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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Triangular Numbers in action
integrated_JRC   29
N an hour ago by Aiden-1089
Source: RMO 2018 P5
Find all natural numbers $n$ such that $1+[\sqrt{2n}]~$ divides $2n$.

( For any real number $x$ , $[x]$ denotes the largest integer not exceeding $x$. )
29 replies
integrated_JRC
Oct 7, 2018
Aiden-1089
an hour ago
Cute property of Pascal hexagon config
Miquel-point   1
N an hour ago by FarrukhBurzu
Source: KoMaL B. 5444
In cyclic hexagon $ABCDEF$ let $P$ denote the intersection of diagonals $AD$ and $CF$, and let $Q$ denote the intersection of diagonals $AE$ and $BF$. Prove that if $BC=CP$ and $DP=DE$, then $PQ$ bisects angle $BQE$.

Proposed by Géza Kós, Budapest
1 reply
Miquel-point
2 hours ago
FarrukhBurzu
an hour ago
Number theory problem
Angelaangie   3
N an hour ago by megarnie
Source: JBMO 2007
Prove that 7p+3^p-4 it is not a perfect square where p is prime.
3 replies
Angelaangie
Jun 19, 2018
megarnie
an hour ago
another n x n table problem.
pohoatza   3
N 2 hours ago by reni_wee
Source: Romanian JBTST III 2007, problem 3
Consider a $n$x$n$ table such that the unit squares are colored arbitrary in black and white, such that exactly three of the squares placed in the corners of the table are white, and the other one is black. Prove that there exists a $2$x$2$ square which contains an odd number of unit squares white colored.
3 replies
pohoatza
May 13, 2007
reni_wee
2 hours ago
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   18
N 2 hours ago by ihategeo_1969
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
18 replies
MarkBcc168
Apr 28, 2020
ihategeo_1969
2 hours ago
Anything real in this system must be integer
Assassino9931   8
N 2 hours ago by Abdulaziz_Radjabov
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
8 replies
Assassino9931
May 9, 2025
Abdulaziz_Radjabov
2 hours ago
Good Permutations in Modulo n
swynca   10
N 2 hours ago by MR.1
Source: BMO 2025 P1
An integer $n > 1$ is called $\emph{good}$ if there exists a permutation $a_1, a_2, a_3, \dots, a_n$ of the numbers $1, 2, 3, \dots, n$, such that:
$(i)$ $a_i$ and $a_{i+1}$ have different parities for every $1 \leq i \leq n-1$;
$(ii)$ the sum $a_1 + a_2 + \cdots + a_k$ is a quadratic residue modulo $n$ for every $1 \leq k \leq n$.
Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.
10 replies
swynca
Apr 27, 2025
MR.1
2 hours ago
Quadratic + cubic residue => 6th power residue?
Miquel-point   0
2 hours ago
Source: KoMaL B. 5445
Decide whether the following statement is true: if an infinite arithmetic sequence of positive integers includes both a perfect square and a perfect cube, then it also includes a perfect $6$th power.

Proposed by Sándor Róka, Nyíregyháza
0 replies
Miquel-point
2 hours ago
0 replies
II_a - r_a = R - r implies A = 60
Miquel-point   0
2 hours ago
Source: KoMaL B. 5421
The incenter and the inradius of the acute triangle $ABC$ are $I$ and $r$, respectively. The excenter and exradius relative to vertex $A$ is $I_a$ and $r_a$, respectively. Let $R$ denote the circumradius. Prove that if $II_a=r_a+R-r$, then $\angle BAC=60^\circ$.

Proposed by Class 2024C of Fazekas M. Gyak. Ált. Isk. és Gimn., Budapest
0 replies
Miquel-point
2 hours ago
0 replies
Cheating effectively in game of luck
Miquel-point   0
2 hours ago
Source: KoMaL B. 5420
Ádám, the famous conman signed up for the following game of luck. There is a rotating table with a shape of a regular $13$-gon, and at each vertex there is a black or a white cap. (Caps of the same colour are indistinguishable from each other.) Under one of the caps $1000$ dollars are hidden, and there is nothing under the other caps. The host rotates the table, and then Ádám chooses a cap, and take what is underneath. Ádám's accomplice, Béla is working at the company behind this game. Béla is responsible for the placement of the $1000$ dollars under the caps, however, the colors of the caps are chosen by a different collegaue. After placing the money under a cap, Béla
[list=a]
[*] has to change the color of the cap,
[*] is allowed to change the color of the cap, but he is not allowed to touch any other cap.
[/list]
Can Ádám and Béla find a strategy in part a. and in part b., respectively, so that Ádám can surely find the money? (After entering the casino, Béla cannot communicate with Ádám, and he also cannot influence his colleague choosing the colors of the caps on the table.)

Proposed by Gábor Damásdi, Budapest
0 replies
Miquel-point
2 hours ago
0 replies
IMO Genre Predictions
ohiorizzler1434   68
N 3 hours ago by Koko11
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
68 replies
ohiorizzler1434
May 3, 2025
Koko11
3 hours ago
Gcd(m,n) and Lcm(m,n)&F.E.
Jackson0423   1
N 3 hours ago by WallyWalrus
Source: 2012 KMO Second Round

Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all positive integers \( m, n \),
\[
f(mn) = \mathrm{lcm}(m, n) \cdot \gcd(f(m), f(n)),
\]where \( \mathrm{lcm}(m, n) \) and \( \gcd(m, n) \) denote the least common multiple and the greatest common divisor of \( m \) and \( n \), respectively.
1 reply
Jackson0423
May 13, 2025
WallyWalrus
3 hours ago
Trigonometric Product
Henryfamz   3
N 3 hours ago by Aiden-1089
Compute $$\prod_{n=1}^{45}\sin(2n-1)$$
3 replies
Henryfamz
May 13, 2025
Aiden-1089
3 hours ago
"Eulerian" closed walk with of length less than v+e
Miquel-point   0
3 hours ago
Source: IMAR 2019 P4
Show that a connected graph $G=(V, E)$ has a closed walk of length at most $|V|+|E|-1$ passing through each edge of $G$ at least once.

Proposed by Radu Bumbăcea
0 replies
Miquel-point
3 hours ago
0 replies
a_i/i sequence
pad   19
N Apr 29, 2025 by ihatemath123
Source: TSTST 2021/2
Let $a_1<a_2<a_3<a_4<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence
\[ \frac{a_1}{1},\frac{a_2}{2},\frac{a_3}{3},\frac{a_4}{4},\ldots.\]
Merlijn Staps
19 replies
pad
Nov 8, 2021
ihatemath123
Apr 29, 2025
a_i/i sequence
G H J
G H BBookmark kLocked kLocked NReply
Source: TSTST 2021/2
The post below has been deleted. Click to close.
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pad
1671 posts
#1 • 4 Y
Y by centslordm, megarnie, jasperE3, rightways
Let $a_1<a_2<a_3<a_4<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence
\[ \frac{a_1}{1},\frac{a_2}{2},\frac{a_3}{3},\frac{a_4}{4},\ldots.\]
Merlijn Staps
This post has been edited 4 times. Last edited by pad, Nov 8, 2021, 5:33 PM
Z K Y
The post below has been deleted. Click to close.
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pad
1671 posts
#2 • 4 Y
Y by PRMOisTheHardestExam, centslordm, dantee, IAmTheHazard
Define an infinte graph $G$ whose vertices are $1,2,3,\ldots$ and connect $i\leftrightarrow j$ if and only if $a_i/i = a_j/j$. Note that $G$ is the disjoint union of cliques, each grouped on the common value of $a_i/i$. Assume for the sake of contradiction that no such clique has size $1$.

