I am curious about the following.


Let $K$ be a number field. For any $a \in \mathcal{O}\_K$ in its ring of integers, let $N(a)$ be zero if there exist elements $b, c \in \mathcal{O}\_K \setminus \mathcal{O}\_K^\*$ such that $a = bc$ and $b$ has at least two different (that is, nonassociate) factorizations into irreducible factors (over $\mathcal{O}\_K$), otherwise let $N(a)$ be the number of different factorizations of $a$ into irreducible factors. Is the set $\{N(a) | a \in \mathcal{O}\_K\}$ unbounded for some $K$? Is it bounded for some $K$ that has nontrivial ideal class group? 


Similarly interesting is the same question with $N(a)$ defined to be zero if and only if $a=bc$ and both $b,c$ have at least two different factorizations. I have asked this question earlier on math.stackexchange forum (<https://math.stackexchange.com/questions/712482/on-number-of-different-factorizations-over-integers-of-a-number-field>), but have not received an answer.



Distinct primitive factorizations over integers of number fields
The quantity in question can be bounded in terms of the class number only. I will show that it is at most the number of partitions of $h^2$, where $h$ is the class number. (This could be improved in several ways.)


Let $a \in \mathcal{O}\_K$. Suppose $N(a) > 1$. Clearly $a$ is neither irreducible nor invertible.


Let $c$ be an irreducible dividing $a$. Then $c^{-1}a$ has an (essentially) unique factorization, since otherwise $N(a)=0$. 


Let $a= c\_1\dots c\_n = c\_1'\dots c\_m'$ be two essentially distinct factorizations, which exist by assumption. By the just made observation we have that $(c\_i) \neq (c\_j')$ for all $i,j$; and this is true for any two distinct factorizations. 


Let $(c\_1) = P\_1 \dots P\_s$ be a factorization into prime ideals. There exists a subset $J$ of $\{1, \dots , m\}$ of cardinality at most $s$ such that $c\_1$ divides $\prod\_{j \in J} c\_j' = d$. (This follows by the uniqueness of the factorization into prime ideals and picking, if necessary, for each $P\_j$ a $c\_j'$ that contains it in its factorization.)


Now, consider a factorization into irreducibles of $d$ that contains $c\_1$, say $d=c\_1c\_2'' \dots c\_k''$. Then $c\_1c\_2'' \dots c\_k'' \prod\_{j \notin J} c\_j'$ is a factorization of $a$. This factorization contains $c\_1$ and is thus (essentially) equal to $c\_1\dots c\_n$. Since $(c\_i) \neq (c\_j')$, this is only possible if $\prod\_{j \notin J} c\_j'$ is empty, that is $J = \{1, \dots , m\}$. Yet this means $m \le s$.


Now, note that the number of prime ideals in a factorization of an irreducible is at most the class number $h$. [See the other answer for details.] 


This tells us two things. First, that $m \le s \le h$, and then that the total number $q$ of prime ideals (with multiplicity) in the factorization into prime ideal of $(a)$ is at most $mh \le h^2$. 


Since there certainly cannot be more essentially distinct factorizations of $a$ than there can be partitions of a set of cardinality $q$ we see that $N(a)$ is at most the number of partitions of $h^2$. 


