Something like : "recursive" harmonic numbers? Where can I read more? In my other thread I discussed a matrix-decomposition; for one matrix (U) I found now a description of its entries, which may best be denoted as "recursive harmonic numbers". However, googling with this term leads to hits on a different thing. What I have is the following.    
Define a recursive type of harmonic-numbers of two arguments:      
$ \qquad \small h_{0,c} =1   \text{      for all } c\gt 0 $
$ \qquad \small h_{r,1} =1   \text{      for all } r\gt 0  $
$ \qquad \small h_{r,c} = h_{r,c-1} + h_{r-1,c}/c \text{      for all } c\gt r $     
Then we have
$\qquad \small h_{0,*} = (1,1,1,1,1,\ldots) $
$\qquad \small h_{1,*} = (1,
                          1+{1 \over 2}, 
                          1+{1 \over 2}+{1 \over 3},
                          1+{1 \over 2}+{1 \over 3}+{1 \over 4},
                          \ldots) $
$\qquad \small h_{2,*} = (1,
                          1+ {1 \over 2}\left( 1+{1 \over 2}\right), 
                          1+ {1 \over 2}\left(1+{1 \over 2}\right)
                            +{1 \over 3}\left(1+{1 \over 2}+{1 \over 3},\right),
                          \ldots) $
where the recursion becomes visible by
$\qquad \small h_{2,*} = (1,
                          1+ {1 \over 2} h_{1,2}, 
                          1+ {1 \over 2} h_{1,2} + {1 \over 3} h_{1,3} 
                          \ldots) $        
Is someone aware of such a generalization/extension? I have a vague idea that I've seen the terms of $\small h_{2,*} $ in some online available article not too long ago (arXiv?) but can't remember exactly. Also with "generalized harmonic numbers" mostly is meant either the summation to fractional bounds of the index and/or higher exponents in the denominator of the harmonic numbers, so this is different...

I've found the arXiv-article. However, the recursive harmonic numbers are only mentioned as intermediate terms (pg 6, eq (6)) for a generalization of the Stirling numbers first kind (which is in the context of my own discussion not surprising). If there are some more references dealing more specific with these, this were very good...
 A: Your recursion can be expressed as 
$$h_{r,c} = \sum_{i=1}^c \frac{h_{r-1,i}}{i},$$
which, unrolled, becomes the nested sum
$$h_{r,c} = \sum_{i_r = 1}^c \frac{1}{i_r} \left(\sum_{i_{r-1} = 1}^{i_r} \frac{1}{i_{r-1}} \left(\cdots \sum_{i_1=1}^{i_2} \frac{1}{i_1} \right)\right) = \sum_{i_r = 1}^c \sum_{i_{r-1} = 1}^{i_r} \cdots \sum_{i_1=1}^{i_2} \frac{1}{i_1 i_2 \cdots i_r}.$$
If we reverse the order of summation, we get
$$h_{r,c} = \sum_{i_1 = 1}^c \sum_{i_2 = i_1}^c \cdots \sum_{i_r=i_{r-1}}^c \frac{1}{i_1 i_2 \cdots i_r},$$
which is almost identical to an expression I asked about  on MO a while back.  A slight modification of Richard Stanley's answer there shows that
$$h_{r,c} = Z_c(H_c, H^{(2)}_c, \ldots, H^{(c)}_c),$$ where $Z_c(x_1, x_2, \ldots, x_c)$ is the $c$th cycle index polynomial of the symmetric group.  In other words, your $h_{r,c}$ has the same expression in terms of the generalized harmonic numbers $H^{(r)}_n$ as my $S(n,k)$ except that $h_{r,c}$ has  all of the terms positive.
For instance, we have 
$$\begin{align}
h_{2,c}&= \frac{1}{2}\left(H_c^2 + H^{(2)}_c \right), \\
h_{3,c} &= \frac{1}{6}\left(H_c^3 + 3H_c H^{(2)}_n + 2 H_c^{(3)}\right),\\
h_{4,c} &= \frac{1}{24}\left(H_c^4 + 6 H^2_c H_c^{(2)} + 3 (H_c^{(2)})^2 + 8 H_c H_c^{(3)}  + 6 H_c^{(4)}\right). \end{align}$$
See also the references in my question.
The $h_{r,c}$ values can also be expressed in terms of the generalized Stirling numbers in the Loeb paper you cite.  See, for example, the recursion (3) in Loeb's paper, which is the same as that for $h_{r,c}$.  It's just the initial conditions that are different.  Specifically, $h_{r,c} = (-1)^r c! s(-c,r)$, in which case Proposition 2.3 in Loeb's paper says that 
$$h_{r,c} = -\sum_{m=1}^c \binom{c}{m} (-1)^m m^{-r}.$$
A: (This is more of an extended comment.)
I'll only note that your $h_{r,c}$ are easily generated as the first column of the product of a diagonal matrix and a lower triangular matrix. More precisely, the matrix that generates your $h_{r,c}$ can be expressed as $\mathbf D\mathbf L^{r+1}$, where $\mathbf D=\mathrm{diag}(1,2,\dots,n)$ and
$$\ell_{j,k}=\begin{cases}\frac1{j}&\text{if }j\geq k\\0&\text{otherwise}\end{cases}$$
$\mathbf L$, which is apparently called the "lower triangular mean matrix", looks a bit like this:
$$\mathbf L=\begin{pmatrix}1\\\frac12&\frac12\\\frac13&\frac13&\frac13\\\vdots&\vdots&\vdots&\ddots\\\frac1{n}&\cdots&\cdots&\cdots&\frac1{n}\end{pmatrix}$$
$h_{2,c}$ is expressible in terms of generalized harmonic numbers: $\dfrac{(H_c)^2+H_c^{(2)}}{2}$, where $H_c^{(k)}=\sum\limits_{j=1}^c\frac1{j^k}$. I haven't been able to find a simpler form for $h_{3,c}$ and others.