Note that all cliques of $G$ have finite size; suppose a clique with vertices $(x_1,x_2,\ldots)$ has infinite size. Then
\[ \frac{a_{x_1}}{x_1}=\frac{a_{x_2}}{x_2}=\cdots, \]but since $(x_i)$ is an unbounded sequence, $(a_{x_i})$ is also unbounded, contradiction.

For each $n\ge 1$, define $n$ to be an ender if $n$ has no edge to an $n'>n$. Define $n$ to be a non-ender if $n$ is not an ender.

Lemma 1. There are infinitely many non-enders.

Proof. Assume for the sake of contradiction that for some finite $N$, all $n\ge N$ are non-enders. Note that no edge connects two numbers $N_1,N_2>N$; else, the lower is a non-ender. Also note that no two $N_1,N_2>N$ connect to the same $n'$, for any $n'<n$; else, $N_1 \leftrightarrow n' \leftrightarrow N_2$, so all are part of the same clique, contradicting $N_1,N_2$ not being an edge. Combining the above two facts implies that each of the values in $(N,\infty)$ has an edge to a unique value in $[1,N]$. This contradicts pigeonhole principle. $\blacksquare$
We construct an infinite sequence $(n_1,n_2,\ldots)$. Define $n_1=1$. Note that $1$ is a non-ender since it must have degree at least $1$ by assumption, and it must connect to a number greater than $1$. For each $i\ge 1$:
  • Define $x_i$ to be the maximal integer for which $n_i\leftrightarrow n_i+x_i$ is an edge. We will show that $n_i$ is a non-ender later, which is necessary for this to be well-defined. Also note that $x_i$ exists since it is finite since cliques have finite size.
  • Define $y_i$ to be the minimal nonnegative integer for which $n_i+x_i+y_i$ is a non-ender. This exists by Lemma 1.
Now let $n_{i+1}=n_i+x_i+y_i$. Note that $n_k=n_1+(x_1+\cdots+x_{k-1})+(y_1+\cdots+y_{k-1})$ for all $k\ge 1$.

Note that the definition of $x_i$ makes sense since we proved $n_1=1$ is a non-ender, and by construction, all later $n_i$ are non-enders.

Lemma 2. [Key combinatorial estimate] We have $y_1+\cdots+y_{\ell} \le x_1+\cdots+x_\ell$ for all $\ell\ge 1$.

Proof. Intuitive diagram: the bottom line is the number line, and the curved lines above are edges of $G$:
https://media.discordapp.net/attachments/752979051370774612/905980143485476864/tstst2021_2_pic.JPG
Define $S_i=[n_i,n_i+x_i)$ and $E_i=[n_i+x_i,n_{i+1})$ for each $i\ge 1$. Everything in $E_i$ is an ender by construction, $n_{i+1}=n_i+x_i+y_i$ is the minimal non-ender after $n_i+x_i$. Consider some fixed $u\in E_i$, for some fixed $i$.

We claim $u$ cannot have an edge to $n_j+x_j$, for any $j\le i$. (Except the trivial case of $j=i$ and $u=n_i+x_i$ so $u$ connects to itself.) Suppose it did. Then $u\leftrightarrow n_j+x+j\leftrightarrow n_j$, but since $u\ge n_i+x+i \ge n_j+x_j$, and one of these inequalities is strict since we threw out the trivial case, this means $n_j+x_j$ was not the maximal value that $n_j$ has an edge to, contradicting the construction of $x_j$.

We further claim that $u$ does not have an edge to any element of $E_j$ for any $j\le i$. We showed above $u$ does not connect to the first element of any $E_j$. If $u$ has an edge to some $v\in E_j \setminus \{n_j+x_j\}$ for some $j\le i$, then $v$ is not an ender since it connects to $u>v$, contradiction.

Finally, we claim no two $u,u' \in E_1\cup\cdots \cup E_i$ connect to the same element $s$ of $S_1\cup \cdots \cup S_i$. If they did, then $u\leftrightarrow s \leftrightarrow u'$, so $u$ and $u'$ are connected, so $\min\{u,u'\}$ is not an ender, contradiction.

Combining the above facts implies that each element of $E_1\cup\cdots\cup E_i$ connects to a unique element of $S_1\cup \cdots \cup S_i$. Hence
\[ |E_1\cup \cdots E_i| \le |S_1\cup \cdots \cup S_i| \implies y_1 + \cdots +y_i \le x_1+\cdots+x_i. \]$\blacksquare$
Now we use the $(n_i)$ sequence to prove that the corresponding subsequence of $(a_i)$ is unbounded. We have for all $i\ge 1$ that
\[
\frac{a_{n_i}}{n_i} = \frac{a_{n_i+x_i}}{n_i+x_i} \implies a_{n_i+x_i} = \frac{n_i+x_i}{n_i}a_{n_i},
\]and since $(a_i)$ is increasing, we have
\[
a_{n_{i+1}}=a_{n_i+x_i+y_i} > a_{n_i+x_i} = \frac{n_i+x_i}{n_i}a_{n_i}.
\]Iterating the above procedure, we get that
\begin{align*}
    \frac{a_{n_{i+1}}}{a_{n_1}} &> \frac{n_i+x_i}{n_i}\cdot \frac{n_{i-1}+x_{i-1}}{n_{i-1}} \cdot \cdots \cdot \frac{n_1+x_1}{n_1} \\
    &=\prod_{\ell=1}^i \frac{n_\ell+x_\ell}{n_\ell} \\
    &= \prod_{\ell=1}^i \frac{1+(x_1+\cdots+x_{\ell-1}+x_\ell)+(y_1+\cdots+y_{\ell-1})}{1+(x_1+\cdots+x_{\ell-1})+(y_1+\cdots+y_{\ell-1})} \\
    &= \prod_{\ell=1}^i \frac{1+X_\ell+Y_{\ell-1}}{1+X_{\ell-1}+Y_{\ell-1}},
\end{align*}where $X_k:=x_1+\cdots+x_k$ and $Y_k:=y_1+\cdots+y_k$ for all $k\ge 1$. We know $Y_\ell \le X_\ell$ for all $\ell$ by Lemma 2. We show this product is unboundedly large as $i$ increases.

The function
\[f(x)=\frac{1+X_\ell+x}{1+X_{\ell-1}+x}\]is decreasing for $x>0$; this can be verified by subtracting $1$ and noting that $X_\ell-X_{\ell-1}=x_{\ell}>0$. In particular, $f(Y_\ell) \ge f(X_\ell)$. Secondly, for $a\ge b \ge 1$, it can be verified that
\[ \frac{\frac12+a}{\frac12+b} \ge \sqrt{\frac{a}{b}}. \]Now, using these two facts and AM-GM:
\begin{align*}
    \frac{1+X_\ell+Y_{\ell-1}}{1+X_{\ell-1}+Y_{\ell-1}} &\ge \frac{1+X_\ell+X_{\ell-1}}{1+X_{\ell-1} + X_{\ell-1}} =\frac{\frac12 + \frac{X_{\ell}+X_{\ell-1}}{2}}{\frac12 + X_{\ell-1}} \\
    &> \sqrt{\frac{\frac{X_\ell+X_{\ell-1}}{2}}{X_{\ell-1}}} \ge \sqrt{\frac{\sqrt{X_\ell X_{\ell-1}}}{X_{\ell-1}}} = \left(\frac{X_\ell}{X_{\ell-1}}\right)^{1/4}. 
\end{align*}Finally,
\begin{align*}
    \frac{a_{n_{i+1}}}{a_{n_1}}> \prod_{\ell=1}^i  \frac{1+X_\ell+Y_{\ell-1}}{1+X_{\ell-1}+Y_{\ell-1}} > \prod_{\ell=2}^i \left(\frac{X_\ell}{X_{\ell-1}}\right)^{1/4} = \left(\frac{X_i}{X_1}\right)^{1/4}. 
\end{align*}(Note that we omitted one term -- it still works since each term is greater than $1$.) Now, we have a contradiction, since $X_i$ grows unboundedly large since $X_{i+1}-X_i = x_{i+1}$, so it grows by at least $1$ at each step.

