An eely function $\mu (n):\;\;\prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (k)} \right)} = 1$ Time ago, dealing with a generalization of the Stirling numbers, I stumbled on the following implicit recurrence
$$
\mu (n):\;\;\prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (k)} \right)}  = 1\quad \left| \matrix{
  \,0 \le n \in Z \hfill \cr 
  \,0 \le \mu (n) \in R \hfill \cr 
  \,\mu (0) = 0 \hfill \cr}  \right.
$$
Using a good CAS it is not difficult to compute the first few values and plot them

Clearly, the sequence is monotonically increasing, and its finite difference is monotonically decreasing (1).
Such a regular behaviour leads me to expect that it might be extended over the reals and that
$\mu (x)$ might be expressible through a combination of conventional functions.
From time to time I am returning to this challenge with some inspiration for a new approach, but
the combination difference & product has frustrated all my attempts.
I couldn't even succeed to establish its asymptotic behaviour (now, thanks to @AMarino answer I know it's logarithmic) .
So I am asking for hints, suggestions.
-- addendum  --
Putting $\rho _{\,n,\,m}  = \mu _{\,n + 1}  - \mu _{\,m} $ , an alternative way to express the problem is
$$
\left\{ \matrix{
  \prod\limits_{0\,\, \le \,k\, \le \,n} {\rho _{\,n,\,k} }  = 1 \hfill \cr 
  \rho _{\,n,\,k}  - \rho _{\,n - 1,\,k}  = \mu _{\,n + 1}  - \mu _{\,n}  \hfill \cr}  \right.
$$
which means to find a family of functions whose product wrt $k$ is $1$
and whose difference wrt $n$ is constant, as shown

My last tentative has been to take two discrete pmf's on the support $[0,n]$ and put
$$
\rho \left( {n,m} \right) = e^{\,h(n)\left( {p(m\,|n) - q(m\,|n)} \right)} 
$$
which by definition gives the unitary product, but cannot go  yet through keeping the difference constant.
-- note 1 --
That the difference is monotonically decreasing comes from being
$$
\eqalign{
  & \prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (k)} \right)}
  = \prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (n - 1) + \mu (n - 1) - \mu (k)} \right)}  =   \cr 
  &  = \prod\limits_{k = 0}^{n - 1} {\left( {x + \left( {\mu (n - 1) - \mu (k)} \right)} \right)}
  = p_n (x)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  p_n (0) = 0 \hfill \cr 
  p_n (x) < p_{n + 1} (x)\quad \left| {\,0 < x} \right. \hfill \cr 
  1 = p_n (\mu (n) - \mu (n - 1)) < p_{n + 1} (\mu (n + 1) - \mu (n)) \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \mu (n + 1) - \mu (n) < \mu (n) - \mu (n - 1) \cr} 
$$
which tells the interesting fact that $\Delta \mu (n-1)$ comes as the root of $p_n(x)=1$, which in turn is added
to its zeros, shifting them to the left as to start from $0$ and making them the zeros of $p_{n+1}(x)$.
 A: This is only a partial answer.
If your claim is correct, i.e. differences are decreasing, then the sequence is sublogarithmic. Indeed, let $s(n) := \mu(n+1) -\mu(n) $. The recurrence relation, using the claim $s(j) > s(n)$ for $j<n$, rewrites as
$$ 1 = \prod_{k=0}^n (\sum_{j=k}^n s(j) ) > \prod_{k=0}^n (n-k+1) s(n) = (n+1)! s(n) ^{n+1} $$
Which implies that
$$ s(n) < \left ( \frac{1}{(n+1)! } \right ) ^{1/(n+1) }$$
The Stirling approximation can be written as $m! \approx (m/e) ^{m}$. Substituting we get that $s(n) < \frac{e}{n+1}$ . Thus we have
$$ \mu(n) = s(n-1) +\ldots+s(0) < $$
$$ e \sum_{k=0}^{n-1} \frac{1}{k+1} \approx e \log(n) $$
We also have a sort of converse to this. Suppose that for all n we have
$$(*) \ \ s(n) > \left ( \frac{1}{(n+2)! } \right ) ^{1/(n+2) }$$
That is, the sequence is also superlogarithmic. We want to show that differences are decreasing. Suppose that at some point $n$ we have for the first time that the difference increase, aka $s(n) \ge s(n-1) $. Then we get
$$ 1 = \prod_{k=0}^n (\sum_{j=k}^n s(j) ) \ge \prod_{k=0}^n (n-k+1) s(n-1) = (n+1)! s(n-1) ^{n+1} $$
From which we get, using again Stirling approximation, the opposite (*) inequality, yielding a contradiction.
This argument can be made precise with inequalities instead of plain Stirling approximation, but since it is not a strong result with respect to your requests, I think it's not worth. Hope it can inspire someone else for a more sophisticated approach :)
Edit. I have an approach for the lower bound. By substituting each factor with the biggest $s$ appearing we get a reversed inequality
$$ 1\le n s(0) \cdot  (n-1)s(1) \ldots 1 \cdot s(n-1) $$
Rearranging terms and taking the logarithms we get
$$ 0 \le \log ns(n-1) + \ldots +\log 1\cdot s(0) $$
Let us define
$$\sigma(n) := \frac{1}{n} \sum_{j=0} ^{n-1} \log (j+1) s(j) $$
Equation above ensures that $\limsup \sigma(n) \ge 0$. Now the plot twist: by Stolz Cesaro the limsup of the general term is greater or equal than the average one, so that
$$ \limsup \log s(n) (n+1) \ge \limsup \sigma(n) \ge 0$$
Which implies that
$$ \limsup s(n) (n+1) \ge 1$$
Which provides a lower bound like estimate. If this can be refined to be a limit, than by summing over n we would get that $\mu(n) \ge \log(n) $ asymptilotically.
