We write $a\xleftrightarrow{} b$ if one of $a$ and $b$ being fantabulous implies that the other is as well. Observe that \[2n+1\xleftrightarrow{} n\]and \[2n\xleftrightarrow{} 4n+1\xleftrightarrow{} 12n+3\xleftrightarrow{} 6n+1\xleftrightarrow{} 3n\xleftrightarrow{} n\implies 2n\xleftrightarrow{} n\] But, this means that we can find a smaller number that is always fantabulous for any given fantabulous number. Thus, $1$ must be fantabulous. Now, by induction all numbers are fantabulous.

Let $F$ be the set of fantabulous numbers, and let $C=\mathbb N\setminus F$. We will write $x\leftrightarrow y$ if $x\in F$ iff $y\in F$. First note that for any even $x\in F$, we have: $$x\leftrightarrow 2x+1\leftrightarrow6x+3\leftrightarrow3x+1\leftrightarrow\frac32x\leftrightarrow\frac12x$$So $\frac12x\in F$. In addition: $$2021\leftrightarrow1010\leftrightarrow505\leftrightarrow252\leftrightarrow126\leftrightarrow63\leftrightarrow31\leftrightarrow15\leftrightarrow5\leftrightarrow2\leftrightarrow1$$so $1\in F$. Consider the minimal $n\in C$. Case 1: $n$ is even Now $\frac n2$ must also be in $C$, since if $\frac n2$ was in $F$, we would also have to have $n$ be in $F$. But $\frac n2<n$, which contradicts our minimality assumption. $\blacksquare$ Case 1: $n$ is odd Similarly, $\frac{n-1}2\in C$. But $n>1$ since $1\in F$, so $\frac{n-1}2\in C$ and $\frac{n-1}2<n$, contradiction. $\blacksquare$ Since such an $n$ cannot exist, all numbers are fantabulous, hence $2021^{2021}$ is fantabulous. $\square$

https://bmos.ukmt.org.uk/ gives the proposing countries: The problems were submitted by Australia, Slovakia, Ukraine, Australia, Austria, and Austria, respectively.

We claim that all positive integers are good. Since there are solutions @above, mentioning the method of acquiring the $n$ is fantabulous $\iff 2n$(we will use this property though) is also, I'm going to show a different method. The idea here is also to show that $1$ is fantabulous and induct upward. We will write $x\leftrightarrow y$ if $x$ and $y$ are in the same set $\{m,2m+1,3m\}$ for some $m\in \mathbb{N}$(to keep the consistency with the other posts). Then we have: $$n\leftrightarrow 2n+1\leftrightarrow 6n+3\leftrightarrow 3n+1$$$$n\leftrightarrow 2n+1 \leftrightarrow 4n+3 \leftrightarrow 12n+9\leftrightarrow 6n+4\leftrightarrow 3n+2$$Therefore we have that if $n$ is good, then $3n,3n+1,3n+2$ are also good (sorry, I can't be bothered to type fantabulous). Therefore: $$2021\leftrightarrow 4043\leftrightarrow 1347\leftrightarrow 449 \leftrightarrow 149 \leftrightarrow 49 \leftrightarrow 16 \leftrightarrow 5\leftrightarrow 1$$Now let's look at the representation of $2021^{2021}$ in base $3$. Let's write $2021^{2021}=\overline{a_{1}a_{2}\cdots a_{k}}$. Then if $a_{i}=0$ we use the $3n$ function, if $a_{i}=1$ we will use the $3n+1$ function and if $a_{i}=2$ we will use the $3n+2$ function. Starting from $1$ and using this algorithm for $i=k,k-1,\cdots , 1$ we get that $2021^{2021}$ is good. Anallogically every number is good by the same algorithm.

The answer is yes. We will show that all positive integers are fantabulous. First, $m$ is fantabulous if and only if $2m$ is fantabulous. This is because $m$ fantabulous $\iff$ $3m$ fantabulous $\iff$ $6m+1$ fantabulous $\iff$ $12m+3$ fantabulous $\iff$ $4m+1$ fantabulous $\iff$ $2m$ fantabulous by considering the sets $\{m, 2m+1, 3m\}, \{3m, 6m+1, 9m\}, \{6m+1, 12m+3, 18m+3\}, \{4m+1, 8m+3, 12m+3\}, \{2m, 4m+1, 6m\}$.

of fantabulous numbers from 2021 to 1. Now inductively with base case $k = 1$, if all positive integers $< 2^k$ are fantabulous, then all positive integers in the interval $[2^k, 2^{k+1})$ are of the form $2m$ or $2m+1$ for some $m < 2^k$, and thus fantabulous. Thus all positive integers, including $2021^{2021}$ are fantabulous.

