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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.

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[*]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
Rational coefficients polynomial
Cats_on_a_computer   0
8 minutes ago
Given a quartic monic polynomial with rational coefficients, show that if the polynomial has exactly 1 real root r, r must be rational.
I solved this somewhat differently (using the division algorithm), but it really seems like Vieta should work here. I haven’t been able to find another workable solution however.
0 replies
Cats_on_a_computer
8 minutes ago
0 replies
JBMO TST Bosnia and Herzegovina 2024 P2
FishkoBiH   1
N 15 minutes ago by Rotten_
Source: JBMO TST Bosnia and Herzegovina 2024 P2
Determine all $x$, $y$, $k$ and $n$ positive integers such that:

$10^x$ + $10^y$ + $n!$ = $2024^k$

1 reply
FishkoBiH
an hour ago
Rotten_
15 minutes ago
JBMO TST Bosnia and Herzegovina 2024 P1
FishkoBiH   2
N 18 minutes ago by grupyorum
Source: JBMO TST Bosnia and Herzegovina 2024 P1
Let $a$,$b$,$c$ be real numbers different from 0 for which $ab$ + $bc$+ $ca$ = 0 holds
a) Prove that ($a$+$b$)($b$+$c$)($c$+$a$)≠ 0
b) Let $X$ = $a$ + $b$ + $c$ and $Y$ = $\frac{1}{a+b}$ + $\frac{1}{b+c}$ + $\frac{1}{c+a}$. Prove that numbers $X$ and $Y$ are both positive or both negative.
2 replies
FishkoBiH
an hour ago
grupyorum
18 minutes ago
1997 Problem
Chinaboy   74
N 23 minutes ago by Schintalpati
Source: 1997 USAMO #2
Let $ABC$ be a triangle. Take points $D$, $E$, $F$ on the perpendicular bisectors of $BC$, $CA$, $AB$ respectively. Show that the lines through $A$, $B$, $C$ perpendicular to $EF$, $FD$, $DE$ respectively are concurrent.
74 replies
Chinaboy
Apr 17, 2005
Schintalpati
23 minutes ago
A second final attempt to make a combinatorics problem
JARP091   0
24 minutes ago
Source: At the time of writing this problem I do not know the source if any
Arthur Morgan is playing a game.

He has $n$ eggs, each with a hardness value $k_1, k_2, \dots, k_n$, where $\{k_1, k_2, \dots, k_n\}$ is a permutation of the set $\{1, 2, \dots, n\}$. He is throwing the eggs from an $m$-floor building.

When the $i$-th egg is dropped from the $j$-th floor, its new hardness becomes
\[
\left\lfloor \frac{k_i}{j+1} \right\rfloor.
\]If $\left\lfloor \frac{k_i}{j+1} \right\rfloor = 0$, then the egg breaks and cannot be used again.

Arthur can drop each egg from a particular floor at most once.
For which values of $n$ and $m$ can Arthur always determine the correct ordering of the eggs according to their initial hardness values?
Note: The problem might be wrong or too easy
0 replies
JARP091
24 minutes ago
0 replies
Hunter rabbit makes a comeback
everythingpi3141592   16
N 25 minutes ago by Atmadeep
Source: India IMOTC 2024 Day 1 Problem 1
A sleeping rabbit lies in the interior of a convex $2024$-gon. A hunter picks three vertices of the polygon and he lays a trap which covers the interior and the boundary of the triangular region determined by them. Determine the minimum number of times he needs to do this to guarantee that the rabbit will be trapped.

