Does a sequence based on hereditary factorisation always terminate? The well-known Goodstein sequences are based on the hereditary base-$b$ notation, where you don't just present the digits in base $b$, but also the corresponding exponents etc.
That lead me to the question what happens when you try to do the same with prime factorisation? 
First, definition of hereditary factorisation: Instead of just writing $n \in \mathbb{N}$ as $\prod{{p_i}^{a_i}}$, continue to do so with the exponents $a_i$ as well, and the exponents of their factors etc.
So, for example, $1024=2^{2*5}$
This is still a unique representation.
Next, definite a Goodstein-like sequence for every starting value $n \in \mathbb{N}$: $h_0(n)=n$, and for each $k>1$, write $h_{k-1}(n)$ in hereditary factorisation, then replace all occurances of the $k$th prime with the $k+1$st prime and substract $1$.
The question is: Does this sequence terminate (i.e. reach 1 in a finite number of steps) for all $n \in \mathbb{N}$?
Intuitively, I would expect the answer to be yes, but I don't really see a simple way to prove it.
These sequences terminating would not be half as surprising as Goodsteins, but of course the proof for his theorem doesn't remotely apply here since the structures change heavily whenever $1$ is substracted.
If the concept isn't clear, a quick example:
The sequence starting at $4$: $2^2$
Replace $2$ with $3$, substract $1$
$3^3-1=26=2*13$ 
Replace $3$ with $5$, substract $1$
$25=5^2$ 
Replace $5$ with $7$, substract $1$
$7^2-1=48=2^4*3$
Replace $7$ with $11$, substract $1$
$47$
Replace $11$ with $13$, substract $1$
$46=2*23$
Replace $13$ with $17$, substract $1$
$45=3^2*5$
Replace $17$ with $19$, substract $1$
$44=2^2*11$
Replace $19$ with $23$, substract $1$
$43$
At this point it is clear that no following prime can be a factor, so the sequence just terminates to $42, 41, 40, ..., 1$
Lastly: Is there any noteworthy known result about hereditary factorisation? While it seems like an obvious generalisation of prime factorisation (and tanget to Goodstein), I couldn't find anything when googling. Perhaps this is known under another term? I even checked OEIS for the sequence starting at 4, but it is not listed. But I can't believe that this is really such a novel idea?
 A: The sequences always terminate, and you can prove it using a method similar to the one used in the proof of Goodstein's Theorem.
Just like in Goodstein's Theorem, if we have a number $n$ and a base $p_i$, write the number $n$ in hereditary base-$p_i$ notation, and then replace each occurrence of $p_i$ with $\omega$.  Call the resulting ordinal $O(n,p_i)$.  Now, this ordinal does not seem relevant at first; for this problem, the relevant representation is hereditary factorization.  The key point is that, when we write $n$ in hereditary base-$p_i$ notation, and then bump each occurrence of $p_i$ to $p_{i+1}$, we will get at least a big a number than if we write $n$ in hereditary factorization, and bump $p_i$ to $p_{i+1}$.  Call the former number $B(n,p_i,p_{i+1})$ and the latter number $P(n,p_i,p_{i+1})$; then $B(n,p_i,p_{i+1}) > P(n,p_i,p_{i+1})-1$.  So $O(B(n,p_i,p_{i+1}),p_{i+1}) > O(P(n,p_i,p_{i+1})-1,p_{i+1})$.  But $O(B(n,p_i,p_{i+1}),p_{i+1}) = O(n,p_i)$, so we have $O(n,p_i) > O(P(n,p_i,p_{i+1})-1,p_{i+1})$; that is, the ordinal associated with our number decreases after each step of the process.  Therefore, the ordinal will reach $0$ in a finite amount of steps, which means our number will reach $0$ as well.
We still need to prove
Claim 1: If $b > a$, $B(n,a,b) \ge P(n,a,b)$.
Proof: First, note that, in the hereditary factorization of $n$, only occurrences of $a$ get replaced; all other factors can therefore be grouped together.  So the hereditary factorization is something like $n_1 a^{n_2 a^{n_3 a}}$, except that we can have any number of terms in the exponential tower.  We will use induction on the number of terms $h$ in the tower.
$h=1$: So $n = n_1 a$.  The hereditary base-$a$ representation will be $c_1 a^{b_1} + \ldots + c_i az^{b_i}$, where the $b_j$ are themselves written in hereditary base-$a$ notation, and $b_j \neq 0$ for all $j$.  That last fact means that replacing $a$ with $b$ will multiply each of the terms by at least $\frac{b}{a}$, so $B(n,a,b) \ge \frac{b}{a}n = n_1 b = P(n,a,b)$.
Assume the claim for $h$, and assume $n$ has a hereditary factorization of height $h+1$.  Then $n = n_1 a^d$, where $d$ has a hereditary factorization of height $h$.  Then $B(n,a,b) \ge B(n_1,a,b) b^{B(d,a,b)} \ge n_1 b^{P(d,a,b)} = P(n,a,b)$.
This completes the proof.