This took me way too long for a P1. The answer is yes. We will prove that all positive integers are fantabulous. We are allowed the following transformations: $m$ to $2m+1$, $3m$, $\frac{m-1}{2}$ (if $m$ is odd) or $\frac{m}{3}$ (if $3 \mid m$). Note that fantabulousness (how is there no squiggly red line underneath) is invariant under these transformations. Write $a \rightarrow b$ if we can change $a$ to $b$ by applying finitely many transformations. Since they are reversible, we must have $b \rightarrow a$ as well. It is sufficient to prove that $m \rightarrow 1$ for all positive integers $m$. By using Strong Induction, it is sufficient to prove that for any $m>1$, there exists a positive integer $n<m$ such that $m \rightarrow n$. If $m$ is odd: $$m \rightarrow \frac{m-1}{2}$$and we are done. Else, $m$ is even, and in that case: $$m \rightarrow 2m+1 \rightarrow 6m+3 \rightarrow 3m+1 \rightarrow \frac{3m}{2} \rightarrow \frac{m}{2}$$and we are done in this case as well. $\blacksquare$

Same stuff as everyone else but with less efficiency. I claim that there is no smallest fantabulous number. We first note that 1 is fantabulous, since we have a chain of fantabulous numbers: \begin{align*} 2021 \rightarrow 6063 \rightarrow 3031 \rightarrow 1515 \rightarrow 757 \rightarrow 378 \rightarrow 126 \rightarrow 42 \rightarrow 14\rightarrow 29 \\ \rightarrow 87\rightarrow 43 \rightarrow 21 \rightarrow 63 \rightarrow 31 \rightarrow 15 \rightarrow 7 \rightarrow 3 \rightarrow 1. \end{align*}Now let $n>1$ be the smallest non-fantabulous number, if it exists. Then there is a number of cases: $n \equiv 1 \pmod{2}$. In this case, $(n - 1)/2$ is also non-fantabulous, which is smaller than $n$. $n \equiv 0 \pmod{6}$. In this case, $n/3$ is also non-fantabulous, which is smaller than $n$. $n \equiv 4 \pmod{6}$. In this case, writing $n = 6k + 4$, we have $6k + 4 \rightarrow 12k + 9 \rightarrow 4k + 3$ all being non-fantabulous, and $4k + 3$ is smaller than $n$. $n \equiv 2 \pmod{6}$. In this case, writing $n = 6k + 2$, we have $6k + 2 \rightarrow 12k +5 \rightarrow 36k + 15 \rightarrow 18k + 7 \rightarrow 9k + 3 \rightarrow 3k + 1$ all being non-fantabulous, and $3k + 1$ is smaller than $n$. Thus in each case we can get another number smaller than $n$ which is non-fantabulous, contradicting the minimality of $n$. So every positive integer is fantabulous, including $2021^{2021}$.

Basically same sol. We claim the answer is yes. Claim 1: $1$ is fantabulous We have $n \leftrightarrow 2n+1$ and $n \leftrightarrow 3n \leftrightarrow 6n+1\leftrightarrow 12n+3 \leftrightarrow 4n+1 \leftrightarrow 2n$. Thus, for all fantabulous numbers $k>1$, there exists $k'<k$ where $k'$ is fantabulous. Thus, $1$ is fantabulous. Thus, working forwards and backward we get $2021 \leftrightarrow \cdots \leftrightarrow 1 \leftrightarrow \cdots \leftrightarrow 2021^{2021}$.

All numbers are fantabulous, actually. Suppose that the number $n$ was "fantabulous". If it is odd, then $\frac{n-1}2$ is also fantabulous, and if $n$ is even, then \[2n+1,6n+1,3n,n\]are all fantabulous. Thus, by descending, we can get $1$ is fantabulous. The rest of the solution follows by induction.

Same solution as well. We prove the following claims: Claim 1: If $n$ is good, so is $2n$. The reverse is obviously also true. Proof: We have $n$->$3n$->$6n+1$->$12n+3$->$4n+1$->$2n$. Claim 2: 1 is reachable from 2021 Proof: We have $2021->1010->505->252->126->63->31->15->5->2->1$. Suppose FTSoC that not all positive integers are reachable from 2021 and that $x$ is the minimal positive integer not reachable from 2021. If $x$ is even, then take $x/2$, contradiction to the minimality. If $x$ is odd, then it is different from 1, so take $(x-1)/2$, contradiction. So all positive integers are fantabulous.