Motivation
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jj_ca888
2726 posts
#3 • 4 Y
Y by YIFANK, PRMOisTheHardestExam, centslordm, Realgauss1
FTSoC suppose numbers appear at least twice. Partition $\mathbb{N}$ into groups $S_1, S_2, \ldots$ so that for every group $S_i$, the value of $\tfrac{a_k}{k}$ is constant across all $k \in S_i$. By the assumption of the opposite of the problem, $|S_i| \geq 2$ for all $i$.

Claim: All groups have finite size.
Proof: Suppose there is an group with infinite size, and let $m$ be its smallest element. Then, we can choose arbitrarily large $M$ in the group for which\[\frac{a_m}{m} = \frac{a_M}{M} \implies a_M = \frac{Ma_m}{m} \geq 1\]which is no good.

Call a number an apex predator if it is the greatest element of some group. Let $M_i$ be the sequence of all apex predators in increasing order. We prove the seemingly random claim:

Claim:\[\frac{M_1}{1}\left(\frac{M_2}{M_1 + 1}\right)\ldots \left(\frac{M_{n+1}}{M_n + 1}\right) \geq \frac{2n+2}{\sqrt{2n+1}}\]Proof: Rewrite this as\[\frac{M_1}{1}\left(\frac{M_2}{M_1 + 1}\right)\ldots \left(\frac{M_{n+1}}{M_n + 1}\right) = M_{n+1} \left(\prod_{i = 1}^n \frac{M_i}{M_i + 1}\right)\]where since each group has size at least $2$, we can conclude that among any first $n$ positive integers there are at most $\tfrac{n}{2}$ apex predators. This also tells us that the $i$-th apex predator must be $\geq 2i$, else there are too many apex predators among the first $M_i$ positive integers. Therefore, it follows that\[M_{n+1} \left(\prod_{i = 1}^n \frac{M_i}{M_i + 1}\right) \geq (2n+2) \left(\prod_{i = 1}^n \frac{2i}{2i + 1}\right) = \frac{(2n+2)!!}{(2n+1)!!}\]which we can induct to prove is $\geq \tfrac{2n+2}{\sqrt{2n+1}}$.

Now with this, we will be able to construct some $a_M \geq 1$. Let $y_1 = 1$ and $M_{r_1}$ be the apex predator in its group. Let $y_2$ be the least number $> M_{r_1}$ that is not an apex predator, then let $M_{r_2}$ be the apex predator in $y_2$'s group, and so on defining the sequence $y_i$ and $r_i$. For any $k$, note\[a_{M_{r_k}} = \left(\frac{M_{r_k}}{y_k}\right)a_{y_k} > \left(\frac{M_{r_k}}{y_k}\right)a_{M_{r_{k-1}}} > \ldots > \left[\left(\frac{M_{r_k}}{y_k}\right)\ldots \left(\frac{M_{r_1}}{y_1}\right)\right]a_1.\]It is actually not hard to see that\[\left(\frac{M_{r_k}}{y_k}\right)\ldots \left(\frac{M_{r_1}}{y_1}\right) > \left(\frac{M_1}{1}\right)\left(\frac{M_2}{M_1 + 1}\right)\ldots \left(\frac{M_{r_k}}{M_{r_{k-1}} + 1}\right)\]mainly due to two things:
  • Note that we actually do get the $y_i$'s successfully on the RHS because if we have consecutive apex predators $M_i, M_{i+1}$, then $\tfrac{M_{i+1}}{M_i + 1}$ is $1$ so it does not matter, so in effect we are skipping past all consecutive apex predators to get the least number that is not an apex predator.
  • If the $r_i$'s skip any apex predators, say $M_{r_i} < M_{\ell_1} < \ldots < M_{\ell_t} < M_{r_{i+1}}$, then\[\frac{M_{r_{i+1}}}{y_{i+1}} > \prod_{i = 1}^{t + 1} \left(\frac{M_{\ell_i}}{M_{\ell_{i-1}} + 1}\right)\]is true.
Hence, by the lemma, we know\[a_{M_{r_k}} > \frac{2r_k}{\sqrt{2r_k - 1}} a_1\]so continuously going to further terms in the sequence we can get arbitrarily large $r_k$ so we can create an arbitrarily large constant $c$ for which $a_{something} = ca_1$, breaking $(0, 1)$, a contradiction so we are done.
This post has been edited 2 times. Last edited by jj_ca888, Nov 8, 2021, 5:05 PM
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TheUltimate123
1740 posts
#4 • 4 Y
Y by PRMOisTheHardestExam, centslordm, VicKmath7, Kobayashi
Assume for contradiction none of \(\frac{a_1}1\), \(\frac{a_2}2\), \(\ldots\) are unique.

Say \(i\) is a peak if for all \(j>i\), we have \(\frac{a_j}j\ne\frac{a_i}i\). Then for each \(j\), there is a peak \(i\ge j\) with \(\frac{a_i}i=\frac{a_j}j\); moreover, for each \(k\), there are at most \(k/2\) peaks among 1, 2, \(\ldots\), \(k\).

Claim: For each \(n\), there is an \(N>n\) with \(a_N\ge1.99a_n\).

Proof. Let \(x_1=n\), and consider the \((x_i)\) and \((y_i)\) defined as follows:
  • Let \(y_i>x_i\) be the peak with \(\frac{a_{y_i}}{y_i}=\frac{a_{x_i}}{x_i}\).
  • Let \(x_{i+1}\) be the smallest non-peak with \(x_{i+1}>y_i\).
Since \(a_{x_{i+1}}>a_{y_i}\), we always have \[a_{y_{i+1}}=\frac{y_{i+1}}{x_{i+1}}\cdot a_{x_{i+1}}>\frac{y_{i+1}}{x_{i+1}}\cdot a_{y_i}.\]It follows that for each \(k\), \[a_{y_k}\ge\frac{y_k}{x_k}\cdot\frac{y_{k-1}}{x_{k-1}}\cdots\frac{y_1}{x_1}\cdot a_n.\]
Let there be \(m\) peaks \(j\le n\). By construction, for each \(i\), each of \(y_i\), \(y_i+1\), \(\ldots\), \(x_{i+1}-1\) is a peak. Thus the number of peaks among 1, \(\ldots\), \(y_k\) is \[m+(x_2-y_1)+(x_3-y_2)+\cdots+(x_k-y_{k-1})+1.\]Recall there are at most \(y_k/2\) peaks among 1, \(\ldots\), \(y_k\), so \begin{align*}         m+(x_2-y_1)+\cdots+(x_k-y_{k-1})+1&\le\tfrac12y_k\\         \implies (y_k-x_k)+\cdots+(y_1-x_1)         &=y_k-x_1-\big[(x_2-y_1)+\cdots+(x_k-y_{k-1})\big]\\         &\ge y_k-n-\big[\tfrac12y_k-m-1\big]\\         &=\tfrac12y_k-(n-m-1).     \end{align*}
Now notice that \begin{align*}         \frac{a_{y_k}}{a_1}&\ge\frac{y_k}{x_k}\cdot\frac{y_{k-1}}{x_{k-1}}\cdots\frac{y_1}{x_1}\\         &=\underbrace{             \left(\frac{y_k}{y_k-1}\cdot\frac{y_k-1}{y_k-2}\cdots\cdot\frac{x_k+1}{x_k}\right)             \cdots             \left(\frac{y_1}{y_1-1}\cdots\frac{x_1+1}{x_1}\right)         }_{\ge\frac12y_k-(n-m-1)\text{ fractions}}\\         &\ge\frac{y_k}{y_k-1}\cdot\frac{y_k-1}{y_k-2}\cdots         \frac{y_k-(\frac12y_k-(n-m-1)-1)}{y_k-(\frac12y_k-(n-m-1))}\\         &=\frac{y_k}{\frac12y_k+(n-m-1)}>1.99     \end{align*}for sufficiently large \(y_k\) (i.e.\ sufficiently large \(k\)). \(\blacksquare\)