Proposed by Anant Mudgal and Rohan Goyal
16 replies
everythingpi3141592
May 31, 2024
Atmadeep
25 minutes ago
JBMO TST Bosnia and Herzegovina 2024 P4
FishkoBiH   0
33 minutes ago
Source: JBMO TST Bosnia and Herzegovina 2024 P4
Let $m$ and $n$ be natural numbers. Every one of the $m*n$ squares of the $m*n$ board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this:
- a white square becomes black;
-a black square becomes blue;
-a blue square becomes white.
For which $m$ and $n$ can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
0 replies
FishkoBiH
33 minutes ago
0 replies
JBMO TST Bosnia and Herzegovina 2023 P1
FishkoBiH   1
N 41 minutes ago by clarkculus
Source: JBMO TST Bosnia and Herzegovina 2023 P1
Determine all real numbers $a, b, c, d$ for which

$ab+cd=6$
$ac+bd=3$
$ad+bc=2$
$a+b+c+d=6$
1 reply
FishkoBiH
2 hours ago
clarkculus
41 minutes ago
JBMO TST Bosnia and Herzegovina 2024 P3
FishkoBiH   0
42 minutes ago
Source: JBMO TST Bosnia and Herzegovina 2024 P3
Let $ABC$ be a right-angled triangle where $ACB$=90°.Let $CD$ be an altitude of that triangle and points $M$ and $N$ be the midpoints of $CD$ and $BC$, respectively.If $S$ is the circumcenter of the triangle $AMN$, prove that $AS$ and $BC$ are paralel.
0 replies
FishkoBiH
42 minutes ago
0 replies
Inspired by 2025 Beijing
sqing   7
N an hour ago by ytChen
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
7 replies
sqing
Yesterday at 4:56 PM
ytChen
an hour ago
Inequality em981
oldbeginner   18
N an hour ago by xzlbq
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
18 replies
+1 w
oldbeginner
Sep 22, 2016
xzlbq
an hour ago
JBMO TST Bosnia and Herzegovina 2023 P2
FishkoBiH   1
N an hour ago by clarkculus
Source: JBMO TST Bosnia and Herzegovina 2023 P2
Determine all non negative integers $x$ and $y$ such that $6^x$ + $2^y$ + 2 is a perfect square.
1 reply
FishkoBiH
2 hours ago
clarkculus
an hour ago
Divisiblity...
TUAN2k8   1
N an hour ago by Natrium
Source: Own
Let $m$ and $n$ be two positive integer numbers such that $m \le n$.Prove that $\binom{n}{m}$ divides $lcm(1,2,...,n)$
1 reply
TUAN2k8
Today at 6:13 AM
Natrium
an hour ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   0
an hour ago
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
0 replies
FishkoBiH
an hour ago
0 replies
Nonnegative integer sequence containing floor(k/2^m)?
polishedhardwoodtable   7
N Apr 19, 2025 by Maximilian113
Source: ELMO 2024/4
Let $n$ be a positive integer. Find the number of sequences $a_0,a_1,a_2,\dots,a_{2n}$ of integers in the range $[0,n]$ such that for all integers $0\leq k\leq n$ and all nonnegative integers $m$, there exists an integer $k\leq i\leq 2k$ such that $\lfloor k/2^m\rfloor=a_i.$

Andrew Carratu
7 replies
polishedhardwoodtable
Jun 21, 2024
Maximilian113
Apr 19, 2025
Nonnegative integer sequence containing floor(k/2^m)?
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G H BBookmark kLocked kLocked NReply
Source: ELMO 2024/4
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polishedhardwoodtable
130 posts
#1 • 3 Y
Y by ehuseyinyigit, OronSH, ihatemath123
Let $n$ be a positive integer. Find the number of sequences $a_0,a_1,a_2,\dots,a_{2n}$ of integers in the range $[0,n]$ such that for all integers $0\leq k\leq n$ and all nonnegative integers $m$, there exists an integer $k\leq i\leq 2k$ such that $\lfloor k/2^m\rfloor=a_i.$

Andrew Carratu
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YaoAOPS
1541 posts
#2 • 3 Y
Y by VicKmath7, shafikbara48593762, MS_asdfgzxcvb
We claim there are $2^n$ such sequences.
Define the $k$-interval as $\{a_k, a_{k+1}, \dots, 2k\}$. Then the condition requires that $k, \left\lfloor \frac{k}{2} \right\rfloor, \dots$ are in the $k$-interval. Call these the lamps for $k$.

