Inverse of a matrix with binomial entries This is closely related to this question: Eigenvalues of a matrix with binomial entries.
We consider the matrix:
$$M_{ij} = 4^{-j}\binom{2j}{i}$$
where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k<0$ or $k>m$. The indices $i,j$ traverse a discrete finite range, $i,j \in \{a, a+1, \dots, b\}$, from $a$ to $b$, where $a,b$ are non-negative integers with $0\le a\le b$. Therefore the matrix $M_{ij}$ has dimensions $(b-a+1) \times (b-a+1)$.
Can we find the inverse matrix, $M^{-1}$? Numerical computations of the determinant suggest that this is a non-singular matrix (for all $0 \le a < b$).
A close-form expression for the determinant could be useful.
Why this question is not a duplicate: Although in principle the eigenvalues and eigenvectors of a matrix are enough to invert it, the other question focuses on the largest positive eigenvalue alone. Moreover the eigenvalues/eigenvectors (not even the largest alone) have not been solved, so maybe finding the inverse of this matrix turns out to be easier. So, if the other question suddenly received a complete response and all the eigenvalues / eigenvectors were found, then yes, this question would be automatically solved. But that does not seem likely to happen.
 A: Let's refer everything to square matrices indexed from $0$ to $h$, that I will denote as
$$
{\bf M}_{\,h}  = \left\| {\;f(n,m)\;} \right\|_{\,h} 
$$
with $n$ being the row index and $m$ the column index.
I will then denote by
$$
\left( {f(n) \circ {\bf I}_{\,h} } \right)
$$
the diagonal matrix whose entries are equal to $f(n)$.
So I write  the matrix you proposed as
$$ \bbox[lightyellow] {  
{\bf M}_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h}  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h} 
 }$$
where $h=b-a$.
That premised, consider that in general
$$
\eqalign{
  & \left( \matrix{
  r\,n + t \cr 
  m + q \cr}  \right) = {{\left( {r\,n + t} \right)^{\,\underline {\,m + q} } } \over {\left( {m + q} \right)!}} = {{\left( {r\,n + t} \right)^{\,\underline {\,q} } \left( {r\,n + t - q} \right)^{\,\underline {\,m} } } \over {\left( {m + q} \right)^{\,\underline {\,q} } \;m^{\,\underline {\,m} } }} =   \cr 
  &  = \left( {r\,n + t} \right)^{\,\underline {\,q} } \left( \matrix{
  r\,n + t - q \cr 
  m \cr}  \right){1 \over {\left( {m + q} \right)^{\,\underline {\,q} } }} \cr} 
$$
where $x^{\,\underline {\,a} } $ denotes the falling factorial ($x^{\overline {\,a\,} } $ the rising) and
where, for the present problem, we consider $q$ to be a non-negative integer,
while $r$ and $t$ could be real (or even complex).
Then we have that we can write the binomial as
$$
\begin{gathered}
  \left( \begin{gathered}
  r\,n + s \\ 
  m \\ 
\end{gathered}  \right) = \frac{1}
{{m!}}\left( {r\,n + s} \right)^{\,\underline {\,m\,} }  = \frac{1}
{{m!}}\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right)} {\left( \begin{gathered}
  m \\ 
  k \\ 
\end{gathered}  \right)s^{\,\underline {\,m - k\,} } \left( {r\,n} \right)^{\,\underline {\,k\,} } }  =  \hfill \\
   = \frac{1}
{{m!}}\sum\limits_{\left\{ \begin{subarray}{l} 
  \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ 
  \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) 
\end{subarray}  \right.} {\left( \begin{gathered}
  m \\ 
  k \\ 
\end{gathered}  \right)s^{\,\underline {\,m - k\,} } \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right]r^{\,j} n^{\,j} }  =  \hfill \\
   = \sum\limits_{\left\{ \begin{subarray}{l} 
  \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ 
  \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) 
\end{subarray}  \right.} {n^{\,j} r^{\,j} \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right]\frac{1}
{{k!}}\left( \begin{gathered}
  s \\ 
  m - k \\ 
\end{gathered}  \right)}  \hfill \\ 
\end{gathered} 
$$
Then in the last line we can replace $n^m$ with
$$
n^{\,m}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left\{ \matrix{
  m \cr 
  k \cr}  \right\}n^{\,\underline {\,k\,} } }  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left( \matrix{
  n \cr 
  k \cr}  \right)k!\left\{ \matrix{
  m \cr 
  k \cr}  \right\}} 
$$
Thus we arrive finally to
$$ \bbox[lightyellow] {  
\eqalign{
  & {\bf M}_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h}  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h}  =   \cr 
  &  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} }  \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{
  2n + a \cr 
  m \cr}  \right)\;} \right\|_{\,h} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}_{\,h} } \right) =   \cr 
  &  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} }  \circ {\bf I}_{\,h} } \right)\;{\bf B}_{\,h} \left( {n! \circ {\bf I}_{\,h} } \right)\;\overline {{\bf St}_{{\bf 2}\,h} } \left( {2^{\,n}  \circ {\bf I}_{\,h} } \right)\;\overline {{\bf St}_{{\bf 2}\,h} } ^{\,{\bf  - }\,{\bf 1}} \left( {n! \circ {\bf I}_{\,h} } \right)^{\,{\bf  - }\,{\bf 1}} \left( {{\bf I}_{\,h}  + \overline {{\bf E}_{\,h} } } \right)^{\,{\bf a}} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}_{\,h} } \right) \cr} 
\tag{1} }$$
with
$$
\eqalign{
  & {\bf B}_{\,h}  = \;\left\| {\;\left( \matrix{
  n \cr 
  m \cr}  \right)\;} \right\|_{\,h} \quad {\bf St}_{{\bf 2}\,h}  = \;\left\| {\;\left\{ \matrix{
  n \cr 
  m \cr}  \right\}\;} \right\|_{\,h} \quad {\bf I}_{\,h}  + {\bf E}_{\,h}  = \;\left\| {\;\left( \matrix{
  1 \cr 
  n - m \cr}  \right)\;} \right\|_{\,h}   \cr 
  & \overline {\bf X}  = transpose({\bf X}) \cr} 
$$
After that the determinant follows easily, since the matrices other than the diagonal ones
have unitary determinant
$$ \bbox[lightyellow] {  
\left| {\,{\bf M}_{\,h} (a)\,} \right| = \left( {\prod\limits_{0\, \le \,n\, \le \;h} {{{\left( {2\,\left( {n + a} \right)} \right)^{\,\underline {\,a} } } \over {2^{\,n + 2a} \left( {n + a} \right)^{\,\underline {\,a} } }}} } \right) = \left( {\prod\limits_{0\, \le \,n\, \le \;h} {{{\left( \matrix{
  2\,\left( {n + a} \right) \cr 
  a \cr}  \right)} \over {2^{\,n + 2a} \left( \matrix{
  n + a \cr 
  a \cr}  \right)}}} } \right)
 \tag{2}}$$
Some notes concerning the inversion of identity (1), and further analysis  you might possibly want perform on that.
