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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
Is this FE is solvable?
ItzsleepyXD   2
N 2 minutes ago by frost23
Source: Own , If not appear somewhere before
Find all function $f : \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in  \mathbb{R}$ . $$f(x+f(y))+f(x+y)=2x+f(y)+f(f(y))$$. Original
2 replies
ItzsleepyXD
5 hours ago
frost23
2 minutes ago
butterfly theorem in finnish high school math competition
parmenides51   4
N 17 minutes ago by arzhang2001
Source: Finland 2018, p3
The chords $AB$ and $CD$ of a circle intersect at $M$, which is the midpoint of the chord $PQ$. The points $X$ and $Y$ are the intersections of the segments $AD$ and $PQ$, respectively, and $BC$ and $PQ$, respectively. Show that $M$ is the midpoint of $XY$.
4 replies
parmenides51
Sep 8, 2019
arzhang2001
17 minutes ago
2^x + 3^y a perfect square, find positive integers x,y
parmenides51   12
N 27 minutes ago by MR.1
Source: JBMO Shortlist 2017 NT3
Find all pairs of positive integers $(x,y)$ such that $2^x + 3^y$ is a perfect square.
12 replies
parmenides51
Jul 25, 2018
MR.1
27 minutes ago
Geo metry
TUAN2k8   4
N 28 minutes ago by TUAN2k8
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
4 replies
TUAN2k8
May 6, 2025
TUAN2k8
28 minutes ago
harmonic quadrilateral
Lukariman   0
38 minutes ago
Given quadrilateral ABCD inscribed in a circle with center O. CA:CB= DA:DB are satisfied. M is any point and d is a line parallel to MC. Radial projection M transforms A,B,D onto line d into A',B',D'. Prove that B' is the midpoint of A'D'.
0 replies
Lukariman
38 minutes ago
0 replies
Functional equation
Nima Ahmadi Pour   99
N an hour ago by youochange
Source: ISl 2005, A2, Iran prepration exam
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]
for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria
99 replies
Nima Ahmadi Pour
Apr 24, 2006
youochange
an hour ago
JBMO 2018. Shortlist NT
Steve12345   14
N an hour ago by MR.1
Find all ordered pairs of positive integers $(m,n)$ such that :
$125*2^n-3^m=271$
14 replies
Steve12345
Jul 7, 2019
MR.1
an hour ago
2025 HMIC-5
EthanWYX2009   1
N an hour ago by EthanWYX2009
Source: 2025 HMIC-5
Compute the smallest positive integer $k > 45$ for which there exists a sequence $a_1, a_2, a_3, \ldots ,a_{k-1}$ of positive integers satisfying the following conditions:[list]
[*]$a_i = i$ for all integers $1 \le i \le 45;$
[*] $a_{k-i} = i$ for all integers $1 \le i \le 45;$
[*] for any odd integer $1 \le n \le k -45,$ the sequence $a_n, a_{n+1}, \ldots  , a_{n+44}$ is a permutation of
$\{1, 2, \ldots  , 45\}.$[/list]
Proposed by: Derek Liu
1 reply
EthanWYX2009
Wednesday at 3:16 PM
EthanWYX2009
an hour ago
JBMO 2018. Shortlist NT
Steve12345   14
N an hour ago by MR.1
Prove that there exist infinitely many positive integers $n$ such that $\frac{4^n+2^n+1}{n^2+n+1}$ is a positive integer.
14 replies
Steve12345
Jul 7, 2019
MR.1
an hour ago
Kosovo MO 2010 Problem 5
Com10atorics   21
N an hour ago by navier3072
Source: Kosovo MO 2010 Problem 5
Let $x,y$ be positive real numbers such that $x+y=1$. Prove that
$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq 9$.
21 replies
Com10atorics
Jun 7, 2021
navier3072
an hour ago
Hard combi
EeEApO   4
N an hour ago by navier3072
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
4 replies
EeEApO
Yesterday at 6:08 PM
navier3072
an hour ago
Problem 4 of Finals
GeorgeRP   1
N an hour ago by Stanleyyyyy
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
The diagonals \( AD \), \( BE \), and \( CF \) of a hexagon \( ABCDEF \) inscribed in a circle \( k \) intersect at a point \( P \), and the acute angle between any two of them is \( 60^\circ \). Let \( r_{AB} \) be the radius of the circle tangent to segments \( PA \) and \( PB \) and internally tangent to \( k \); the radii \( r_{BC} \), \( r_{CD} \), \( r_{DE} \), \( r_{EF} \), and \( r_{FA} \) are defined similarly. Prove that
\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]
1 reply
GeorgeRP
Sep 10, 2024
Stanleyyyyy
an hour ago
FE on positive reals with a surprise
MarkBcc168   5
N 2 hours ago by NuMBeRaToRiC
Source: 2019 Thailand Mathematical Olympiad P3
Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ such that $f(x+yf(x)+y^2) = f(x)+2y$ for every $x,y\in\mathbb{R}^+$.
5 replies
MarkBcc168
May 22, 2019
NuMBeRaToRiC
2 hours ago
Both a and a+1997 are roots of P, Q(P(x))=1 has no solutions
WakeUp   2
N 2 hours ago by Rohit-2006
Source: Baltic Way 1997
Let $P$ and $Q$ be polynomials with integer coefficients. Suppose that the integers $a$ and $a+1997$ are roots of $P$, and that $Q(1998)=2000$. Prove that the equation $Q(P(x))=1$ has no integer solutions.
2 replies
WakeUp
Jan 28, 2011
Rohit-2006
2 hours ago
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)?
G H J
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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