Claim: We induct on the following claims:
  1. Each $k$-interval consists of all integers $\{0, 1, \dots, k\}$.
  2. $\{a_{2k-1}, a_{2k}\} = \{a_{k-1}, k\}$
  3. $a_{k-1}$ is a lamp for $k$.
Proof. We prove this inductively. The base case of $k = 0, 1$ works.
Now, we show that $a_{k-1}$ is a lamp for $a_k$. Suppose that $k$ is even. Then we get that $\{a_{k-1}, a_{k}\} = \{a_{\frac{k}{2}-1}, \frac{k}{2}\}$, and note that $a_{\frac{k}{2}-1}$ is a lamp for $\frac{k}{2}$ which is a lamp for $\frac{k}{2}$ which implies the result.
Likewise, if $k$ is odd, we consider $\{a_{k-1}, a_{k-2}\} = \{\frac{k-1}{2}, a_{\frac{k-1}{2}-1}\}$ which are both lamps for $k$. As such, since the $a_{k-1}, \dots, a_{2k-2}$ contains $a_k$ exactly once, we get that $a_k, \dots, a_{2k-2}$ doesn't contain $a_k$. It also can't contain $k$. Since $a_k$ and $k$ are lamps for $k$, $a_k, \dots, a_{2k}$ must contain them, which implies that $a_{2k-1, a_{2k}} = \{a_{k-1}, k\}$. Then we get that $\{a_k, \dots, a_{2k}\} = \{a_{k-1}, \dots, a_{2k-2}, k\} = \{0, \dots, k\}$ for the third claim. $\blacksquare$
Notably, we also have that if $\{a_{2k-1}, a_{2k}\} = \{a_{k-1}, k\}$ and $a_0, \dots, a_{2k-2}$ is a valid sequence, then $a_0, \dots, a_{2k}$ is a valid sequence.
This constraint gives us $2^n$ total options for building up $a_0, \dots, a_{2k}$ by first choosing $\{a_1, a_2\}$ and so forth.
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Usernumbersomething
62 posts
#3
Y by
See https://artofproblemsolving.com/community/c6h44479p281572 for something similar. I feel like this problem is like a more accessible version of the 2005 IMO problem.
This post has been edited 3 times. Last edited by Usernumbersomething, Jun 23, 2024, 4:03 PM
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GrantStar
821 posts
#4 • 1 Y
Y by Rajukian
Answer: $2^n$. The key claim is the following:

Claim: $a_k, \dots, a_{2k}$ is a permutation of $(0,1, \dots, k)$ for all $k\leq n$.
Proof. We strong induct on $n$. The base cases of $k=1$ and $k=2$ can be checked.
Now, we know that $a_{k-1}, \dots, a_{2k-2}$ is a permutation of $(0,\dots, k-1)$. Thus $a_k, \dots, a_{2k-2}$ are not $k=\lfloor k/2^0\rfloor$, so $a_{2k-1}$ or $a_{2k}$ is $k$. It suffices to show that $a_{k-1}$ appears in $a_k, \dots, a_{2k}$ or $a_{k-1}=a_{2k-1},a_{2k}$. By our inductive hypothesis and the reasoning above, $\{a_{2l-1}, a_{2l}\} = \{a_{l-1},l\}$ for $l<k$.

Let $j=\lfloor k/2\rfloor$. Thus $\{a_{2j-1},a_{2j}\}=\{a_{j-1},j\}$. Also, $k-1=2j-1$ or $2j$ by the floor definition of $k$. Thus $a_{k-1}=a_{j-1}$ or $k$. If $a_{k-1}=j$, then since $\lfloor k/2\rfloor = a_j$ must appear in $a_k, \dots, a_{2k}$ we conclude. Else, $a_{k-1}=a_{j-1}$. We can repeat the same argument to find $a_{j-1}=\lfloor j/2\rfloor$ or $a_{\lfloor j/2\rfloor -1}$. Repeat this inductively to get $a_{k-1}=\lfloor k/2^m\rfloor$ for some $m$. Thus $a_{k-1}$ is in $a_k, \dots, a_{2k}$ and the claim is proven. $\blacksquare$

