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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.

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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
Points in general position
AshAuktober   3
N 12 minutes ago by blackbluecar
Source: 2025 Nepal ptst p1 of 4
Shining tells Prajit a positive integer $n \ge 2025$. Prajit then tries to place n points such that no four points are concyclic and no $3$ points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?

(Prajit Adhikari, Nepal and Shining Sun, USA)
3 replies
AshAuktober
Mar 15, 2025
blackbluecar
12 minutes ago
Interesting inequalities
sqing   2
N 18 minutes ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc\geq\sqrt{k}$$$$ a+b+kc^2\geq\frac{3\sqrt[3]{k}}{4}$$Where $ k>0. $
$$ a+b+c\geq1$$$$ a+b+4c\geq2$$$$ a+b+c^2\geq\frac{3}{4}$$$$ a+b+8c^2\geq\frac{3}{2}$$
2 replies
sqing
Yesterday at 12:23 PM
sqing
18 minutes ago
IMO 2014 Problem 1
Amir Hossein   132
N 34 minutes ago by maromex
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
132 replies
Amir Hossein
Jul 8, 2014
maromex
34 minutes ago
2-var inequality
sqing   0
40 minutes ago
Source: Own
Let $ a,b,c\geq 0    $. Prove that
$$ \frac{a }{a^2+2b^2+1}+ \frac{b }{b^2+2a^2+1}\leq \frac{1}{\sqrt{3}} $$$$   \frac{a }{2a^2+ b^2+2ab+1}+ \frac{b }{2b^2+ a^2+2ab+1}  \leq \frac{1}{\sqrt{5}} $$$$ \frac{a }{2a^2+ b^2+ ab+1}+ \frac{b }{2b^2+ a^2+ ab+1} \leq \frac{1}{2} $$$$\frac{a }{a^2+2b^2+2ab+1}+ \frac{b }{b^2+2a^2+2ab+1}\leq \frac{1}{2} $$
0 replies
1 viewing
sqing
40 minutes ago
0 replies
IMO Genre Predictions
ohiorizzler1434   31
N 41 minutes ago by ohiorizzler1434
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
31 replies
ohiorizzler1434
Saturday at 6:51 AM
ohiorizzler1434
41 minutes ago
weird FE
tobiSALT   10
N an hour ago by NicoN9
Source: Pan American Girls' Mathematical Olympiad 2024, P5
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(f(x+y) - f(x)) + f(x)f(y) = f(x^2) - f(x+y),$

for all real numbers $x, y$.
10 replies
tobiSALT
Nov 27, 2024
NicoN9
an hour ago
2015 solutions for quotient function!
raxu   49
N an hour ago by blueprimes
Source: TSTST 2015 Problem 5
Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$.

Proposed by Iurie Boreico
49 replies
raxu
Jun 26, 2015
blueprimes
an hour ago
something...
SunnyEvan   0
an hour ago
Source: unknown
Try to prove : $$ \sum csc^{20} \frac{2^{i} \pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^{k} \pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
0 replies
SunnyEvan
an hour ago
0 replies
Cosine of polynomial is polynomial of cosine
yofro   1
N an hour ago by yofro
Source: 2025 HMIC #2
Find all polynomials $P$ with real coefficients for which there exists a polynomial $Q$ with real coefficients such that for all real $t$, $$\cos(P(t))=Q(\cos t).$$
1 reply
yofro
2 hours ago
yofro
an hour ago
Problem 1
randomusername   72
N 2 hours ago by blueprimes
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
72 replies
randomusername
Jul 10, 2015
blueprimes
2 hours ago
Permutation with no two prefix sums dividing each other
Assassino9931   2
N 2 hours ago by Assassino9931
Source: Bulgaria Team Contest, March 2025, oVlad
Does there exist an infinite sequence of positive integers $a_1, a_2 \ldots$, such that every positive integer appears exactly once as a member of the sequence and $a_1 + a_2 + \cdots + a_i$ divides $a_1 + a_2 + \cdots + a_j$ if and only if $i=j$?
2 replies
Assassino9931
3 hours ago
Assassino9931
2 hours ago
Sum of Good Indices
raxu   25
N 5 hours ago by cj13609517288
Source: TSTST 2015 Problem 1
Let $a_1, a_2, \dots, a_n$ be a sequence of real numbers, and let $m$ be a fixed positive integer less than $n$. We say an index $k$ with $1\le k\le n$ is good if there exists some $\ell$ with $1\le \ell \le m$ such that $a_k+a_{k+1}+...+a_{k+\ell-1}\ge0$, where the indices are taken modulo $n$. Let $T$ be the set of all good indices. Prove that $\sum\limits_{k \in T}a_k \ge 0$.

Proposed by Mark Sellke
25 replies
raxu
Jun 26, 2015
cj13609517288
5 hours ago
f(x^2+y^2+z^2)=f(xy)+f(yz)+f(zx)
dangerousliri   7
N 5 hours ago by jasperE3
Source: FEOO, Shortlist A4
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}$ such that for any positive real numbers $x, y$ and $z$,
$$f(x^2+y^2+z^2)=f(xy)+f(yz)+f(zx)$$
Proposed by ThE-dArK-lOrD, and Papon Tan Lapate, Thailand
7 replies
dangerousliri
May 31, 2020
jasperE3
5 hours ago
foldina a rectangle paper 3 times
parmenides51   1
N 6 hours ago by TheBaiano
Source: 2023 May Olympiad L2 p4
Matías has a rectangular sheet of paper $ABCD$, with $AB<AD$.Initially, he folds the sheet along a straight line $AE$, where $E$ is a point on the side $DC$ , so that vertex $D$ is located on side $BC$, as shown in the figure. Then folds the sheet again along a straight line $AF$, where $F$ is a point on side $BC$, so that vertex $B$ lies on the line $AE$; and finally folds the sheet along the line $EF$. Matías observed that the vertices $B$ and $C$ were located on the same point of segment $AE$ after making the folds. Calculate the measure of the angle $\angle DAE$.
IMAGE
1 reply
parmenides51
Mar 24, 2024
TheBaiano
6 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
1540 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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