anser wrote: The number 2021 is fantabulous. For any positive integer $m$, if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous? We can prove that all the positive integers are fantabulous . Claim 1: $1$ is fantabulous. Proof: By the condition of the problem we will use the fantabulous transform to pass from $2021$ to $1$: $2021 \rightarrow 6063 \rightarrow 3031 \rightarrow 1515 \rightarrow 505 \rightarrow 252 \rightarrow 84 \rightarrow 28 \rightarrow 57 \rightarrow 19 \rightarrow 9 \rightarrow 3 \rightarrow 1$ Proved! . Claim 2: If $x$ is fantabulous then $2x$ is also fantabulous. Proof: Again by the problem condition we will use the fantabulous transform to pass from $x$ to $2x$. $x \rightarrow 3x \rightarrow 6x+1 \rightarrow 12x+3 \rightarrow 4x+1 \rightarrow 2x$ Proved . Now using Claim 1 and Claim 2 We are done becuase: $1$ is fantabulous and hence: If $n$ is fantabulous then $2n, 2n+1, 3n$ are fantabulous. replace $n=1$ to get. $(1) \rightarrow (2, 3) \rightarrow (4, 5, 6, 7) \rightarrow (8, 9, 10, 11, 12, 13, 14, 15) \rightarrow \cdots$. We see that for every fantabulous you get $2$ new fantabulous, and if you start form $1$ you get the positive integers secuence becuase the first term of the n-th conjunte of fantabulous numbers is $2^n$ and the last term is $2^{n+1}-1$ thus we are done .

The answer is positive. Since I don't like the word fantabulous, we use the word fantastic. Let $\mathbb{F}$ be the set of all fantastic numbers. Now notice that if $m \in \mathbb{F}$, then we under the conditions stated by the problem we must have that the reverse must hold true. So we have that: $$m \rightarrow 3m \rightarrow 6m+1 \rightarrow 12m+3 \rightarrow 4m +1 \rightarrow 2m \in \mathbb{F}$$meaning that if $m$ is even and in $\mathbb{F}$ we get that $\frac{1}{2}m \in \mathbb{F}$. So we can always get a minimum value. Let $\mathbb{R}_{\mathbb{F}} = \mathbb{N} \backslash \mathbb{F}$. Let $c$ be the minimum value of $\mathbb{R}_{\mathbb{F}}$. If $c$ is odd then we must have that $\frac{c-1}{2} \in \mathbb{R}_{\mathbb{F}}$, but that contradicts the minimality of $c$. If $c$ is even then we must have that $\frac{c}{2} \in \mathbb{R}_{\mathbb{F}}$, but again we get a contradiction on the minimality of $c$. So that means that $\mathbb{R}_{\mathbb{F}}$ must be a null set. Meaning that $\mathbb{F} = \mathbb{N}$, meaning that $2021^{2021} \in \mathbb{F}$

Solution with lilavati_2005 (probably similar to multiple solutions here) Claim: From any given fantabulous number $n > 1$, we can find another fantabulous number that is smaller than it, which implies that $1$ is fantabulous. Further, it implies the other direction, that given all numbers smaller than $n$ are fantabulous, then $n$ is fantabulous. Proof. If $n$ is odd, trivially $\tfrac{n-1}{2}$ is fantabulous. If $n$ is divisible by $3$, then $\tfrac{n}{3}$ is fantabulous. So assume $n$ is even, and also not divisible by $3$. Let $n = 2m$ and consider the following set of transformations $$n \to 2n+1 \to 6n + 3 \to 3n + 1 \to 3 \cdot \left( \tfrac{n}{2} \right) = 3m \to m$$This shows that for any fantabulous $n$ which is even, we have that $\tfrac{n}{2}$ is fantabulous and also from any fantabulous $m$, $2m$ is fantabulous. $\blacksquare$ From the claim, clearly $1$ is fantabulous. We proceed by induction. Assume that $1, 2, 3, \cdots k$ are all fantabulous. If $k+1$ is even, then we have that $k+1$ is fantabulous from $\tfrac{k+1}{2}$ being fantabulous. If $k+1$ is odd, we have $k+1$ is fantabulous from $\tfrac{k}{2}$ being fantabulous. Hence, every positive integer is fantabulous, which implies the conclusion. $\blacksquare$