To finish, repeatedly applying the claim gives $a_0=0$ and $i \in \{a_{2i-1}, a_{2i}\}$. Now, I claim the answer is $2^n$. This is from choosing which of $a_{2i-1}, a_{2i}$ is $i$. It suffices to show each choice gives a unique sequence. This is true by applying the claim on $1$ through $n$ to get the other number in each pair.
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v_Enhance
6877 posts
#5 • 2 Y
Y by Amkan2022, Rajukian
Solution from Twitch Solves ISL:

The answer is $2^n$. We note $a_0=0$ and ignore it going forward, focusing only on $a_i$ for $i \ge 1$.
In what follows, for each positive integer $t$ we let \[ D(t) \coloneqq \left\{ t, \left\lfloor t/2 \right\rfloor, \left\lfloor t/4 \right\rfloor, \dots \right\}. \]For example, $D(13) = \{13, 6, 3, 1, 0\}$. Then the problem condition is equivalent to saying that every element of $D(k)$ appears in $\{a_k, \dots, a_{2k}\}$.
We prove the following structure claim about all the valid sequences.
Claim: In any valid sequence, for each $0 \le k \le n$,
  • $a_{2k-1}$ and $a_{2k}$ are elements of $D(k)$; and
  • $a_k$, \dots, $a_{2k}$ consist of all the numbers from $0$ to $k$ each exactly once.
Proof. We proceed by induction; suppose we know it's true for $k$ and want it true for $k+1$. By induction hypothesis:
  • $\{a_k, \dots, a_{2k}\}$ contains each of $0$ to $k$ exactly once;
  • $a_k$ is an element of $D(\left\lceil k/2 \right\rceil)$;
  • We also know $\{a_{k+1}, \dots, a_{2k+2}\}$ contains all elements of $D(k+1)$ by problem condition.
However, note that \[ D\left( \left\lceil k/2 \right\rceil \right) \subseteq D(k+1) \]so that means either \[ (a_{2k+1} = a_k \text{ and } a_{2k+2} = k+1) \quad\text{OR}\quad (a_{2k+1} = k+1 \text{ and } a_{2k+2} = a_k). \]$\blacksquare$
We return to the problem of counting the sequences. It suffices to show that if $(a_0, \dots, a_{2n})$ is a valid sequence, there are exactly two choices of ordered pairs $(x,y) \in \{0, \dots, n+1\}$ such that $(a_0, \dots, a_{2n}, x, y)$ is a valid sequence. However, the structure claim above implies that $\{x,y\} = \{a_n, n+1\}$, so there are at most two choices. Moreover, both of them work by the structure claim again (because $k=n=1$ is the only new assertion when augmenting the sequence, and it holds also by the structure claim). This completes the proof.

Remark: Here are some examples to follow along with. When $n=4$ the $16$ possible values of $(a_4, a_5, a_6, a_7, a_8)$ are \[ \begin{array}{cc} (2,1,3,4,0) & (2,0,3,4,1) \\ (2,1,3,0,4) & (2,0,3,1,4) \\ (0,1,3,4,2) & (1,0,3,4,2) \\ (0,1,3,4,2) & (1,0,3,4,2) \\ (0,1,3,2,4) & (1,0,3,2,4) \\ (2,3,1,4,0) & (2,3,0,4,1) \\ (2,3,1,0,4) & (2,3,0,1,4) \\ (0,3,1,4,2) & (1,3,0,4,2) \\ (0,3,1,2,4) & (1,3,0,2,4) \\ \end{array} \]Now the point is that when moving to $n=5$, the element $a_4 \in \{0,1,2\} = D(2) \subseteq D(5)$ is chopped-off, and $a_9$ and $a_{10}$ must be $5$ and the chopped-off element in some order. So each of these sequences extends in exactly two ways, as claimed.
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CANBANKAN
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The answer is $2^n$.

