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I will show that the answer to the second question is "no". Note that if the answer to the first question is "no", we are done. Hence assume that the answer is "yes" and for every number $n$, the chain eventually turns into a cycle. Take any number $n$ and consider a sequence of numbers $n, f(n)(n-1), f(f(n)(n-1))(n-2),...$, that is $a_1=n, a_{m+1}=f(a_m)(n-m)$, for every $m=1,...,n-1$. Let $A=a_n$. Let $p$ be a prime that is bigger than any number that occurs in the chain for $A$. Consider $p^n$. Then the chain for $p^n$ looks like $p^n\to a_1p^{n-1}\to ...\to a_n\to...$ and the chain for $A=a_n$ follows. That chain will never hit the first $n$ of the numbers in the chain for $p^n$, hence the chain for $p^n$ does not go into a cycle before the step number $n$. Since $n$ was arbitrary, we are done. This answers the second question. The first question seems more difficult.

Edit. As @DanielSoltész pointed out, I answered a harder question about bounding the length of a pre-cycle in the chain $m\to f(m)\to\ldots$. If we want to show that there is no bound for the number of different elements in a chain, then assuming $p>2^{2^n-1}$ is enough. This leads to another reasonable question about bounding the lengths of cycles $m\to...\to m$. That question is open.

I will show that the answer to the second question is "no". Note that if the answer to the first question is "no", we are done. Hence assume that the answer is "yes" and for every number $n$, the chain eventually turns into a cycle. Take any number $n$ and consider a sequence of numbers $n, f(n)(n-1), f(f(n)(n-1))(n-2),...$, that is $a_1=n, a_{m+1}=f(a_m)(n-m)$, for every $m=1,...,n-1$. Let $A=a_n$. Let $p$ be a prime that is bigger than any number that occurs in the chain for $A$. Consider $p^n$. Then the chain for $p^n$ looks like $p^n\to a_1p^{n-1}\to ...\to a_n\to...$ and the chain for $A=a_n$ follows. That chain will never hit the first $n$ of the numbers in the chain for $p^n$, hence the chain for $p^n$ does not go into a cycle before the step number $n$. Since $n$ was arbitrary, we are done. This answers the second question. The first question seems more difficult.

Edit. As @DanielSoltész pointed out, I answered a harder question about bounding the length of a pre-cycle in the chain $m\to f(m)\to\ldots$. If we want to show that there is no bound for the number of different elements in a chain, then assuming $p>n^{2^n-1}$ is enough. This leads to another reasonable question about bounding the lengths of cycles $m\to...\to m$. That question is open.

I will show that the answer to the second question is "no". Note that if the answer to the first question is "no", we are done. Hence assume that the answer is "yes" and for every number $n$, the chain eventually turns into a cycle. Take any number $n$ and consider a sequence of numbers $n, f(n)(n-1), f(f(n)(n-1))(n-2),...$, that is $a_1=n, a_{m+1}=f(a_m)(n-m)$, for every $m=1,...,n-1$. Let $A=a_n$. Let $p$ be a prime that is bigger than any number that occurs in the chain for $A$. Consider $p^n$. Then the chain for $p^n$ looks like $p^n\to a_1p^{n-1}\to ...\to a_n\to...$ and the chain for $A=a_n$ follows. That chain will never hit the first $n$ of the numbers in the chain for $p^n$, hence the chain for $p^n$ does not go into a cycle before the step number $n$. Since $n$ was arbitrary, we are done. This answers the second question. The first question seems more difficult.

I will show that the answer to the second question is "no". Note that if the answer to the first question is "no", we are done. Hence assume that the answer is "yes" and for every number $n$, the chain eventually turns into a cycle. Take any number $n$ and consider a sequence of numbers $n, f(n)(n-1), f(f(n)(n-1))(n-2),...$, that is $a_1=n, a_{m+1}=f(a_m)(n-m)$, for every $m=1,...,n-1$. Let $A=a_n$. Let $p$ be a prime that is bigger than any number that occurs in the chain for $A$. Consider $p^n$. Then the chain for $p^n$ looks like $p^n\to a_1p^{n-1}\to ...\to a_n\to...$ and the chain for $A=a_n$ follows. That chain will never hit the first $n$ of the numbers in the chain for $p^n$, hence the chain for $p^n$ does not go into a cycle before the step number $n$. Since $n$ was arbitrary, we are done. This answers the second question. The first question seems more difficult.

Edit. As @DanielSoltész pointed out, I answered a harder question about bounding the length of a pre-cycle in the chain $m\to f(m)\to\ldots$. If we want to show that there is no bound for the number of different elements in a chain, then assuming $p>2^{2^n-1}$ is enough. This leads to another reasonable question about bounding the lengths of cycles $m\to...\to m$. That question is open.

I will show that the number of steps before a cycle is unbounded. Take any natural number $n$. Suppose that $p > n^{2^n}$, a prime. Then the chain starting at $p^n$ looks like: $p^n\to np^{n-1}\to f(n)(n-1)p^{n-2}\to f(f(n)(n-1))(n-2)p^{n-3}...$, so the chain hits a cycle after at least $n$ steps. This answers the second question. The first question seems more difficult.

I will show that the answer to the second question is "no". Note that if the answer to the first question is "no", we are done. Hence assume that the answer is "yes" and for every number $n$, the chain eventually turns into a cycle. Take any number $n$ and consider a sequence of numbers $n, f(n)(n-1), f(f(n)(n-1))(n-2),...$, that is $a_1=n, a_{m+1}=f(a_m)(n-m)$, for every $m=1,...,n-1$. Let $A=a_n$. Let $p$ be a prime that is bigger than any number that occurs in the chain for $A$. Consider $p^n$. Then the chain for $p^n$ looks like $p^n\to a_1p^{n-1}\to ...\to a_n\to...$ and the chain for $A=a_n$ follows. That chain will never hit the first $n$ of the numbers in the chain for $p^n$, hence the chain for $p^n$ does not go into a cycle before the step number $n$. Since $n$ was arbitrary, we are done. This answers the second question. The first question seems more difficult.