The heart of this prove lies in the fact that if we can show that from any number $k,$ we can always use the operations in the problem to get to a lower number, then eventually, after repeating the process a certain number of times, we will be able to get that $1$ is achievable. Thus, if $1$ is achievable, then we can reverse this process and get that all of the positive integers are achievable. (Notice that there is no way that we can decrease down to something $\leq 0$ because $\frac{k-1}{2}\leq 0$ iff $k$ had been equal to $1$ at some previous point already, and we could have just ended the process at that point). We first prove that $1$ is fantabulous. This can be seen through the following chain: $2021\to 6063\to 3031\to 1015\to 507\to 253\to 126\to 42\to 14\to 29\to 87\to 43\to 21\to 7\to 3\to 1$. Now that we know $1$ is fantabulous, we can use the method in the first paragraph to do the proof. We handle the cases base on the residue of $k\pmod 3$. Case 1: $k\equiv 0\pmod 3$ Divide by $3$. Case 2: $k\equiv 1\pmod 3$ Multiply by $2$ and add $1$, then divide by $3$. For $k\geq 2,$ this clearly decreases the value. Case 3: $k\equiv 2\pmod 3$ For this case, we use strong induction and suppose all the numbers before $k$ we know are fantabulous. We proceed with cases on $k\pmod 6$: Case 1: $k\equiv 2\pmod 6$ In this case, let $k=6n+2,$ and we provide the following algorithm to get to a fantabulous number, from which we can reverse the process: $6n+2\to 18n+6\to 36n+13\to 12n+6\to 4n+2$ Case 2: $k\equiv 5\pmod 6$ For $k\equiv 5\pmod 6,$ we can perform that following set of moves, where $k=6n+5$: $6n+5\to 18n+15\to 36n+31\to 72n+63\to 8n+7,$ and $8n+7$ is obviously achievable by repeatedly applying $2m+1$ to $n$ multiple times, and $n$ by our inductive hypothesis, is fantabulous. Ok now I realize I phrased this argument really weirdly and basically what the first paragraph is saying is that this is just strong induction. :p

I think this is one of these problems where finding the answer might be the most difficult part. We will prove that all positive integers are fantabulous (short fanta). We write $m \rightarrow n$ if $m$ fanta directly implies that $n$ is fanta. Claim: 1 is fanta Indeed, we have \begin{align*} 2021 &\rightarrow 1010 \rightarrow 3030\rightarrow 6061\rightarrow 12123\rightarrow 4041\rightarrow 1347\rightarrow 449\rightarrow 224\\ &\rightarrow 672\rightarrow 1345\rightarrow 2691\rightarrow 299\rightarrow 149\rightarrow 74\rightarrow 222\rightarrow 445\rightarrow 891\\ &\rightarrow 99\rightarrow 11\rightarrow 5\rightarrow 2\rightarrow 6\rightarrow 13\rightarrow 27 \rightarrow 1\hspace{38mm} \square \end{align*} We will now prove by induction that $1,2,...,n-1$ fanta implies $n$ fanta. Note that $1,2$ are fanta by the claim above. We will have to distinguish between the following cases: $\bullet$ $n$ is odd: Then $n=2m+1$ for some $m\leq n-1$ and so $n$ is fanta. $\bullet$ $n=6m$ for some$m\geq 1$: Then $n=3\cdot(2m)$ and $2m\leq n-1$. Thus $n$ is fanta. $\bullet$ $n=6m+4$ for some $m\geq 0$: Note that $4m+3\leq n-1$. We obtain the following chain: $$4m+3\rightarrow 12m+9\rightarrow 6m+4$$$\bullet $ $n=6m+2$ for some $m\geq 1$: Note that $4m+1\leq n-1$. We have $$4m+1\rightarrow 8m+3\rightarrow 72m+27\rightarrow 36m+13\rightarrow 18m+6\rightarrow 6m+2.$$This completes the induction.

It's interesting that both P1 and P6 are number theory questions where the answer is yes.

The answer is $\boxed{yes}$. Claim: $m$ is fantabulous if $2m+1$ is and $2m+1$ is fantabulous if $m$ is. Proof: Trivial. Claim: $m$ is fantabulous if $2m$ is and $2m$ is fantabulous if $m$ is. Proof: Consider the chain $2m\to 4m+1\to 12m+3\to 6m+1\to 3m\to m$, which shows that $m$ is fantabulous if $2m$ is, and consider the reverse chain to show that $2m$ is fantabulous if $m$ is. Thus, for any integer $n$ greater than $1$, if $n$ is even, we can just divide by $2$ to find another fantabulous number, and if $n$ is odd, we can just subtract by $1$ and then divide by $2$ to find another fantabulous number. So we can keep decreasing until $1$, which implies $1$ is fantabulous.

it's obvious that every positive integer is fantabulous.

