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

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

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

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[*]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
Something nice
KhuongTrang   33
N 2 minutes ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
33 replies
+1 w
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
2 minutes ago
Geometry hard
Lukariman   2
N 2 minutes ago by Primeniyazidayi
Given triangle ABC inscribed in circle (O). The bisector of angle A intersects (O) at D. Let M, N be the midpoints of AB, AC respectively. OD intersects BC at P and AD intersects MN at S. The circle circumscribed around triangle MPS intersects BC at Q different from P. Prove that QA is tangent to (O).
2 replies
Lukariman
43 minutes ago
Primeniyazidayi
2 minutes ago
help!!!!!!!!!!!!
Cobedangiu   3
N 4 minutes ago by sqing
help
3 replies
Cobedangiu
Mar 23, 2025
sqing
4 minutes ago
Inequality
nguyentlauv   1
N 4 minutes ago by NguyenVanHoa29
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
1 reply
nguyentlauv
Yesterday at 12:19 PM
NguyenVanHoa29
4 minutes ago
Great similarity
steven_zhang123   3
N 17 minutes ago by Lil_flip38
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
3 replies
steven_zhang123
44 minutes ago
Lil_flip38
17 minutes ago
AD=BE implies ABC right
v_Enhance   117
N 2 hours ago by cj13609517288
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
117 replies
v_Enhance
Apr 10, 2013
cj13609517288
2 hours ago
Geometry
gggzul   6
N 3 hours ago by Captainscrubz
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
6 replies
gggzul
Yesterday at 8:22 AM
Captainscrubz
3 hours ago
Geometry
Lukariman   5
N 3 hours ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
5 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
3 hours ago
Aime type Geo
ehuseyinyigit   4
N 5 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
4 replies
ehuseyinyigit
Monday at 9:04 PM
ehuseyinyigit
5 hours ago
n variables with n-gon sides
mihaig   1
N 5 hours ago by mihaig
Source: Own
Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$$
1 reply
mihaig
Apr 25, 2025
mihaig
5 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   3
N Today at 5:48 AM by EeEeRUT
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
3 replies
BR1F1SZ
Monday at 9:45 PM
EeEeRUT
Today at 5:48 AM
area of O_1O_2O_3O_4 <=1, incenters of right triangles outside a square
parmenides51   2
N Today at 4:35 AM by Solilin
Source: Thailand Mathematical Olympiad 2012 p4
Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.
2 replies
parmenides51
Aug 17, 2020
Solilin
Today at 4:35 AM
Geo metry
TUAN2k8   3
N Today at 4:34 AM 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.
3 replies
TUAN2k8
Yesterday at 10:33 AM
TUAN2k8
Today at 4:34 AM
China South East Mathematical Olympiad 2013 problem 2
s372102   3
N Today at 2:11 AM by AGCN
$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$
3 replies
s372102
Aug 10, 2013
AGCN
Today at 2:11 AM
Easy Number Theory
jj_ca888   12
N Apr 27, 2025 by joshualiu315
Source: SMO 2020/1
The sequence of positive integers $a_0, a_1, a_2, \ldots$ is recursively defined such that $a_0$ is not a power of $2$, and for all nonnegative integers $n$:

(i) if $a_n$ is even, then $a_{n+1} $ is the largest odd factor of $a_n$
(ii) if $a_n$ is odd, then $a_{n+1} = a_n + p^2$ where $p$ is the smallest prime factor of $a_n$

Prove that there exists some positive integer $M$ such that $a_{m+2} = a_m $ for all $m \geq M$.

Proposed by Andrew Wen
12 replies
jj_ca888
Aug 28, 2020
joshualiu315
Apr 27, 2025
Easy Number Theory
G H J
G H BBookmark kLocked kLocked NReply
Source: SMO 2020/1
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jj_ca888
2726 posts
#1
Y by
The sequence of positive integers $a_0, a_1, a_2, \ldots$ is recursively defined such that $a_0$ is not a power of $2$, and for all nonnegative integers $n$:

(i) if $a_n$ is even, then $a_{n+1} $ is the largest odd factor of $a_n$
(ii) if $a_n$ is odd, then $a_{n+1} = a_n + p^2$ where $p$ is the smallest prime factor of $a_n$

Prove that there exists some positive integer $M$ such that $a_{m+2} = a_m $ for all $m \geq M$.