Now inductively, for each \(k\) there is an \(N_k\) with \[a_{N_k}\ge1.99^k\cdot a_1.\]Taking \(1.99^k>\frac1{a_1}\) gives \(a_{N_k}>1\), contradiction.
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VicKmath7
1390 posts
#5 • 1 Y
Y by megarnie
Wait, I'm probably being really really dumb right now (sorry if this is the case), but @above can you tell me why there are no peak indices between $x_i$ and $y_i$ ($x_i<y_i$) for $i=1,2,..., k$? Or I should assume that $m+(x_2-y_1)+(x_3-y_2)+\cdots+(x_k-y_{k-1})+1.$ is just a lower bound of the peak indices in $1,2,..., y_k$ (well, probably it would be better if it is clarified, if this is the case, because now one assumes that this is the exact count; not that it changes anything though)
This post has been edited 4 times. Last edited by VicKmath7, Nov 9, 2021, 11:14 AM
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txc
12 posts
#7 • 4 Y
Y by rama1728, Mathematicsislovely, PRMOisTheHardestExam, centslordm
Lemma. No real number can occur infinitely many times in the sequence.
Proof. Suppose such a number $c$ exists. Then $\frac{a_k}{k}=c\Rightarrow k=\frac{a_k}{c}<\frac{1}{c}$ is a contradiction.

Let $b_k=\frac{a_k}{k}$. Define a sequence $1=p_1<q_1<p_2<q_2<\dots$ recursively. For each $k\geq 1$, let $q_k$ be the largest positive integer with $b_{q_k}=b_{p_k}$, and let $p_{k+1}$ be the smallest positive integer greater than $q_k$ such that there exists some positive integer $q>p_{k+1}$ with $b_q=b_{p_{k+1}}$.

Now $a_{q_k}=\frac{q_k}{p_k}\cdot a_{p_k}>\frac{q_k}{p_k}\cdot a_{q_{k-1}}>\frac{q_k}{p_k}\cdot\frac{q_{k-1}}{p_{k-1}}\cdot a_{q_{k-2}}>\dots>\frac{q_k}{p_k}\cdot\frac{q_{k-1}}{p_{k-1}}\cdot\dots\cdot\frac{q_2}{p_2}\cdot a_{q_1}=\frac{q_k}{p_k}\cdot\frac{q_{k-1}}{p_{k-1}}\cdot\dots\cdot\frac{q_1}{p_1}\cdot a_1$.
It suffices to show that $\prod\limits_{i=1}^{\infty}\frac{q_i}{p_i}$ diverges as there will be some large enough $k$ where $a_{q_k}$ will be forced to be greater than $1$.

We now colour the positive integers as such: $p_i$ are green, $q_i$ are red, any $r$ with $p_i<r<q_i$ are yellow, any $r$ with $q_i<r<p_{i+1}$ are blue. Consider a blue integer $q_i<r<p_{i+1}$. By condition $\exists s\neq r, b_s=b_r$. If $s>r$, the minimality of $p_{i+1}$ is contradicted. If $s=p_j$ for some $j\leq i$, the maximality of $q_j$ is contradicted. If $s=q_j$, we can pick $s=p_j$ instead. If $s<r$ is blue, where $q_j<s<p_{j+1}$, the minimality of $p_{j+1}$ is contradicted.

Thus $s$ must be a yellow number smaller than $r$. No two $r_1<r_2$ can map to the same $s$ since then $r_2$ can map to $r_1$, a contradiction. In other words, there is an injection mapping each blue number to a smaller yellow number.

Let $x_1<x_2<\dots$ be the sequence of green and yellow numbers. We have $$\prod\limits_{i=1}^{\infty}\frac{q_i}{p_i}=\prod\limits_{i=1}^{\infty}\left(1+\frac{q_i-p_i}{p_i}\right)\geq 1+\sum\limits_{i=1}^{\infty}\frac{q_i-p_i}{p_i}\geq 1+\sum\limits_{i=1}^{\infty}\left(\frac{1}{p_i}+\frac{1}{p_i+1}+\dots+\frac{1}{q_i-1}\right)=1+\sum\limits_{i=1}^{\infty}\frac{1}{x_i}.$$But now consider the positive integers less than some $x_i$. There are exactly $i-1$ green and yellow numbers. There are not more blue numbers than yellow numbers by the above injection. There are also not more red numbers than green numbers. Thus there are at most $2i-2$ positive integers less than $x_i$, so $x_i\leq 2i-1$, so the above series diverges. $\Box$
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AforApple
134 posts
#8 • 3 Y
Y by PRMOisTheHardestExam, centslordm, brandone
My solution seems to be easier than the rest of the ones in this thread, but I think it's still valid, so here it is...

Claim: Every value of $\frac{a_k}{k}$ only occurs finitely many times.

Proof: Notice that $\frac{a_n}{n} < \frac{1}{n}$ and $\frac{1}{n}$ always decreases and gets arbitrarily close to $0$, so for any fixed value of $k$ there exists some constant $M$ so that for every $n > M$ we have $\frac{a_k}{k} > \frac{1}{n}$, so $\frac{a_k}{k} = \frac{a_n}{n}$ can only be possible for integers $n$ up to $M$, which is finite, as desired $\square$

Now suppose FTSoC every distinct value of $\frac{a_k}{k}$ appears at least twice. Now let $k_1$ be the largest integer such that $\frac{a_1}{1} = \frac{a_{k_1}}{k_1}$. Then it follows that $a_{k_1} = a_1 \cdot \frac{k_1}{1}$

Now let $k_2$ be the largest integer such that $\frac{a_{k_1+1}}{k_1+1} = \frac{a_{k_2}}{k_2}$. Then we must have $$a_{k_2} = a_{k_1 + 1} \cdot \frac{k_2}{k_1 + 1} > a_{k_1} \cdot \frac{k_2}{k_1 + 1} = a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1}$$
We can continue this inductively: suppose $$a_{k_n} > a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1} \cdot ... \cdot \frac{k_n}{k_{n-1} + 1}$$and let ${k_{n+1}}$ be the largest integer such that $\frac{a_{k_n+1}}{k_n+1} = \frac{a_{k_{n+1}}}{k_{n+1}}$. Then we must have $$a_{k_{n+1}} = a_{k_n+1} \cdot \frac{k_{n+1}}{k_n+1} > a_{k_n} \cdot \frac{k_{n+1}}{k_n+1} > a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1} \cdot ... \cdot \frac{k_{n+1}}{k_{n} + 1}$$as desired.