Denote $\longleftrightarrow$ by the following definition: if $a$ is fantabulous if and only if $b$ is fantabulous then $a\longleftrightarrow b$. Note that $$m\longleftrightarrow 2m+1 \longleftrightarrow 3m\longleftrightarrow 6m+1 \longleftrightarrow 12m+3 \longleftrightarrow 4m+1 \longleftrightarrow 2m.$$Thus, $m\longleftrightarrow 2m$ and $2m \longleftrightarrow 2m+1$. $$n\longleftrightarrow\frac{n}{v_2(n)}\longleftrightarrow \frac{n}{v_2(n)}-1$$ We can do the above process again and again until these numbers eventually hit $1$. Hence $$2021 \longleftrightarrow 1 \longleftrightarrow 2021^{2021}.$$

We will write $a \xleftrightarrow{} b$ to mean that fantabulousness is transferred from $a$ to $b$. We also note that fantabulousness is transitive, so if $a \xleftrightarrow{} b$ and $b \xleftrightarrow{} c$, then $a \xleftrightarrow{} c$. We have $m \xleftrightarrow{} 3m \xleftrightarrow{} 6m+1 \xleftrightarrow{} 12m+3 \xleftrightarrow{} 4m+1 \xleftrightarrow{} 2m$ , so $m \xleftrightarrow{} 2m$. We are given that $m \xleftrightarrow{} 2m+1$. Claim: $1$ is fantabulous Proof: The idea is that we can go down with $2x+1 \xleftrightarrow{} x$ and $2x \xleftrightarrow{} x$ for all $x$. We then have $2021 \xleftrightarrow{} 1010 \xleftrightarrow{} 505 \xleftrightarrow{} 252 \xleftrightarrow{} 126 \xleftrightarrow{} 63 \xleftrightarrow{} 31 \xleftrightarrow{} 15 \xleftrightarrow{} 7 \xleftrightarrow{} 3 \xleftrightarrow{} 1$ Claim: All positive integers are fantabulous. Proof: We use strong induction to prove that if $1,2,3,\cdots,n$ are all fantabulous then $n+1$ is also fantabulous. The base case $n=1$ was proved above. Then, if $n+1$ is even you have $\frac{n+1}{2} \xleftrightarrow{} n+1$ and if $n+1$ is odd you have $\frac{(n+1)+1}{2}-1 \xleftrightarrow{} n+1$, so all positive integers are fantabulous. Therefore $2021^{2021}$ is indeed fantabulous. $\blacksquare$

We say $a$ is fantabulous to $b$ and write $a \leftrightarrow b$ if one being fantabulous implies other being the same as well. We claim that all integers are fantabulous. First we show that $1$ is fantabulous. Since $2021$ is fantabulous, we get: $$ 2021 \leftrightarrow 6063 \leftrightarrow 3031 \leftrightarrow 1515 \leftrightarrow 757 \leftrightarrow 2271 \leftrightarrow 1135 \leftrightarrow 567 \leftrightarrow 283 \leftrightarrow 141 \leftrightarrow 423 \leftrightarrow 211\leftrightarrow 105 \leftrightarrow 315 \leftrightarrow 157 \leftrightarrow 235 \leftrightarrow 117\leftrightarrow 351 \leftrightarrow 175 \leftrightarrow 87 \leftrightarrow 43\leftrightarrow 21 \leftrightarrow 63 \leftrightarrow 31 \leftrightarrow 15\leftrightarrow 7 \leftrightarrow 3 \leftrightarrow 1 $$as desired. Now, consider the following algorithm: $\bullet$ For an odd $k$, keep on subtracting $1$ and dividing by $2$ to obtain a new number and continue until we reach an even number. $\bullet$ After we reach an even number, go back to previous number (note that number must be odd), and multiply that number by $3$ and repeat the first process. Note that, the moves in our algorithm obeys the "fantabulatic operations" and we claim that for any positive integer $n$ by applying the moves in the algorithm, we can always reach to $1$. If $n$ is odd, then write $n = 2^{\alpha} m + 1 $ for odd $m$. Now, applying the operations from the algorithm for $n$, we get $$ 2^{\alpha} m + 1 \leftrightarrow 3.2^{\alpha} m + 3 \leftrightarrow 3.2^{\alpha - 1} m + 1 \leftrightarrow \dots \leftrightarrow 3^{\alpha -1}.2m + 1 \leftrightarrow 3^{\alpha}m$$ Now, at this stage, keep on dividing $3^{\alpha}m$ by $3$, $\alpha$ times. Note that, this again preserves the fantabulatic operations. Hence at the end, we have: $$ 2^{\alpha}m + 1 \leftrightarrow m$$ Since $m$ is odd we can decrease $m$ again using the same algorithm and so on until we reach minimum odd number which is $1$. For even $n$ write $n = 2^{\alpha} m$ and change $n$ to $2n+1$ as $n \leftrightarrow 2n+1$. Now analogously we can show it works for even $n$ as well. Hence all integers are fantabulous implying $2021^{2021}$ being fantabulous as well.