Proposed by Andrew Wen
This post has been edited 5 times. Last edited by jj_ca888, Aug 28, 2020, 6:39 PM
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i3435
1350 posts
#2 • 1 Y
Y by Mango247
Sol
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gnoka
245 posts
#3
Y by
Excuse me, what is the full name of SMO?
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brianzjk
1201 posts
#4 • 1 Y
Y by gnoka
gnoka wrote:
Excuse me, what is the full name of SMO?

Summer Math Olympiad
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gnoka
245 posts
#5
Y by
brianzjk wrote:
gnoka wrote:
Excuse me, what is the full name of SMO?

Summer Math Olympiad

thanks very much
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rcorreaa
238 posts
#6
Y by
Assume WLOG $a_1$ is odd. Observe that if $p_1$ is the smallest prime factor $a_1 \implies p_1|a_2=a_1+p_1^2 \implies p_1|a_n \quad \forall n \in \mathbb{N}.$ Hence if $p_n$ is the smallest odd prime factor of $a_n$, we have that $p_0 \geq p_1 \geq p_2 \geq ... \implies$ since this sequence is bounded below, we have that it is eventually constant, i.e., there exists an $N$ such that for all $n \geq N, p_{n+1}=p_n. \quad (*)$

Now, observe that if $a_n$ is odd and composite, $\implies a_{n+1}=a_n+p_n^2 \leq a_n + (\sqrt{a_n}^2)=2a_n \implies a_{n+2}= \frac{a_{n+1}}{2^{v_2(a_{n+1})}} < a_n \implies a_{n+2} < a_n$. Thus, if $a_n$ is composite for all odd $a_n$, $a_n$ is eventually negative, which is clearly a contradiction. Hence, there exists $m_0$ such that $a_{m_0}=q_0$, where $q_0$ is an odd prime number.
$\implies a_{m_0+1}=q_0(q_0+1) \implies a_{m_0+2}= \frac{q_0(q_0+1)}{2^{v_2(q_0+1)}}$.

If $\frac{q_0+1}{2^{v_2(q_0+1)}} >1 \implies$ we find $m_1>m_0$ such that $a_{m_1}$ is an odd prime, and defining $q_n$ in a similar way as previously, keep doing this until $\frac{q_n+1}{2^{v_2(q_n+1)}}=1$ or $m_n \geq N$ (from $(*)$). In both cases we are done! :-D

$\blacksquare$
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amuthup
779 posts
#7
Y by
Suppose FTSOC that such an $M$ does not exist, and assume $a_0$ is odd. Let $b_i=a_{2i}$ for all $i$, and let $p_i$ denote the smallest prime factor of $b_i.$

$\textbf{Claim: }$ For all $i,$ there exists $j>i$ such that $p_i>p_j.$

$\emph{Proof: }$ If $b_i=p_i,$ then we have $$b_{i+1}=\frac{p_i(p_i+1)}{2^{\nu_2(p_i+p_{i}^2)}}.$$
  • If $p_i+1$ is a power of $2,$ then $b_{i+1}=b_{i},$ contradiction.
  • Otherwise, $p_i+1$ has a prime factor smaller than $p_i.$.

On the other hand, if $b_i$ is composite, write $b_i=kp_i$ for some $k>p_i$ (check that $k=p_i$ is impossible). Then, note that $$b_{i+1}=\frac{kp_{i}+p_{i}^2}{2^{\nu_2(kp_{i}+p_{i})^2}}\le\frac{kp_{i}+p_{i}^2}{2}<kp_{i}=b_i.$$Continuing in this manner, we will eventually reach $b_{i'}$ which is at most $p_i^2.$
  • If $b_{i'}=p_{i}^2,$ then $b_{i'}=b_{i'+2},$ contradiction.
  • If $b_{i'}=p_i,$ we are done by our previous work.
  • Otherwise, $b_{i'}$ has a prime factor smaller than $p_i,$ as desired.
$\blacksquare$

Since the set of odd primes has a minimum, we are done.
This post has been edited 3 times. Last edited by amuthup, Apr 3, 2021, 3:26 AM
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mathleticguyyy
3217 posts
#8 • 1 Y
Y by centslordm
Let $A_{x}=a_{2x}$.