For the Binomial
$$
{\bf B}_{\,h} ^{\,{\bf r}}  = \;\left\| {\;r^{\,n - m} \left( \matrix{
  n \cr 
  m \cr}  \right)\;} \right\|_{\,h}  = \left( {r^n  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \;\left( {r^n  \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}} \quad \;\left| {\;r \in R,C} \right.
$$
where the second expression for $r=0$ is understood to be taken in the limit.
So
$$
{\bf B}_{\,h} ^{\, - \,{\bf 1}}  = \left( {\left( { - 1} \right)^n  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \;\left( {\left( { - 1} \right)^n  \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}}  = \left( {\left( { - 1} \right)^n  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \;\left( {\left( { - 1} \right)^n  \circ {\bf I}_{\,h} } \right)
$$
For the Stirling Numbers,  1st and 2nd kind are related by
$$
{\bf St}_{{\bf 2}\,h} ^{\, - \,{\bf 1}}  = \;\left\| {\;\left( { - 1} \right)^{\,n - m} \left[ \matrix{  n \cr   m \cr}  \right]\;} \right\|_{\,h}  =
 \left( {\left( { - 1} \right)^n  \circ {\bf I}_{\,h} } \right)\;{\bf St}_{{\bf 1}\,h} \;\left( {\left( { - 1} \right)^n  \circ {\bf I}_{\,h} } \right)
$$
${\bf E}$ is the "shift", "first off-diagonal", .. matrix, i.e:
$$
{\bf E}_{\,h}  = \left\| {\;\left[ {1 = n - m} \right]\;} \right\|_{\,h}  = \left\| {\;\left( \matrix{  0 \cr 
  n - m - 1 \cr}  \right)\;} \right\|_{\,h} 
$$
(where $[P]$ is the Iverson bracket)   
then
$$
\eqalign{
  & {\bf E}_{\,h} ^{\,{\bf q}}  = \left\| {\;\left[ {q = n - m} \right]\;} \right\|_{\,h}  = \left\| {\;\left( \matrix{
  0 \cr 
  n - m - q \cr}  \right)\;} \right\|_{\,h} \quad \;\left| {\;0 \le q \in Z} \right.  \cr 
  & \left( {{\bf I}_{\,h}  + {\bf E}_{\,h} } \right) = \left\| {\;\left[ {0 \le n - m \le 1} \right]\;} \right\|_{\,h}  = \left\| {\;\left( \matrix{
  1 \cr 
  n - m \cr}  \right)\;} \right\|_{\,h}   \cr 
  & \left( {{\bf I}_{\,h}  + {\bf E}_{\,h} } \right)^{\,{\bf r}}  = \sum\limits_{0\, \le \,k\,\left( { \le \;h} \right)} {\left( \matrix{
  r \cr 
  k \cr}  \right){\bf E}_{\,h} ^{\,{\bf k}} }  = \left\| {\;\left( \matrix{
  r \cr 
  n - m \cr}  \right)\;} \right\|_{\,h} \quad \;\left| {\;r \in R,C} \right. \cr} 
$$
and finally that ${\bf B}$ and ${\bf {I+E}}$ are tied by the similarity
$$
{\bf B}_{\,h} \; = \left( {{\bf St}_{\,{\bf 2}\,h} \left( {n! \circ {\bf I}_{\,h} } \right)} \right)\left( {{\bf I}_{\,h}  + {\bf E}_{\,h} } \right)\left( {{\bf St}_{\,{\bf 2}\,h} \left( {n! \circ {\bf I}_{\,h} } \right)} \right)^{{\bf  - 1}} 
$$
and by a bunch of other relations, among which
$$
\left( {{\bf I}_{\,h}  + {\bf E}_{\,h} } \right)^{\,{\bf  - q}} \quad \left| {\;0 \le {\rm integer }q} \right.\quad  = \left( {{\bf B}_{\,h} \left( {n^{\,\underline {\, - q\,} }  \circ {\bf I}_{\,h} } \right)} \right)^{\,{\bf  - 1}} \;\left( {\left( {n^{\,\underline {\, - q\,} }  \circ {\bf I}_{\,h} } \right)\;\;{\bf B}_{\,h} } \right)
$$