Claim: 1 is fantabulous. Observe that $$2021\rightarrow 1010\rightarrow 3030\rightarrow 6061 \rightarrow 12123\rightarrow 4041$$$$\rightarrow 1347\rightarrow 449\rightarrow 224\rightarrow 672\rightarrow 1345\rightarrow 2691\rightarrow 897$$$$\rightarrow 299\rightarrow 149\rightarrow 74\rightarrow 222\rightarrow 445\rightarrow 891\rightarrow 297\rightarrow 99\rightarrow 33\rightarrow 11$$$$\rightarrow 5\rightarrow 15\rightarrow 7\rightarrow 3\rightarrow 1$$ We claim that for any positive integer $n$, we can reach down to 1 in some number of moves by each move, either replacing $n$ with $2n+1$ or $3n$ or $(n-1)/2$ or $n/3$ (as long as the result is an integer). We will use induction. This can be easily verified for $n\leq 7$. Assume $n\geq8$ from now on. Take cases on $n$ mod 6. If $n$ is odd, then we can obviously induct down to $(n-1)/2$. Similarly if $n$ is a multiple of 3. If $n$ is 4 mod 6, then we can make it $(2n+1)/3$ in two moves, which is also less than $n$ so we can also induct down. This leaves only $n\equiv 2\pmod 6.$ However, in this case, we can triple it, so it becomes 6 mod 18. Then, do the $2n+1$ move twice, so it becomes 9 mod 18. Then, note that this is odd. Divide by 3 twice, and then do the $(n-1)/2$ move. This series of moves turns $n$ into $$\frac{\frac{\frac{2(2(3n)+1))+1}{3}}{3}-1}{2}=\frac{2n-1}{3},$$so we can also induct down. Therefore, by induction, all $n$ can reach 1 in this manner. Therefore, there exists such a chain that takes $2021^{2021}$ all the way down to 1. Since 1 is fantabulous, $2021^{2021}$ also is, so we are done.

Answer. $2021^{2021}$ is indeed "fantabulous". Proof. Define an "aloof" as follows - An aloof is a string consisting of $a$'s, $b$'s, $c$'s, and $d$'s - We do an "aloofing" of the aloof $x$ on an integer $y$ by plugging $x$ into the function defined by the first character of the aloof, and plug the resulting number into the function defined by the next character of the aloof $x$, and so on and so forth. - For example, if the aloof $x$ is $abcd$, then an aloofing of $x$ on $y$ would be in the form of $d(c(b(a(y))))$. - The functions $a$, $b$, $c$, and $d$ are defined as $a(x)=\frac{x}{3}$, $b(x)=\frac{x-1}{2}$, $c(x)=3x$, and $d(x)=2x+1$. Notice that a number $y$ is "fantabulous" iff there exists an aloof $z$ such that if we do an aloofing of $z$ onto $x$ for some fantabulous integer $x$, we get $y$. Claim 1: $1$ is "fantabulous". Define the aloof called $x$ to be the following \[x=cbbacbbaaadcbbaccbbacbba. \]Notice that if we take an aloofing of $x$ onto $2021$, we are left with $1$, meaning that $1$ is indeed fantabulous. Claim 2: All positive integers are "fantabulous". Notice that if an aloof $x$ exists such that an aloofing of $x$ onto $y$ maps $y$ to $z$, there also must exist an aloof $w$ such that an aloofing of $w$ on $z$ maps $z$ to $y$, since $a$ is the inverse function of $c$ and $b$ is the inverse function of $d$. Therefore, in order to prove that all positive integers are fantabulous, we only have to prove that there always exists and aloof that maps the number to $1$. We can do this by creating a function that creates such an aloof(written in C++). string makeAloof(int z) { string x = ""; int y = z; while (y>1) { if () { y=y/3; x+="a"; } else if () { y=3y; y=(y-1)/2; y=(y-1)/2; y=y/3; x+="cbba"; } else if () { y=2y+1; y=y/3; x+="da"; } else if () { y=2y+1; y=3y; y=(y-1)/2; y=(y-1)/2; y=y/3; x+="dcbba"; } else { y=3y; y=3y; y=(y-1)/2; y=(y-1)/2; y=y/3; x+="ccbba"; } } return x; } Now all that's left to prove is that the loop eventually terminates. We can prove this by proving that for each iteration of the loop where $y>1$, $y$ will decrease. Case 1: $y$ is a multiple of $3$ Here, we have that $y$ is just divided by $3$, which pretty obviously shows that $y$ decreases. Case 2: $y$ is either $1$ or $5$ mod $12$ Case 2a: $y=12k+1$, $k\geq{}1$ \[12k+1\rightarrow{}36k+3\rightarrow{}18k+1\rightarrow{}9k\rightarrow{}3k\]Case 2b: $y=12k+5$, $k\geq{}0$ \[12k+5\rightarrow{}36k+15\rightarrow{}18k+7\rightarrow{}9k+3\rightarrow{}3k+1\] Case 3: $y$ is $10$ or $4$ mod $12$ Case 3a: $y=12k+10$, $k\geq{}0$ \[12k+10\rightarrow{}24k+21\rightarrow{}8k+7\]Case 3b: $y=12k+4$, $k\geq{}0$ \[12k+4\rightarrow{}24k+9\rightarrow{}8k+3\] Case 4: $y$ is $2$ or $8$ modulo $12$ Case 4a: $y=12k+2$, $k\geq{}0$ \[12k+2\rightarrow{}24k+5\rightarrow{}72k+15\rightarrow{}36k+7\rightarrow{}18k+3\rightarrow{}6k+1\]Case 4a: $y=12k+8$, $k\geq{}0$ \[12k+8\rightarrow{}24k+17\rightarrow{}72k+51\rightarrow{}36k+25\rightarrow{}18k+12\rightarrow{}6k+4\] Case 5: $y$ is $7$ or $11$ mod $12$ Case 5a: $y=12k+7$, $k\geq{}0$ \[12k+7\rightarrow{}108k+63\rightarrow{}54k+31\rightarrow{}27k+15\rightarrow{}9k+5\]Case 5a: $y=12k+11$, $k\geq{}0$ \[12k+11\rightarrow{}108k+99\rightarrow{}54k+49\rightarrow{}27k+24\rightarrow{}9k+8\] Therefore the loop always terminates, thus proving that all positive integers are fantabulous. Thus $2021^{2021}$ is also fantabulous, and we are done. $\blacksquare$