$p$ represents any generic prime. Note that the smallest prime divisor of the sequence cannot increase. Also, if $A_i$ isn't $p$ or $p^2$, $A_{i+1}$ strictly decreases. Hence, we will eventually reach a number of the form $p$ or $p^2$; in the latter case we are done, and in the former case, notice that the smallest prime factor will strictly decrease, and if we never hit $p^2$, we will eventually get to $A_i=p=2^k-1$ since $3=2^2-1$, at which we achieve the periodicity.
This post has been edited 1 time. Last edited by mathleticguyyy, Jan 15, 2022, 3:19 AM
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IAmTheHazard
5001 posts
#9
Y by
Obviously the smallest odd prime factor of $a_n$ is nonincreasing, hence eventually constant. Suppose that for every $n>N_0$, the least prime divisor of $a_n$ is a fixed prime $p$. If $a_n>p^2$, then advancing two moves starting with (ii) (so then we do (i), and go back to (ii), etc.) will go from $a_n$ to at most $\tfrac{a_n+p^2}{2}$, hence $a_n$ will strictly decrease. Suppose that for all $n>N_1>N_0$ we have $a_n \leq p^2$. Then $a_n/p \in \{1,p\}$ for size reasons, else $p$ is not the minimal prime divisor. If $a_n/p=1 \implies a_n=p$, then we go from $p \to p^2 \to \tfrac{p(p+1)}{2^{\nu_2(p+1))}}$. Since this final fraction is strictly less than $p^2$, we must have $\tfrac{p+1}{2^{\nu_2(p+1)}}=1$, else this term has a smaller prime divisor than $p$, in which case our fraction just equals $p$ so we have the desired periodicity. If $a_n/p=p \implies a_n=p^2$ (the easy case) we go $p^2 \to 2p^2 \to p^2$ so we're done here as well. $\blacksquare$
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GrantStar
821 posts
#10 • 1 Y
Y by OronSH
Solved with OronSH

Note that the smallest prime factor of each term is non increasing. Thus, after some point, it remains constant, say this is after the term $a_n=p\cdot k$ with $p$ being said factor (WLOG $k$ is odd, otherwise (i) is applied). We take $3$ cases.
  • If $k=1$, then $a_{i+1}=p(p+1)$. By assumption, $p+1$ thus must be a power of $2$, hence $a_{i+2}=p$ again.
  • If $k=p$ then It oscillates between $p^2$ and $2p^2$
  • If $k>p$, then $a_{i+2}\leq \frac{p(p+k)}{2}<pk=a_i$. Thus the sequence is decreasing and thus either hits $p$ or $p^2$.
This post has been edited 1 time. Last edited by GrantStar, Jan 8, 2024, 12:07 AM
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HamstPan38825
8860 posts
#11
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For each $n$, let $p_n$ denote the smallest prime factor of $a_n$. Observe that the $\{a_n\}$ alternate parity, so we will assume WLOG that $a_n \equiv n \pmod 2$.

Claim. $\{p_n\}$ is nondecreasing.

Proof. Clearly $p_n \neq 2$, so if $a_n$ is even, then $p_n \mid a_{n+1}$. If $a_n$ is odd, then $a_{n+1} = a_n + p_n^2$ is still a multiple of $p_n$ too. $\blacksquare$

It follows that there is some $N$ such that for all $n > N$, $p_n = p$ is constant.

Claim. There is some $n > N$ such that $a_n \leq p^2$.

