Inverting the binomial convolution of sequences let $\left \{ a_{n} \right \}_{n=0}^{\infty}$ and $\left \{ b_{n} \right \}_{n=0}^{\infty}$ be two sequences. The binomial convolution of the two sequences is given by : 
$$\left(a\star b\right)_{n}=c_{n}=\sum_{k=0}^{n}\binom{n}{k}a_{k}b_{n-k}$$ 
Is there a way to recover $a_{n}b_{n}$ via something akin to :
$$a_{n}b_{n}=\sum_{k=0}^{n}d_{n,k}c_{k}$$ 
 A: Edit: this does not answer the question (see comments.)
Let $X$ be the space of all (real or complex) sequences $a=\{a_n\}_0^\infty$ and define $T\colon X\to X$ as
$$
(Ta)_n=\sum_{k=0}^n\binom{n}{k}a_k.
$$
What you are asking is if $a$ can be recovered from $Ta$. The answer is yes, and it can be done recursively. First of all, it is clear that $a_0=(Ta)_0$. Now, if $a_0,\dots,a_{n-1}$ have been found, we have
$$
a_n=(Ta)_n-\sum_{k=0}^{n-1}\binom{n}{k}a_k.
$$
A: Given
$$
c_{\,n}  = \,\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\binom{n}{k}  a_{\,k} \,b_{\,n - k} } 
$$
then notice that the e.g.f.'s are in the following relation
$$
\eqalign{
  & C(z) = \sum\limits_{0\, \le \,n\,} {{{c_{\,n} } \over {n!}}\,z^{\,n} }
  = \,\sum\limits_{0\, \le \,n\,} {{1 \over {n!}}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\binom{n}{k}
a_{\,k} \,b_{\,n - k} \,z^{\,n} } }  =   \cr 
  &  = \,\sum\limits_{0\, \le \,n\,} {{1 \over {n!}}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\binom{n}{k}
 a_{\,k} z^{\,k} \,b_{\,n - k} z^{\,n - k} \,} }  =   \cr 
  &  = \,\sum\limits_{0\, \le \,n\,} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {{{a_{\,k} } \over {k!}}z^{\,k}
  \,{{b_{\,n - k} } \over {\left( {n - k} \right)!}}z^{\,n - k} \,} }  =   \cr 
  &  = \,\left( {\sum\limits_{0\, \le \,n\,} {{{a_{\,k} } \over {k!}}z^{\,k} } } \right)\left( {\sum\limits_{0\, \le \,j\,} {\,{{b_{\,j} } \over {j!}}z^{\,j} \,} } \right) =   \cr 
  &  = \,A(z)B(z) \cr} 
$$
That premised, and passing for simplicity to the ordinary g.f.
$$
C(z) = \sum\limits_{0\, \le \,n\,} {c_{\,n} \,z^{\,n} }
  = \,\sum\limits_{0\, \le \,n\,} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {a_{\,k} \,b_{\,n - k} \,z^{\,n} } }  = A(z)B(z)
$$
if I understand properly your question and comment, you need to compute $D(z)$
$$
D(z) = \sum\limits_{0\, \le \,n\,} {a_{\,n} b_{\,n} \,z^{\,n} } 
$$
knowing $A(z)$ and $B(z)$ .
Then, assuming that these are convergent in a domain of non-null measure, you can use the 
duality of the Convolution Theorem , same as for the Fourier series, which for the z-Transform 
reads as
$$
D(z) = {1 \over {i2\pi }}\oint\limits_C {A(z)B(z/v){{dv} \over v}} 
$$
refer to last entry of the table in this Wikipedia Article
A: Let 


*

*$a_n=1$, 

*$b_n=1$, 

*$\tilde a_n=2^n$, and 

*$\tilde b_0=1$, $\tilde b_n=0$ for all $n>0$.


Note that $$a_n\star b_n=\tilde a_n\star \tilde b_n=2^n$$ have the same binomial convolution, but $$a_nb_n=1\neq \tilde b_n=\tilde a_n\tilde b_n.$$ Therefore, the binomial convolution is insufficient to recover the Hadamard product. 