Now note that the last expression can be rewritten as $$a_1 \cdot \frac{k_1}{k_1 + 1} \cdot \frac{k_2}{k_2 + 1} \cdot ... \cdot \frac{k_n}{k_n + 1} \cdot k_{n+1}$$and the expression $\frac{n}{n+1} = 1 - \frac{1}{n+1}$ is minimized when $n$ itself is minimized, so we are left to minimize each of $k_1$ through $k_{n+1}$.

Notice that the minimum possible value of $k_1$ is $2$ otherwise the value of $\frac{a_1}{1}$ would be unique. This forces $k_2$ to be at least $4$ otherwise the value of $\frac{a_3}{3}$ would be unique as $\frac{a_1}{1} = \frac{a_2}{2} \neq \frac{a_3}{3}$. Then we see that the minimum possible value of $k_n$ is at least $2n$ as every term before has already been "paired up" and we need to "pair up" the next term as well or else it would be unique. This means $$a_{k_{n+1}} > a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1} \cdot ... \cdot \frac{k_{n+1}}{k_{n} + 1} \geq a_1 \cdot \frac{2}{1} \cdot \frac{4}{3} \cdot ... \cdot \frac{2n+2}{2n+1} = a_1 \bigg(1 + \frac{1}{1} \bigg)\bigg(1 + \frac{1}{3} \bigg)\bigg(1 + \frac{1}{5} \bigg)\cdots\bigg(1 + \frac{1}{2n+1}\bigg) > a_1 \cdot \bigg(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n+1}\bigg)$$
But since the original sequence is infinite and it can be derived that $\sum_{n=0}^{\infty }\frac{1}{2n+1}$ does not converge from the diverging summand $\sum_{n=1}^{\infty }\frac{1}{n}$, then the final expression in our inequality chain can be arbitrarily large for arbitrarily large $n$, contradicting the fact that it is less than $a_{k_{n+1}}$ and thus less than one, as desired $\blacksquare$
This post has been edited 14 times. Last edited by AforApple, Dec 3, 2021, 9:40 PM
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GeronimoStilton
1521 posts
#9 • 1 Y
Y by centslordm
Suppose otherwise. Then each $\frac{a_n}{n}$ is equal to some other $\frac{a_m}{m}$.

Let $i_1 = 1$. Define the sequence $i_1,i_2,\cdots$ by letting, for any $i_k$, $i_{k+1}$ be $1$ more than the maximal $i\ge i_k$ with $\frac{a_i}{i} \ge \frac{a_{i_k}}{i_k}$. Such maximal $i$ exists because $\frac{a_i}{i} \ge \frac{a_{i_k}}{i_k}$ implies $1>a_i \ge i \cdot \frac{a_{i_k}}{i_k}$. This way, we get the bound
\[a_{i_{k+1}} > \frac{i_{k+1}-1}{i_k} \cdot a_{i_k}.\]Now, write
\[a_{i_{k+1}} > a_1\prod_{j=1}^k \frac{i_{j+1}-1}{i_j} = a_1\prod_{j=1}^k \prod_{i_j < t < i_{j+1}} \frac{t}{t-1}.\]We now find a lower bound for this product. Suppose there was a particular minimum $j$ for which $i_{j+1} = 1+i_j$. Observe $j>1$, since $\frac{a_1}{1}$ is equal to another $\frac{a_n}{n}$ with $n>1$. The condition $i_{j+1} = 1+i_j$ implies that $\frac{a_{i_j}}{i_j}$ is greater than $\frac{a_i}{i}$ for all $i>i_j$, so there is some $i'<i_j$ with $\frac{a_{i_j}}{i_j} = \frac{a_{i'}}{i'}$. This $i'$ has $a_{i'}/i'$ not equal to any $a_{i_{j'}}/i_{j'}$ for $j'\ne j$. Thus for any $j\ge 1$,
\[i_{j+1}-i_j-1 \ge 1+\#\{j'>j: i_{j'+1} = i_{j'}+1\}.\]This means that the following iterated algorithm will eventually yield $i_{j+1}-i_j = 2$ for all $j<k$ and $i_{k+1}-i_k \ge 2$: If $j<k$ and $i_{j+1}-i_j-1 \ge 2$, replace $i_{j+1}$ with $i_{j+1}-1$, which has the effect of replacing $\frac{i_{j+1}-1}{i_{j+1}-2}$ with $\frac{i_{j+1}}{i_{j+1}-1}$ and thus reducing the overall product value. Thus the product is at least
\[\frac 21\cdot \frac 43\cdot \cdots \cdot \frac{2k}{2k-1} = \prod_{s=1}^k \left(1+\frac{1}{2s-1}\right) > \sum_{s=1}^k \frac{1}{2s-1}.\]Note that this sum is unbounded as $k\to \infty$. Thus for any $N>0$, there exists $k$ for which $1>a_{i_{k+1}} > Na_1$. This is absurd because we cannot have $\frac 1N > a_1$ for all $N>0$.
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dgrozev
2463 posts
#10 • 1 Y
Y by centslordm
Most probably, the same core idea as in the above posts, but I still would put down the following solution hoping it's a bit different.
Let $A:=\{a_n/n: n\in\mathbb N\}.$ Define $f:\mathbb{N}\to A$ as $f(n)=a_n/n$. We must prove the set $f^{-1}(a)$ consists of only one element for some $a\in A$ . For the sake of contradiction, assume $f^{-1}(a)$ is a set of at least two elements for every $a\in A$ Note that the family of sets $\{f^{-1}(a): a\in A\}$ is a disjoint cover of $\mathbb{N}$. Let $I_a$ be the closed interval with left and right ends respectively the minimal and maximal element of $f^{-1}(a), a\in A.$ It is straightforward to see that $f^{-1}(a)$ consists of only finitely many elements, so the definition is correct. Further, $\mathcal{I}:= \{I_a : a\in A\}$ is also a cover of $\mathbb{N}$, though it may not be disjoint. It's not difficult to be seen that the family $\mathcal{I}$ contains a subfamily (to avoid clumsy notations we denote it again by $\mathcal{I}$) which also is a cover of $\mathbb{N}$, and can be further split into two families $\mathcal{I}_1, \mathcal{I}_2,$ each one of disjoint sets. For any $I\in \mathcal{I}, I=[a,b]$ we define $g(I):= b/a$ (the fact that $b>a$ is essential for the following arguments). It follows
$$\prod_{I\in \mathcal{I}_i} g(I)$$is finite for $i=1,2$ because otherwise the sequence $\{a_i\}_{i\in\mathbb N}$ would be unbounded.It means
$$\prod_{I\in \mathcal{I}} g(I)<\infty\qquad(1)$$Let $I\in \mathcal{I}, I=[a,b].$
\begin{align*}
	g(I)=\frac{b}{a}&=1+\frac{b-a}{a}\\
	&\ge 1+\frac{1}{a}+\frac{1}{a+1}+\dots +\frac{1}{b-1}\\
	&\ge 1+\frac{1}{2}\left(\frac{1}{a}+\frac{1}{a+1}+\dots +\frac{1}{b-1}+\frac{1}{b} \right)\\
	&=1+\frac{1}{2}\sum_{n\in I\cap\mathbb{N}}\frac{1}{n}  
	\end{align*}Denote $\displaystyle g_1(I):=\frac{1}{2}\sum_{n\in I\cap\mathbb{N}}\frac{1}{n}.$ Further, we get
\begin{align*}
	\prod_{I\in \mathcal{I}} g(I) &= \prod_{I\in \mathcal{I}} \left(1+g_1(I)\right)\\
	&>\sum_{I\in\mathcal{I}} g_1(I)\\
	&\ge \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}=\infty
	\end{align*}which contradicts $(1)$. (the fact that $\mathcal{I}$ is a family of (closed) intervals that cover $\mathbb{N}$ was used in the last inequality)
This post has been edited 2 times. Last edited by dgrozev, Dec 19, 2021, 11:48 AM
Reason: see #12
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AwesomeYRY
579 posts
#11 • 2 Y
Y by centslordm, Doru2718
Let a season be a set of $a_i$ with the same value of $\frac{a_i}{i}$, and the finale be the last $a_i$ with this common value. AFTSOC that all seasons have size $\geq 2$.