Proof. If otherwise, then note that $a_{2n+1} \leq \frac 12 a_{2n}$ as $a_{2n}$ is always even, so
\[a_{2n+1} \leq \frac 12 a_{2n} = \frac 12 (a_{2n-1} + p^2) < a_{2n-1}\]meaning that $\{a_{2n+1}\}$ is strictly decreasing. So it must drop below $p^2$ at some point, contradiction. $\blacksquare$

Consider the $a_n$ described by the claim. Clearly $p \mid a_n$, but if $a_n = pk$ for $2 \leq k \leq p-1$, then there is a smaller prime factor of $a_n$ than $p$, contradiction. Hence either $a_n = p$ or $a_n = p^2$.

If $a_n = p^2$, then $a_{n+1} = 2p^2$ and $a_{n+2} = p^2$, and the conclusion follows immediately. If $a_n = p$, either $p+1$ is a power of two, or it has an odd prime factor less than $p$. The latter case yields a contradiction, and the former case yields $a_{n+2} = p$, so again $\{a_n\}$ is periodic with period $2$. This finishes the problem.
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bin_sherlo
719 posts
#12
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This is a similar one.
Let $p_i$ be the smallest odd prime divisor of $a_i$. First, if $a_n$ is even then $p_n|a_{n+1}$ hence $p_{n+1}\leq p_n$. Also if $a_n$ is odd, then $a_{n+1}=p_n(p_n+\frac{a_n}{p_n})$ which implies $p_{n+1}\leq p_n$ again. Since $\{p_i\}$ is nondecreasing and is a set of primes, this seuqence must be constant evantually. Let this constant prime be $p$. Let $p_n=p$ for $n\geq T$. Consider some constant $C>a_i$ for $i\geq T^T$. If $a_l$ is even, then $a_{l+1}<a_l$ and if $a_l$ is odd, then $a_{l+1}=p^2+a_l<2C$ and $a_{l+2}\leq \frac{a_{l+1}}{2}=\frac{p^2+a_l}{2}<C$ hence $a_i<2C$ for all $i$. Note that there is an $a_k$ repeating at least twice (actually for infinitely many times) and $a_k=a_l$ causes $a_{k+i}=a_{l+i}$ hence this sequence becomes periodic. Let $a_i=pb_i$ for $i\geq T$.
\[2^{r_1}pk_1\rightarrow pk_1\rightarrow 2^{r_2}pk_2\rightarrow pk_2\rightarrow \dots \rightarrow 2^{r_m}pk_m\rightarrow pk_m\rightarrow 2^{r_1}pk_1\]Also we have the equations $p(p+k_i)=2^{r_{i+1}}pk_{i+1}$ or $p+k_i=2^{r_{i+1}}k_{i+1}$. By adding these we get $pm\geq k_1+\dots + k_m$ and since $k_i=1$ or $k_i\geq p+1$, we observe that there must exist some $k_i=1$. WLOG $k_1=1$. However, $p+1=2^{r_2}k_2\geq 2k_2$ thus, $k_2=1$. Similarily, $k_1=k_2=\dots =k_m=1$. So the length of the period must be $2$ where $p$ is a prime with $p+1=2^n$ as desired.$\blacksquare$
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joshualiu315
2534 posts
#13
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Begin by noticing that we only need $a_{i+2} = a_i$ for any $i$. Take an arbitrary term $a_k = px$ where $a_k$ is odd and $p$ is the smallest prime factor of $a_k$. Note that $a_{k+2} = \tfrac{p(x+p)}{2} > a_k$ unless $x=1$ or $x=p$. Hence, the sequence is decreasing until one of the terms is $p^2$ or $p$.

If $a_k = p^2$, then $a_{k+2}=a_k=p^2$. If $a_k=p$, then $a_{k+1} = p(p+1)$; we have $a_{k+2} < a_k$ unless $p+1$ is a power of $2$, at which point the sequence satisfies $a_{k+2} = a_k$. Thus, the primes decrease until $p=3$, at which point $p+1$ is a power of $2$, implying periodicity. $\blacksquare$
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