Clearly $a_0=0$, and $\{a_1,a_2\} = \{0,1\}$.

The key structural claim is that $\{a_{2k-1},a_{2k}\} = \{a_{k-1},k\}$, call $P(k)$. It is clearly true for $k=1$.

Note that one can show $P(1),P(2),\cdots,P(k)$ imply $(a_k,\cdots,a_{2k})$ is a permutation of $\{0,\cdots,k\}$ via induction. Furthermore, $P(1),\cdots,P(k-1)$ and $\{a_k,\cdots,a_{2k}\}$ being permutation of $\{0,\cdots,k\}$ imply that $\{a_{2k-1},a_{2k}\} = \{a_{k-1},k\}$, because I remove $a_{k-1}$, add $a_{2k},a_{2k-1}$, and end up just adding $k$ to the set.

I will use $P(1),\cdots,P(k-1)$ to show $\{a_k,\cdots,a_{2k}\}\supset\{0,\cdots,k\}$, which suffices. Let $x \in \{0,\cdots,k\}$. If $x=k$ we are done, since the problem condition tells us that $$\{a_k,\cdots,a_{2k}\} \supset \left\{x,\lfloor \frac x2\rfloor, \cdots, \lfloor \frac{x}{2^m}\rfloor, \cdots, 1,0\right\}. (*)$$
Henceforth assume $x<k$. Then by $P(x)$, we have $x \subset \{a_{2x+1},a_{2x+2}\}$.

By $P(2x+1),P(2x+2)$, if $2x+2 < k$ we have $\{a_{2x+1},a_{2x+2} \} \subset \{a_{4x+3},\cdots,a_{4x+6}\}$.

(If $2x+1\ge k$ then we also have $\{a_{2x+1},a_{2x+2}\}\subset \{a_k,\cdots,a_{2k}\}$ since $x<k$. If $2x+2=k$ then we have $\{a_{2x+1},a_{2x+2} \} = \{a_k, a_{2k-1},a_{2k}\}$ by $P(k-1)$)

We iterate the inductive step. Set a counter $e=2$. At each $e$ we have a set $\{a_{2^ex + (2^e-1)}, \cdots, a_{2^ex + 2^{e+1}-2}\}$ containing $x$, and we want to show it is contained in $\{a_k,\cdots,a_{2k}\}$. If $2^ex + 2^{e+1}-2 < k$ then by inductive hypothesis on $2^ex+(2^e-1)$ to $2^ex + (2^{e+1}-2)$ we have

$$ \{a_{2^ex + (2^e-1)}, \cdots, a_{2^ex + 2^{e+1}-2}\} \subset \{  a_{2(2^ex + (2^e-1))+1}, a_{2(2^ex + (2^e-1))+2}, \cdots, a_{2(2^ex + (2^{e+1}-2))+1}, a_{2(2^ex + (2^{e+1}-2))+2} \} = \{a_{2^{e+1}x + (2^{e+1}-1)}, \cdots, a_{2^{e+1}x + 2^{e+2}-2}\} $$
Thus we set $e \leftarrow e+1$. Otherwise, $2^ex + 2^{e+1}-2\ge k$, so $ 2^ex + (2^e-1) > \frac 12 (2^ex + 2^{e+1}-2) \ge \frac k2$. For everything from $2^ex + 2^e-1$ to $k-1$ (possibly there is nothing there) I apply inductive hypothesis, so this ends up giving us a subset of $\{a_k,\cdots,a_{2k}\}$. This proves $P(k)$, as desired.

Since the only restrictions we have are $\{a_{2k-1},a_{2k}\} = \{a_{k-1},k\}$, given any $a_0,\cdots,a_{2k-2}$ there are 2 choices for $(a_{2k-1},a_{2k})$. We make $n$ such choices so the answer is $2^n$.
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awesomeming327.
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Let the $k$-set be the set of all nonnegative integers that can be the result of
\[\left\lfloor\frac{k}{2^m}\right\rfloor\]for some nonnegative integer $m$. For example, the $5$-set is $\{5,2,1,0\}$.

