Beaver Math Olympiad: Difference between revisions

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== Solved problems ==
== Solved problems ==
=== {{TM|1RB0RB3LA4LA2RA_2LB3RA---3RA4RB}} and {{TM|1RB1RB3LA4LA2RA_2LB3RA---3RA4RB}} ===
Let <math>v_2(n)</math> be the largest integer <math>k</math> such that <math>2^k</math> divides <math>n</math>. Let <math>(a_n)_{n \ge 0}</math> be a sequence such that
<math display="block">a_n = \begin{cases}
2 & \text{if } n=0 \\
a_{n-1}+2^{v_2(a_{n-1})+2}-1 & \text{if } n \ge 1
\end{cases}</math>
for all non-negative integers <math>n</math>. Is there an integer <math>n</math> such that <math>a_n=4^k</math> for some positive integer <math>k</math>?
Link to Discord discussion: https://discord.com/channels/960643023006490684/1084047886494470185/1252634913220591728
=== {{TM|1RB3RB---1LB0LA_2LA4RA3LA4RB1LB}} ===
Bonnie the beaver was bored, so she tried to construct a sequence of integers <math>\{a_n\}_{n \ge 0}</math>. She first defined <math>a_0=2</math>, then defined <math>a_{n+1}</math> depending on <math>a_n</math> and <math>n</math> using the following rules:
* If <math>a_n \equiv 0\text{ (mod 3)}</math>, then <math>a_{n+1}=\frac{a_n}{3}+2^n+1</math>.
* If <math>a_n \equiv 2\text{ (mod 3)}</math>, then <math>a_{n+1}=\frac{a_n-2}{3}+2^n-1</math>.
With these two rules alone, Bonnie calculates the first few terms in the sequence: <math>2, 0, 3, 6, 11, 18, 39, 78, 155, 306, \dots</math>. At this point, Bonnie plans to continue writing terms until a term becomes <math>1\text{ (mod 3)}</math>. If Bonnie sticks to her plan, will she ever finish?


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Revision as of 03:54, 24 July 2024

Beaver Mathematical Olympiad (BMO) is an attempt to re-formulate the halting problem for some particular Turing machines as a mathematical problem in a style suitable for a hypothetical math olympiad.

The purpose of the BMO is twofold. First, statements where every non-essential details (e.g. related to tape encoding, number of steps, etc) are discarded are more suitable to be shared with mathematicians who perhaps are able to help. Second, it's a way to jokingly highlight how a hard question could appear deceptively simple.

Unsolved problems

1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE

Let (an)n1 and (bn)n1 be two sequences such that (a1,b1)=(1,2) and

(an+1,bn+1)={(anbn,4bn+2)if anbn(2an+1,bnan)if an<bn

for all positive integers n. Does there exist a positive integer i such that ai=bi?

Hydra and Antihydra

Let (an)n0 be a sequence such that an+1=an+an2 for all non-negative integers n.

  1. If a0=3, does there exist a non-negative integer k such that the list of numbers a0,a1,a2,,ak have more than twice as many even numbers as odd numbers? (Hydra)
  2. If a0=8, does there exist a non-negative integer k such that the list of numbers a0,a1,a2,,ak have more than twice as many odd numbers as even numbers? (Antihydra)

Solved problems

1RB0RB3LA4LA2RA_2LB3RA---3RA4RB (bbch) and 1RB1RB3LA4LA2RA_2LB3RA---3RA4RB (bbch)

Let v2(n) be the largest integer k such that 2k divides n. Let (an)n0 be a sequence such that

an={2if n=0an1+2v2(an1)+21if n1

for all non-negative integers n. Is there an integer n such that an=4k for some positive integer k?

Link to Discord discussion: https://discord.com/channels/960643023006490684/1084047886494470185/1252634913220591728

1RB3RB---1LB0LA_2LA4RA3LA4RB1LB (bbch)

Bonnie the beaver was bored, so she tried to construct a sequence of integers {an}n0. She first defined a0=2, then defined an+1 depending on an and n using the following rules:

  • If an0 (mod 3), then an+1=an3+2n+1.
  • If an2 (mod 3), then an+1=an23+2n1.

With these two rules alone, Bonnie calculates the first few terms in the sequence: 2,0,3,6,11,18,39,78,155,306,. At this point, Bonnie plans to continue writing terms until a term becomes 1 (mod 3). If Bonnie sticks to her plan, will she ever finish?