Triangle of denominators of the square of a certain lower triangular matrix. Disclaimer: This may sound like a noob question.
I am new to matrices, but I have recently ran into some math involving them while searching in the Online Encyclopedia Of Integer Sequences. More specifically, the sequence is A119948.



The formula provided is as follows:     
$$a(i,j) = denominator(r(i,j))$$ 
with
 $$r(i,j):=(A^2)[i,j]$$
where the matrix $A$ has elements 
$$a[i,j] = \left\{         \begin{array}{ll}             \frac{1}{i} & \quad j \leq i \\             0 & \quad j > i         \end{array}     \right.$$
(lower triangular)




While I have found explanations as to how this works, I can't find a source that I can comprehend without hurting my brain.
My trouble with the formula mainly stems from the following:


*

*How can $r(i,j)$ return a fraction?

*The whole thing about matrix $A$ having elements $a[i,j] = \left\{         \begin{array}{ll}             \frac{1}{i} & \quad j \leq i \\             0 & \quad j > i         \end{array}     \right.$, i.e., how can $a[i,j]$ equal a real number?

*Aside from the comprehension side of things, I'm also confused about the exact process to calculate $a(i, j)$ even if I understood the workings behind it.
 A: The matrix A is
$$
A = \left( {\matrix{
   1 & 0 &  \cdots  & 0  \cr 
   {1/2} & {1/2} &  \cdots  & 0  \cr 
    \vdots  &  \vdots  &  \ddots  &  \vdots   \cr 
   {1/n} & {1/n} &  \cdots  & {1/n}  \cr 
 } } \right)
$$
its square is
$$
A^{\,2}  = \left( {\matrix{
   1 & 0 & 0 & 0  \cr 
   {3/4} & {1/4} & 0 & 0  \cr 
   {11/18} & {5/18} & {1/9} & 0  \cr 
    \vdots  &  \vdots  &  \vdots  &  \ddots   \cr 
 } } \right)
$$
The denominators of the elements of $A^2$ read by rows (returning before the zeros) are
$1,4,4,18,18,9, ..$ which is the sequence given.
Applying the general matrix product rule, to the case of lower triangular ones, and then tour specific case we get
$$
\eqalign{
  & a^2 _{i,\,j}  = \sum\limits_{1\, \le \,l\, \le \,n} {a_{i,\,l} \,a_{l,\,j} }  = \sum\limits_{j\, \le \,l\, \le \,i} {a_{i,\,l} \,a_{l,\,j} }  =   \cr 
  &  = \sum\limits_{j\, \le \,l\, \le \,i} {{1 \over i}\,{1 \over l}}  = {1 \over i}\sum\limits_{j\, \le \,l\, \le \,i} {\,{1 \over l}}
  = {1 \over i}\left( {H_i  - H_{j - 1} } \right) \cr} 
$$
where
$$
H_i  = \sum\limits_{1\, \le \,l\, \le \,i} {\,{1 \over l}} 
$$
is the Harmonic Number. There is no "closer way" to express it, but it is related to other various "numbers",
and it has a known asymptotic relation with $\ln i$.
In particular, the diagonals will be expressed by
$$
a^2 _{j + m,\,j}  = {1 \over {j + m}}\left( {H_{j + m}  - H_{j - 1} } \right)
$$
However, the sequence you are interested in, reports the denominators of the reduced fraction
obtained from the relation above, and I do not know a practical way to express the denominators alone.