I claim that all positive integers are fantabulous. First make the observation that \[m \rightarrow 3m \rightarrow 6m + 1 \rightarrow 12m + 3 \rightarrow 4m + 1 \rightarrow 2m.\]Therefore $m$ is fantabulous iff $2m$ is fantublous, iff $2m + 1$ is fantabulous. Now \[2021 \rightarrow 1010 \rightarrow 505 \rightarrow 252 \rightarrow 126 \rightarrow 63 \rightarrow 31 \rightarrow 15 \rightarrow 7 \rightarrow 3 \rightarrow 1.\]We now make the following claim: Claim: All integers are fantabulous. Proof: We prove by strong induction. The base case for $N = 1$ is trivial. Assume all $1, 2 \ldots, N$ are fantabulous. We split into two cases according to the parity of $N$: Case 1: ($N$ is odd). Consider $(N + 1)/2 < N$. Then as $(N + 1)/2$ is fantabulous, so must $2[(N + 1)/2] = N + 1$ also be fantabulous. Case 2: ($N$ is even). Consider $N/2 < N$. Then as $N/2$ is fantabulous, so must $2[(N/2)] + 1 = N + 1$ also be fantabulous. $\square$ Therefore we have all $N \geq 1$ are fantabulous so that $2021^{2021}$ is also indeed fantabulous. $\blacksquare$

Claim: $1$ is fantabulous. Proof. Omitted $\blacksquare$ Now, INDUCT If $a\equiv 3,6 \pmod 6,$ divide by $3$ If $a\equiv 1,4 \pmod{6}$, then use $\frac{2a+1}{3}$ If $a\equiv 5 \pmod 6$, then do $\frac{a-1}{2}$ If $a\equiv 2\pmod 6$, then we have \[a\rightarrow 3a \rightarrow 6a+1 \rightarrow 12a+3 \rightarrow 4a+1 \rightarrow \frac{4a+1}{3} \rightarrow \frac{2a-1}{3}\]

Consider a connected simple graph $G$ with each vertex denoting a positive integer. An integer $m$ is connected to all positive integers from the set \[\left\{2m+1, 3m, \frac m3, \frac {m-1}2\right\}\]The problem is equivalent to showing that the graph is connected. Claim: $\{k,6k,6k+1,6k+2,6k+3,6k+4,6k+5\}$ are all reachable. Proof. Consider the following paths: $6k+1, 12k+3, 4k+1, 2k, 6k.$ $6k+3, 2k+1, k, 3k, 6k+1.$ $6k+2, 18k+6, 36k+13, 72k+27, 24k+9, 8k+3, 4k+1, 2k, 6k.$ $6k+4, 12k+9, 4k+3, 2k+1, k, 3k, 6k+1.$ $6k+5, 3k+2, 9k+6, 18k+13, 36k+27, 12k+9, 4k+3, 2k+1, k, 3k, 6k+1.$ Inducting on $k$, we are done. $\blacksquare$

Pretty cool. The main idea is to show that $n$ being fantabulous is true iff $1$ being fantabulous is true, which is a matter of strong induction.

