I have a matrix $A\in R^{n,n}$.
$A= \begin{bmatrix} 1&0&-2&0&0&\dots&0\\ 0&1&0&-6&0&\dots&0\\ 0&0&1&0&-12&\dots&0\\ \vdots&\vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&0&1&0&-(n-2)(n-1)\\ 0&0&0&0&0&1&0\\ 0&0&0&0&0&0&1 \end{bmatrix}$
$A$ is a matrix with $1$ on diagonal and two spots to the right is $-i(i+1)$, where $i$ is the line number.
It has a characteristic polynomial $p_A(t)=(1-t)^n$.
So jordan's form of this matrix has $1$ on diagonal, and $0$ or $1$ above it.
EDIT:
$J= \begin{bmatrix} J_1 & 0 & 0 &\dots& 0\\ 0&J_2&0&\dots&0\\ \vdots & \vdots &\ddots & \dots &0\\ 0&0&0&\dots&J_n \end{bmatrix}$
Where $J_i= \begin{bmatrix} 1 & 1 & 0&0 & \dots & 0\\ 0 &1 &1&0&\dots&0\\ 0&0&1&1&\dots&0\\ \vdots&\vdots&\dots&\ddots&\ddots&\vdots\\ 0&0&0&0&1&1\\ 0&0&0&0&0&1 \end{bmatrix}$
How do I know how many blocks $J_i$(and of what size) are used in $J$ matrix?
Simple answer: you cannot with minimal polynomial, which would tell you what is the biggest block. However without further details, you cannot go any further.
also note that Jordan's matrix is not unique, so i don't quite know what 'is it 0 or 1' above each entry means.
Edit:
let me give you an example, take n = 5,
I can only tell you the possible block size combinations are
$(1,1,1,1,1)$, $(2,1,1,1)$, $(2,2,1)$, $(3,1,1)$, $(3,2)$, $(4,1)$, $(5)$
(the first one mean 5 blocks of size 1, etc)
without knowing the minimal polynomial. Basically, it is the possible combinations are ways you can break up $n$ into sum of integers.
EDIT: for your given matrix, there are 2 jordan blocks
$J_(n-1)$ and $J_1$
To show this:
show $A^k \neq 0$ for $k < n$, so $q(t) = (1-t)^{n-1}$ is the minimal polynomial, the biggest block is size $n-1$
The only possible Jordan block decomposition is $J_{n-1}$ and $J_1$, so there are $(n-2)$ 1's and one 0 above the diagonal.