Checkboard matrix, brand new or old? Ok so what I found was a square matrix of order $n×n$ where $n$ follows $2m+1$ and $m$ is a natural number
the pattern these matrices follow is as follows:
for a $3×3$ matrix:
$$ A = \left( \begin{array}{ccc}
b & a & c \\
a & a+b+c & a \\
c & a & b \end{array} \right)$$
and the cool thing is that
$$ A^x = \left( \begin{array}{ccc}
q & p & r \\
p & p+q+r & p \\
r & p & q \end{array} \right)$$
where $ x $ is any natural number
Now for $5×5$ it goes like
$$ A = \left( \begin{array}{ccc}
b & b & a & c & c\\
b & b & a & c & c\\
a & a & 2b+a+2c & a & a\\
c & c & a & b & b\\
c & c & a & b & b\end{array} \right)$$
and again
$$ A^x = \left( \begin{array}{ccc}
q & q & p & r & r\\
q & q & p & r & r\\
p & p & 2q+p+2r & p & p\\
r & r & p & q & q\\
r & r & p & q & q\end{array} \right)$$
Once more for a $7×7$ matrix we have
$$ A = \left( \begin{array}{ccc}
b & b & b & a & c & c & c\\
b & b & b & a & c & c & c\\
b & b & b & a & c & c & c\\
a & a & a & 3b+a+3c & a & a & a\\
c & c & c & a & b & b & b\\
c & c & c & a & b & b & b\\
c & c & c & a & b & b & b\\
\end{array} \right)$$
and then 
$$ A^x = \left( \begin{array}{ccc}
q & q & q & p & r & r & r\\
q & q & q & p & r & r & r\\
q & q & q & p & r & r & r\\
p & p & p & 3q+p+3r & p & p & p\\
r & r & r & p & q & q & q\\
r & r & r & p & q & q & q\\
r & r & r & p & q & q & q\\
\end{array} \right)$$
and so on and so forth, I didn't find whether this has already been observed and neither a name for this so I am calling this checkboard matrix, because it looks like that.
I just have one question, has this been found already, if yes please give details.
Addendum: Thanks to Robert Israel for pointing out the error,  I have made some changes,  please check into it.  The pattern for the middle element is $ m (b+c)+a $

Here's my attempt to explain the pattern with stuff already known:
We can express the matrix as
$$
A = \pmatrix{
bX & ax & cX\\
ax^T & (a+b+c) & ax^T\\
cX & ax & bX}
$$
Where $x = (1,\dots,1)^T \in \Bbb R^n$, and $X = xx^T$.  Note that $Xx = \|x\|^2x = nx$.  
Extend $u_1 = x/\sqrt{n}$ into an orthonormal basis $u_1,\dots,u_n$.  Let $U$ be the $n \times n$ matrix with columns $u_i$.
Apply the change of basis $S^{-1}AS = S^TAS$ where
$$
S = \pmatrix{U\\&1\\&&U}
$$
We find
$$
S^{-1}AS = 
\pmatrix{
nbE_{11} & \sqrt n a e_1 & cn E_{11}\\
a\sqrt n e_1^T & (a+b+c) & a\sqrt n e_1^T\\
cnE_{11} & a \sqrt n e_1 & b n E_{11}
}
$$
where $e_1 = (1,0,\dots,0)^T$ is the first standard basis vector, and $E_{11} = e_1e_1^T$.
You should find that this new matrix has the same properties, and quite a bit less of the mystery.
So, in the $5 \times 5$ case, we're left with
$$
S^{-1}AS = \left( \begin{array}{ccc}
2b        & 0 & \sqrt 2 a & 2c & 0\\
0         & 0 &          0 & 0 & 0\\
\sqrt 2 a & 0 & a+b+c      & \sqrt 2a & 0\\
2c        & 0 & \sqrt 2a  & 2b & 0\\
0         & 0 & 0          & 0 & 0\end{array} \right)
$$
