Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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)!

Be sure to mark your calendars for the following upcoming events:
[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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   2
N a minute ago by lolsamo
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If $M$ is the foot of the perpendicular from $O$ to $AD$, prove that $\angle MAD = \angle EAL$.

Proposed by Strahinja Gvozdić
2 replies
OgnjenTesic
21 minutes ago
lolsamo
a minute ago
Consecutive squares are floors
ICE_CNME_4   1
N 10 minutes ago by ICE_CNME_4

Determine how many positive integers \( n \) have the property that both
\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]are consecutive perfect squares.
1 reply
ICE_CNME_4
3 hours ago
ICE_CNME_4
10 minutes ago
JBMO TST- Bosnia and Herzegovina 2022 P1
Motion   5
N 14 minutes ago by justaguy_69
Source: JBMO TST 2022 Bosnia and Herzegovina P1
Let $a,b,c$ be real numbers such that $$a^2-bc=b^2-ca=c^2-ab=2$$. Find the value of $$ab+bc+ca$$and find at least one triplet $(a,b,c)$ that satisfy those conditions.
5 replies
Motion
May 21, 2022
justaguy_69
14 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   0
15 minutes ago
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
0 replies
OgnjenTesic
15 minutes ago
0 replies
No more topics!
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
This post has been edited 1 time. Last edited by v_Enhance, Oct 26, 2024, 1:20 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
CANBANKAN
1301 posts
#6 • 1 Y
Y by GeoKing
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 $2^n$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
awesomeming327.
1727 posts
#7
Y by
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.
This post has been edited 1 time. Last edited by awesomeming327., Jan 27, 2025, 4:23 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#8
Y by
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.
Z K Y
N Quick Reply
G
H
=
a