Now, we generate the sequence $f_i$ where $f_0=1$, and $f_{k+1}$ is the finale of $f_k+1$. (This is fine if $f_{k+1} = f_k+1$). Additionally, $f_k$ is well-defined because an infinitely long season clearly gives us $a_i>1$. Then, we have that
\[a_{f_{n+1}} = \frac{f_{n+1}}{f_n+1} a_{f_n+1} >\frac{f_{n+1}}{f_n+1} a_{f_n}  \]Thus, we have that
\[a_{f_n} \geq \prod_{i=2}^{n} \frac{f_{n}}{f_{n-1}+1} a_{f_1}\]
$\textbf{Remark: }$ Now, note that if any point we have $f_{k+1}\geq f_k+4$, then we can weaken the bound by instead going from $f_k\to f_k+2\to f_{k+1}$ instead of $f_k\to f_{k+1}$. This is because $\frac{f_{k+1}}{f_{k}+1} \cdot \frac{f_k}{f_{k-1}+1}>\frac{f_{k+1}}{f_k+3}\cdot \frac{f_k+2}{f_k+1}\cdot \frac{f_k}{f_{k-1}+1}$. Thus, we may assume that for all $k$ we have $f_{k+1}\leq f_k+3$, and thus, $f_k\leq 1+3k$.

Back to the solution. The old bound becomes
\[a_{f_n} \geq \prod_{i=2}^{n} \frac{f_{n}}{f_{n-1}+1} a_{f_1}\]Note that we have $f_n\geq 2n$ because if $f_n\leq 2n-1$, then there are $n$ seasons that end in the first $2n-1$ terms, which is absurd because each season has at least two episodes. Thus, we now have
\[a_{f_n} \geq f_n \cdot \prod_{i=1}^{n-1} \frac{f_i}{f_i+1} a_1\geq 2n \cdot \prod_{i=1}^{n-1} \frac{2i}{2i+1} a_1\]Now, note that
\[(\prod_{i=1}^{n-1} \frac{2i+1}{2i})^2\leq \prod_{i=1}^{n-1} \frac{2i+1}{2i}\cdot \prod_{i=1}^{n-1} \frac{2i}{2i-1} = 2n-1\]Thus,
\[a_{f_n} \geq 2n\cdot \frac{1}{\sqrt{2n-1}} a_1\]so no matter how small $a_1$, we can still find some $n$ such that $\frac{2n}{\sqrt{2n-1}}> \frac{1}{a_1}$ so $a_{f_n} > 1$, a contradiction. Thus, the original hypothesis was incorrect, and there exists a season with size $<2$ and we're done. $\blacksquare$.
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shalomrav
330 posts
#12 • 1 Y
Y by dgrozev
dgrozev wrote:
. It's not difficult to be seen the family $\mathcal{I}$ can be split into two subfamilies $\mathcal{I}_1, \mathcal{I}_2$, each one of disjoint sets.

Why that is true?
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dgrozev
2463 posts
#13 • 1 Y
Y by centslordm
@shalomrav: Sorry, my bad, it's not written as it should be.
I meant that the family $\mathcal{I}$ contains a subfamily (to avoid clumsy notations we denote it again by $\mathcal{I}$) which is also a cover of $\mathbb{N}$, and can be split into two families, each one of disjoint sets.
We start with an interval $I_1\in \mathcal{I}$ which left end is $1$, that is $I_1=[1,i_1].$ Further, we choose $I_2$ to be the interval with the greatest right end which covers $i_1+1$. There is only one like that, since no two intervals in $\mathcal{I}$ can share a common end (the family of sets $\{f^{-1}(a): a\in A\}$ is a disjoint cover of $\mathbb{N}$). Let the right end of $I_2$ be $i_2$. We proceed by choosing $I_3$ be the interval with the greatest right end that covers $i_2+1$ and so on. Obviously $I_j,j=1,2,\dots$ is a cover of $\mathbb{N}$ and each of the families $\mathcal{I}_1:=\{I_j : j \text{ is odd}\} $ and $\mathcal{I}_2 :=\{I_j : j\text{ is even}\}$ consists of disjoint intervals.
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Justpassingby
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#14
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I really like this problem. This feels like the converse of problems which are mainly combinatorics but the main step is a casual AM-GM or some algebraic manipulation, which just feel like some sort of toxic relationship lol.
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SHZhang
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#15 • 1 Y
Y by centslordm
Let $b_i = \frac{a_i}{i}$. Suppose for the sake of contradiction that every value in $b$ appears at least twice. Call an index $i$ terminal if there is no $j > i$ such that $b_j = b_i$, and nonterminal otherwise.

Lemma 1: Let $s_1 < s_2 < \dots < s_k$ be nonterminal indices. Then there exists $m$ such that \[a_m \ge a_1 \prod_{i=1}^k \frac{s_i + 1}{s_i}.\]Proof: We perform the following algorithm that generates a set $S$ of disjoint intervals.

There is a counter $c$ initially set to 1. Do the following:
1. Advance $c$ to the next nonterminal index in $s$, possibly doing nothing if $c$ is already on such an index.
2. Now there exists $d > c$ such that $b_d = b_c$ since $c$ is a nonterminal index. Insert $[c, d-1]$ into $S$.
3. Advance $c$ to $d$.
4. If $c > s_k$, then stop. Otherwise return to step 1.

Let $S = \{[l_1, r_1], [l_2, r_2], \dots, [l_n, r_n]\}$ be the intervals of $S$ listed in increasing order. Then for all $i$, $b_{l_i} = b_{r_i + 1}$ by looking at step 2 of the algorithm. Therefore $a_{r_i + 1} = \frac{r_i + 1}{l_i} a_{l_i}$. Also, $a_{r_i + 1} \le a_{l_{i+1}}$ since $a$ is increasing and $r_i + 1 \le l_{i+1}$. Chaining these relations, between $a_1 \le a_{l_1}, a_{r_1 + 1}, a_{l_2}, a_{r_2 + 1}, \dots, a_{l_n}, a_{r_n + 1}$, together gives that \[a_{r_n + 1} \ge a_1 \prod_{i=1}^n \frac{r_i + 1}{l_i}.\]Let $T$ be the set of integers that lie in the union of the intervals in $S$. Now this relation can be rewritten as \[a_{r_n + 1} \ge a_1 \prod_{x \in T} \frac{x + 1}{x}.\]But notice that $T \subseteq \{s_1, s_2, \dots, s_k\}$ since the only time indices can be skipped in the algorithm is in step 1, which cannot skip over anything in $s$. Therefore we have \[a_{r_n + 1} \ge a_1 \prod_{x \in T} \frac{x + 1}{x} \ge a_1 \prod_{i=1}^k \frac{s_i + 1}{s_i},\]as desired. $\square$

Lemma 2: For any positive integer $n$, at least half of $\{1, 2, \dots, n\}$ are nonterminal.
Proof: Since we assume each value in $b$ appears at least twice, there is a map sending any terminal index $x \le n$ to the greatest nonterminal index $y < x$ such that $b_y = b_x$. This map is injective because $b_x$ is pairwise distinct as $x$ varies over terminal indices. Therefore there are at least as many nonterminal indices at most $n$ as terminal indices, as desired. $\square$

