Beaver Math Olympiad

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 non-essential details (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

1. 1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE (bbch)

Let $(a_n{)}_{n\ge 1}$ and $(b_n{)}_{n\ge 1}$ be two sequences such that $(a_1,b_1)=(1,2)$ and

$$(a_{n+1},b_{n+1})=\{\begin{array}{ll}(a_n-b_n,4b_n+2)& \text{if }{a}_n\ge b_n\\ (2a_n+1,b_n-a_n)& \text{if }{a}_n<b_n\end{array}$$

for all positive integers $n$. Does there exist a positive integer $i$ such that $a_i=b_i$?

The first 10 values of $(a_n,b_n)$ are $(1,2),(3,1),(2,6),(5,4),(1,18),(3,17),(7,14),(15,7),(8,30),(17,22)$.

2. Hydra and Antihydra

Let $(a_n{)}_{n\ge 0}$ be a sequence such that $a_{n+1}=a_n+\lfloor \frac{a_n}{2}\rfloor$ for all non-negative integers $n$.

1. If $a_0=3$, does there exist a non-negative integer $k$ such that the list of numbers $a_0,a_1,a_2,\dots ,a_k$ have more than twice as many even numbers as odd numbers? (Hydra)
2. If $a_0=8$, does there exist a non-negative integer $k$ such that the list of numbers $a_0,a_1,a_2,\dots ,a_k$ have more than twice as many odd numbers as even numbers? (Antihydra)

5. 1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE (bbch)

Let $(a_n{)}_{n\ge 1}$ and $(b_n{)}_{n\ge 1}$ be two sequences such that $(a_1,b_1)=(0,5)$ and

$$(a_{n+1},b_{n+1})=\{\begin{array}{ll}(a_n+1,b_n-f(a_n))& \text{if }{b}_n\ge f(a_n)\\ (a_n,3b_n+a_n+5)& \text{if }{b}_n<f(a_n)\end{array}$$

where $f(x)=10\cdot 2^x-1$ for all non-negative integers $x$.

Does there exist a positive integer $i$ such that $b_i=f(a_i)-1$?

6. 1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD (bbch)

Let $f(b)=b+k+3a$, where $k$ and $a$ are non-negative integers satisfying $b=(2a+1)\cdot 2^k$.

Now consider the iterated application of the function $f^{n+1}(b)=f(f^n(b)))$. Does there exist a non-negative integer $n$ such that $f^n(6)$ equals a power of 2?

Solved problems

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

Let $v_2(n)$ be the largest integer $k$ such that $2^k$ divides $n$. Let $(a_n{)}_{n\ge 0}$ be a sequence such that

$$a_n=\{\begin{array}{ll}2& \text{if }n=0\\ a_{n-1}+2^{v_2(a_{n-1})+2}-1& \text{if }n\ge 1\end{array}$$

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

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

4. 1RB3RB---1LB0LA_2LA4RA3LA4RB1LB (bbch)

Bonnie the beaver was bored, so she tried to construct a sequence of integers $\{a_n{\}}_{n\ge 0}$. She first defined $a_0=2$, then defined $a_{n+1}$ depending on $a_n$ and $n$ using the following rules:

• If $a_n\equiv 0\text{ (mod 3)}$, then $a_{n+1}=\frac{a_n}{3}+2^n+1$.
• If $a_n\equiv 2\text{ (mod 3)}$, then $a_{n+1}=\frac{a_n-2}{3}+2^n-1$.

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

Solution

How to guess the closed-form solution: Firstly, notice that $a_n\approx \frac{3}{5}\times 2^n$. Secondly, calculate the error term $a_n-\frac{3}{5}\times 2^n$. The error term appears to have a period of 4. This leads to the following guess:

$$a_n=\frac{3}{5}\{\begin{array}{ll}{2}^n+\frac{7}{3}& \text{if }n\equiv 0(\mathrm{mod}4)\\ 2^n-2& \text{if }n\equiv 1(\mathrm{mod}4)\\ 2^n+1& \text{if }n\equiv 2(\mathrm{mod}4)\\ 2^n+2& \text{if }n\equiv 3(\mathrm{mod}4)\\ \end{array}$$

This closed-form solution can be proven correct by induction. Unfortunately, the induction may require a lot of tedious calculations.

For all $k$, we have $a_{4k}\equiv 2\text{ (mod 3)}$ and $a_{4k+1}\equiv a_{4k+2}\equiv a_{4k+3}\equiv 0\text{ (mod 3)}$. Therefore, Bonnie will never finish.