1, a_2, \dots\) be an infinite sequence of positive integers with the following properties:
- \(a_0\) is a given positive integer;
- For each integer \(n \geq 1\), \(a_n\) is the smallest integer greater than \(a_{n-1}\) such that \(a_n + a_{n-1}\) is a perfect square.
For example, if \(a_0 = 3\), then \(a_1 = 6\), \(a_2 = 10\), \(a_3 = 15\), and so on.
(a) Let \(T\) be the set of numbers of the form \(a_k - a_l\), with \(k \geq l \geq 0\) integers.
Prove that, regardless of the value of \(a_0\), the number of positive integers not in \(T\) is finite.
(b) Calculate, as a function of \(a_0\), the number of positive integers that are not in \(T\).