The answer is yes. We will prove the stronger statement that if any number $k$ is fantabulous, all positive integers are also fantabulous. Define $x\iff y$ to mean that $y$ is fantabulous if and only if $x$ is fantabulous. Note that $$k\iff 3k\iff 6k+1\iff 12k+3\iff 4k+1\iff 2k.$$But we also have $k\iff 2k+1$, so $k$ is fantabulous if and only if $\lfloor \frac{k}{2}\rfloor$ is fantabulous. We can keep applying this process to find that $1$ is fantabulous. But now every number has a way of being reached from $1$ due to its binary expansion, done.

Feels pretty NT-flavoured for some reason Not a very unique solution, but doing this for storage. So I don't think anyone immediately jumps to the conclusion that perhaps entirety of $\mathbb{Z^{+}}$ is fantabulous however that intuition comes when we try to reach the "smallest" number we possibly can from $2021$ as this feels like some inverse transformation for Fermat's Method of Infinite Descent as demonstrated in many Number Theoretic problems or in combinatorics "extremal element" as we like to find some minimum characterization and it seems to create branches below that number. My intuition was to consider inverse operations as specified in the above set. Basically given any odd number $n$ notice that $\exists x : 2x+1=n$ and basically then we have that $\frac{n-1}{2}$ is also fantabulous. Similarly we have $x \to \frac{x}{3}$ for divisble by $3$. We claim that we can reach $1$ from $2021$ using a combination of initial and further inverse operations. The key idea is that we can utilize initial odd to reach some odd/even. But an even won't allow us to apply our crux inverse operation and thus we use a little trick that $x \equiv 1 \ (mod \ 4)$ in which case $\frac{x-1}{2}$ is even, we have $\frac{3x-1}{2}$ is odd since now it is $3x-1 \equiv 2 \ (mod \ 4)$. Basing on this we have the following map : $$2021\leftrightarrow6063\leftrightarrow3031\leftrightarrow1515\leftrightarrow757\leftrightarrow378\leftrightarrow126\leftrightarrow42\leftrightarrow14\leftrightarrow29\leftrightarrow87\leftrightarrow43\leftrightarrow21\leftrightarrow63\leftrightarrow31\leftrightarrow15\leftrightarrow7\leftrightarrow3\leftrightarrow1$$and now we can see methods to reach $2,3,\dots,7$ and so on pretty easily. This is where the intuition comes from. Although notice that $x \to 2x+1$ helps create odd numbers easily, we also need some transformation $x \to 2x$ which doesn't seem apparent, but the map $2 \leftrightarrow 7$ allows us to note a few important lemmas that leads to our solution. Lemma 1: The map $x \leftrightarrow 3x+1$ is possible. Proof: Consider that $x \leftrightarrow 2x+1 \leftrightarrow 3(2x+1)=6x+3 \leftrightarrow \frac{6x+3-1}{2}=3x+1$ which uses the inverse operation along with the fact that $6x+3 \equiv 1 \ (mod \ 2)$. Lemma 2: The map $x \leftrightarrow 2x$ is possible. Proof: Consider that $x \leftrightarrow 3x \leftrightarrow 2(3x)+1=6x+1 \leftrightarrow \frac{6x+1-1}{3}=2x$ using the earlier lemma. Finally, we shall introduce the inductive key idea, wherein we have $1 \leftarrow 2(1)=2$ due to lemma $2$ acting as our base case. Next, we have suppose hypothesis that upto $k$ we have every number as fantabulous. If $k+1$ is odd then we have $k=2k_1$ for some $k_1 \in \mathbb{Z^{+}}$ and thus $k_1 \le k$ belongs to our set, we just apply $k_1 \leftrightarrow 2k_1+1$ to end. Otherwise if $k+1$ is even then we have $\frac{k+1}{2} \in \mathbb{Z^{+}}$ and $\frac{k+1}{2} \le k$ and thus it also belongs to our set and by applying lemma $2$ we are done. Q.E.D.

Say $x\rightarrow y$ iff $x$ being fanta implies that $y$ is fanta. Clearly $x\rightarrow y \iff y\rightarrow x$. Thus it suffices to prove that $x\rightarrow 1$ for all integers $x$. But this is trivial by strong induction on $x$, which finishes and shows that the answer is yes

is incredibly nonoptimal but it doesnt matter). Now strong induction finishes. If we have $1$ to $n$ as fantabulous. Then if $n+1$ is even we have $\frac{n+1}{2}$, but if it is odd then we have $\frac{n-1}{2}$ which finishes.