To finish, consider the following blocks of indices: \[[1, 1], [2, 4], [5, 16], \dots, [4^n + 1, 4^{n+1}], \dots.\]For any block $[4^n + 1, 4^{n+1}]$, by Lemma 2 on the prefix up to $4^{n+1}$, at least $4^n$ indices in the block are nonterminal (there are at least $2 \cdot 4^n$ in the prefix, subtracting at most $4^n$ that should be excluded because they are before $4^n + 1$). Furthermore, in each block, the product of $\frac{x+1}{x}$ with $x$ ranging over all nonterminal indices in the block is at least \[\frac{4^{n+1} - 4^n + 2}{4^{n+1} - 4^n + 1} \frac{4^{n+1} - 4^n + 3}{4^{n+1} - 4^n + 2} \frac{4^{n+1} - 4^n + 4}{4^{n+1} - 4^n + 3} \dots \frac{4^{n+1} + 1}{4^{n+1}} = \frac{4^{n+1} + 1}{4^{n+1} - 4^n + 1},\]which is approximately $\frac{4}{3}$ for large $n$. Putting the nonterminal indices of sufficiently many blocks together and applying Lemma 1 lets us force $a_m$ to be arbitrarily large, contradicting the fact that $a_i < 1$ for all $i$.
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IAmTheHazard
5001 posts
#16 • 1 Y
Y by centslordm
Does this work? Strange feeling solution that looks at the beginning of chains (more or less) instead of the ending


Suppose otherwise. Separate the positive integers into families based on the value of $\frac{a_i}{i}$. Call an integer $i$ sweet if $\frac{a_i}{i}<\frac{a_1}{1},\ldots,\frac{a_{i-1}}{i-1}$. Obviously each sweet integer is the smallest element of a family.

Claim: There are infinitely many sweet integers (therefore families).
Proof: $\frac{i}{a_i}$ is unbounded above. $\blacksquare$

Claim: Let $C=\tfrac{1}{a_1}>1$. For each family $\mathcal{F}$, we have $C\min_{\mathcal{F}} i \geq \max_{\mathcal{F}} i$.
Proof: Otherwise, if $x=\min_{\mathcal{F}} i$ and $y=\max_{\mathcal{F}} i$, then
$$\frac{a_x}{x}=\frac{a_y}{y} \implies a_y=\frac{ya_x}{x}>\frac{Cxa_x}{x}>Ca_1=1,$$contradiction. $\blacksquare$

Now let $1=i_1<i_2<\cdots$ be the sweet integers. We prove a key estimate.

Claim: We have $a_{i_j} \leq \prod_{k=j}^\infty \frac{i_k}{i_{k+1}-1}$.
Proof: We will prove by induction on $a \geq 0$ that $a_{i_j} \leq \prod_{k=j}^{j+a-1} \frac{i_k}{i_{k+1}-1}$ for all $j$, with the base case being trivial (empty product equals $1$). For the inductive step, if $a_{i_{j+1}} \leq \prod_{k=j+1}^{j+a}\frac{i_k}{i_{k+1}-1}$, then
$$\prod_{k=j+1}^{j+a} \frac{i_k}{i_{k+1}-1}\geq a_{i_{j+1}}>a_{i_{j+1}-1}\geq \frac{i_{j+1}-1}{i_j}a_{i_j},$$with the last inequality from the fact that $i_j$ is sweet and no sweet integers exist in $(i_j,i_{j+1})$. Rearranging implies the desired conclusion. $\blacksquare$

We will also need a crucial "independent" lemma.

Lemma: Let $S \subseteq \mathbb{Z}^+$ have positive upper density. Then $\prod_{x \in S} \frac{x+1}{x}$ diverges.
Proof: Suppose it converged. Then,
$$\prod_{x \in S} \left(1+\frac{1}{x}\right)>1+\sum_{x \in S} \frac{1}{x},$$but the sum is well-known to diverge for any $S$ with nonzero density. $\blacksquare$

This sets up the main claim of the problem.

Claim: The set of sweet $i$ has density $1$ (equivalently, the set of non-sweet $i$ has density $0$).
Proof: By our previous claim, we should have
$$a_1 \leq \prod_{j=1}^\infty \frac{i_j}{i_{j+1}-1} \leq \prod_{j=1}^k \frac{i_j}{i_{j+1}-1}$$for all $k$. Since we are assuming convergence issues don't exist (otherwise we would get issues already), we can perform algebraic manipulations, writing
$$a_1 \leq \frac{i_1}{i_{k+1}-1}\prod_{j=2}^k \frac{i_j}{i_j-1} \leq \frac{1}{i_k}\prod_{j=2}^k \frac{i_j}{i_j-1}.$$On the other hand, we have
$$\frac{1}{i_k}\prod_{j=i_2}^{i_k} \frac{j}{j-1}=\frac{1}{i_2-1},$$which upon division means that $\prod_{\substack{i_2 \leq j \leq i_k\\j\text{ non-sweet}}} \frac{j}{j-1}$ is bounded above by $\frac{1}{a_1(i_2-1)}$, i.e. $\prod_{j\text{ non-sweet}} \frac{j}{j-1}$ converges, by sending $k \to \infty$ (the reason for proving the first claim is to ensure we can do this). Using the lemma on $S=\{a-1 \mid a\text{ non-sweet}\}$ implies the desired result. $\blacksquare$

Now, this result implies that for any (fixed) $0<r<1$, there exists some constant $N_r$ such that for all $N>N_r$, at least $rN$ of the integers in $[1,N]$ are sweet, else the lower density of sweet integers is at most $r$. But by the problem condition, there is an injection sending each sweet integer $i$ to a non-sweet integer (in the same family) which is at most $Ci$ according to our second claim, hence at least $rN$ of the integers in $[1,CN]$ are non-sweet, so we should have $\frac{C-r}{C}\geq r$, which is false as $r \to 1^-$, implying the desired contradiction. $\blacksquare$
This post has been edited 7 times. Last edited by IAmTheHazard, Aug 29, 2023, 9:48 PM
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YaoAOPS
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Redacted
This post has been edited 2 times. Last edited by YaoAOPS, Apr 29, 2025, 12:51 AM
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mathhotspot
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REDACTED
This post has been edited 1 time. Last edited by mathhotspot, Jul 27, 2024, 3:43 PM
Reason: wrong answer
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john0512
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#19
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We will show that if there is no isolated value, then the sequence is unbounded.

We say that an index $i$ is a portal if there exists an index $j>i$ for which $\frac{a_i}{i}=\frac{a_j}{j}$.

Claim: Portals have positive density. Partition the indices into "groups" based on the value of $\frac{a_i}{i}$. Each group consists of at least two indices, and in each group, all indices except for the last (if finite) are portals, which shows the claim (in fact, this argument shows $\rho \geq \frac12$).

Now, we will employ the following algorithm to generate unbounded $a_i$:

Start with $i=1$. Then, for each step in the algorithm,

-if $i$ is not a portal, increase $i$ by $1$, which at least does not decrease $a_i$.

-if $i$ is a portal, jump to the smallest $j>i$ such that $\frac{a_i}{i}=\frac{a_j}{j}$. This multiplies $a_i$ by $\frac{j}{i}$.

Note that in order to show that the product of such $\frac{j}{i}$ diverges, it suffices to show that the numbers traversed via portals (as opposed to traversed through incrementing) have positive density, since the $a_i$ multiplies by $\frac{j}{i}$ when we use a portal.