Let $f(n)$ be the desired answer. We will show that $f(n)=2^n$. First, note that $f(0)=1$ is clear because the $0$-set, $\{0\}$, is a subset of $\{a_0\}$, implying that $a_0=0$. $f(1)=2$ because the $1$-set, $\{1,0\}$, is a subset of $\{a_1,a_2\}$ on top of the fact that $a_0=0$, so both $(0,0,1)$ and $(0,1,0)$ work.

We now work recursively, showing that $f(n)=2f(n-1)$ for all $n\ge 2$. Drop the condition that the sequence's terms are in $[0,n]$. We will inductively show that this is forced anyway.

It suffices to show that there are exactly $2$ ways to select $a_{n-1}$ and $a_n$, because $\{a_{2n-1},a_{2n}\}$ is determined, for any selection of $a_1$, $a_2$, $\dots$, $a_{2n-2}$ that satisfies the conditions for $0\le k\le n-1$.

Claim 1:
  • For any sequence that is a solution to the $n-1$ problem, $a_1$, $a_2$, $\dots$, $a_{2n-2}$, we have that the values $a_{n},a_{n+1},\dots,a_{2n-2}$ contain all but two values from the $n$-set.
  • $a_{2n-1}$ and $a_{2n}$ are both in the $n$-set.
  • $a_{2n-1}$ and $a_{2n}$ are different from everything in $a_{n},a_{n+1},\dots,a_{2n-2}$.

We proceed with strong induction, where the base case has already been proved above. Clearly, since $n$ is not in any $k$-set for $k<n$, $a_{n},a_{n+1},\dots,a_{2n-2}$ does not contain $n$. Since $a_{n-1}$ is in the $\lfloor n/2\rfloor$-set, it is in the $n$-set. By the Inductive Hypothesis, $a_{n},a_{n+1},\dots,a_{2n-2}$ are all different from $a_{n-1}$. Thus, the first part of the claim is proved. The next two parts follow naturally because $a_{2n-1},a_{2n}$ must be the two remaining values.
We are done.
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Maximilian113
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For some fixed $k$ let the set $f(x)$ be the range of values $\left \lfloor \frac{k}{2^m} \right \rfloor$ can attain as $m$ varies. A key observation is that $f(\lfloor x/2 \rfloor) = f(x) /\{x\}.$ Note that $a_0=0.$ We prove the following proposition by induction on $k$:

For all $k \geq 1,$ $\{a_k, a_{k+1}, \cdots, a_{2k}\}$ is a permutation of $\{0, 1, 2, \cdots, k\},$ and $a_{2k}, a_{2k-1} \in f(k).$

The base case $k=1$ is trivial. Now assume that the proposition holds for some $k.$ Then for $k+1,$ observe that as $\{a_k, a_{k+1}, \cdots, a_{2k}\}$ is a permutation of $\{0, 1, 2, \cdots, k\}$ one of $a_{2k+1}, a_{2k+2}$ equals $k+1 \in f(k+1).$ Meanwhile $$a_{k} \in f(\lceil k/2 \rceil) = f(\lfloor (k+1)/2 \rfloor) = f(k+1)/\{k+1\}.$$so as $a_{k+1}, a_{k+2}, \cdots, a_{2k}$ are distinct from $a_k$ we have that one of $$a_{2k+1}, a_{2k+2}=a_k \in f(k+1).$$Thus as $a_k \neq k+1$ it follows that $$\{a_{2k+1}, a_{2k+2}\} = \{k+1, a_k\} \subseteq f(k+1).$$We can also see from here that $$\{a_{k+1}, a_{k+2}, \cdots, a_{2k+2}\}$$is a permutation of $\{0, 1, 2, \cdots, k+1\}.$ So the induction is complete.

Now from above we can induct to show that the answer is $\boxed{2^n},$ essentially for every next $\{a_{2k+1}, a_{2k+2}\} = \{a_k, k+1\}$ there are two possible ways to assign them.
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