Due to the "greedy" nature of our algorithm, each portal is either used or skipped when travelling through another portal (they cannot be "stepped on but unused"). Thus, since all of the portals have positive density, this means that either used portals or skipped portals have positive density.

In the former case, we are done, since each time we use a portal we traverse at least one index. In the second case, we are also done as skipped portals is a subset of numbers traversed via portal.
This post has been edited 5 times. Last edited by john0512, May 10, 2024, 7:13 PM
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GrantStar
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#20
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For an integer $n$, let $f(n)$ be the largest integer $m$ for which $a_n/n = a_m/m$, which clearly exists by boundedness. An index $n$ is called a ``stopper" if $f(n)=n$, and note that for any $m \in \mathbb Z$, the index $f(m)$ is a stopper.

Define positive integers $n_1, n_2, \dots$ and $k_1, k_2, \dots$ such that
  • $n_1=f(1)$.
  • For each $i\geq 1$, $k_i$ is the smallest positive integer such that $n_i+k_i$ is not a stopper.
  • For each $i \geq i$, we have $n_{i+1} = f(n_i+k_i)$.
By construction, note that $n_i, n_i+1, \dots, n_i+k_i-1$ are all stoppers. There for each $i$, there are at least $k_i$ stoppers from $n_{i-1}+k_{i-1}$ to $n_i+k_i$, so there are at least $k_1 + k_2 + \dots + k_i$ stoppers at most $n_i+k_i$ for each $i$. However, assuming the problem to be false, for each stopper $l$ there exists $m$ with $f(m)=l$. Thus as each ``bunch" in the arrows graph of $f$ has size at least $2$ and ends at a stopper, we get that $2(k_1 + \dots + k_i) \leq n_i+k_i$.

Now, notice that by definition and increasing-ness of the sequence, we have that $\frac{a_{n_i+k_i}}{n_i+k_i} = \frac{a_{n_{i+1}}}{n_{i+1}}$ and $a_{n_i+k_i}>a_{n_i}$. Repeatedly applying this gives that \[a_{n_m} > a_1 n_1 \prod_{i=1}^{m-1} \left(\frac{n_{i+1}}{n_i+k_i}  \right).\]We turn this into a purely algebraical problem now.

Claim: Let $k_1, k_2, \dots$ and $n_1, n_2, \dots$ be positive integers such that $2(k_1 + \dots + k_i) \leq n_i+k_i$. Then, the product \[ \prod_{i=1}^{m-1} \left(\frac{n_{i+1}}{n_i+k_i}  \right)\]goes to infinity as $m$ goes to infinity.
Proof. Note that if we decrease $n_i$ we also decrease the product since $\frac{n_i}{n_i+k_i}$ decreases. Thus, we can say \[ \prod_{i=1}^{m-1} \left(\frac{n_{i+1}}{n_i+k_i}  \right) \geq \prod_{i=1}^{m-1} \left(\frac{k_{i+1} + 2\sum_{j=1}^i k_j}{2\sum_{j=1}^i k_j} \right).\]We again manipulate this product as \[\prod_{i=1}^{m-1} \left(\frac{k_{i+1} + 2\sum_{j=1}^i k_j}{2\sum_{j=1}^i k_j} \right) = \prod_{i=1}^{m-1} \left(1+ \frac{k_{i+1}}{2\sum_{j=1}^i k_j} \right) > \sum_{i=1}^{m-1} \frac{k_{i+1}}{2\sum_{j=1}^i k_j}.\]It thus suffices to show that $\sum_{i=1}^{\infty} \frac{k_{i+1}}{\sum_{j=1}^i k_j}$ diverges.

Suppose that for some $i>1$ we have $k_i > k_{i+1}$; we claim that swapping these two indices decreases the sum. Let $a = \sum_{j=1}^{i-1}k_j$ and $b=k_1, c=k_{i+1}$. It suffices to show that $\frac{b}{a}+\frac{c}{a+b} > \frac{c}{a}+\frac{b}{a+c}$ or equivalently that \[\frac{b-c}{a} > \frac{(a+b+c)(b-c)}{(a+b)(a+c)} \iff (a+b)(a+c) > a(a+b+c)\]which is true.

Thus (ignoring details about infinite sums) we may swap numbers and decrease the sum into at least \[\sum_{i=1}^{\infty} \frac{k_{i+1}}{\sum_{j=1}^i k_j} \geq \sum_{i=1}^{\infty} \frac{k_{i+1}}{k_1+(i-1)k_{i+1}}\]which diverges by comparison to the harmonic series. $\blacksquare$

The above claim then implies that $a_{n_i}$ is unbounded, a contradiction.
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ihatemath123
3448 posts
#21
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Solved with help from GrantStar16.

Call indices $i$ and $j$ friends if $\tfrac{a_i}{i} = \tfrac{a_j}{j}$. Call an index $i$ overshadowed if it has a friend greater than itself. Let the friend interval of two friends $i$ and $j$ denote the integers in the range $[i,j]$.

We will prove the contrapositive: if every index has a friend, the sequence $a_i$ is unbounded. Construct an infinite set $S$ of pairs of friends as follows:
  1. First, add the friend-pair of $1$ and any of its friends.
  2. Let $I$ denote the union of all friends interval in $S$. Let $x$ be the smallest overshadowed integer not in $I$, and let $y$ be a friend of $x$ with $y > x$. Add the friend pair $(x,y)$ to $S$.
  3. Repeat step 2 infinitely many times.
It is clear that all friend intervals in $S$ are pairwise disjoint. Also, every overshadowed integer belongs in $S$.

Claim: For each integer $n$, at least half of the integers from $1$ to $n$ lie in a friend interval of $S$.
Proof: Note that every overshadowed integer is in an interval of $S$, due the nature of our process. Every "friend group" (maximal set of integers that are pairwise friends) has exactly one non-overshadowed friend, so there is at most one non-overshadowed friend of this friend group less than $n$. So, in each friend group, there are at least as many overshadowed as non-overshadowed indices less than $n$. This proves the claim.

Claim: The sequence $a_i$ is unbounded.
Proof: It suffices to show that if $S = \{ (x_i, y_i) \}_{i=1}^\infty$ with $y_i > x_i$, then the product $\tfrac{y_1}{x_1} \cdot \tfrac{y_2}{x_2} \cdots$ diverges. Suppose FTSOC that it converges. Let $T$ denote the set of integers that lie in the interval $(x_i, y_i]$ for some $i$. Then,
\[\prod_{i=1}^\infty \frac{y_i}{x_i} = \prod_{i \in T} \frac{i}{i-1}.\]Because of our previous claim, for every integer $n$, at least a quarter of the numbers from $1$ to $n$ are in $T$ (not half, because $T$ does not contain anything of the form $x_i$). So, for any integer $k$, at least $\tfrac{1}{16}$ of the integers in $[5^k + 1,5^{k+1}]$ are in $T$. Clearly,
\[\prod_{i \in T_k} \frac{i}{i-1} \geq \frac{5^{k+1}}{4.75 \cdot 5^k} > 1.05.\]In particular, $1.05$ doesn't depend on $k$. Now we get a contradiction because
\[\prod_{i \in T} \frac{i}{i-1} = \left( \prod_{i \in T_0} \frac{i}{i-1} \right) \left( \prod_{i \in T_1} \frac{i}{i-1} \right) \left( \prod_{i \in T_2} \frac{i}{i-1} \right) \cdots > (1.05)(1.05)(1.05) \cdots,\]contradiction.
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