Krdytkyu

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 :
$$left(astar bright)_{n}=c_{n}=sum_{k=0}^{n}binom{n}{k}a_{k}b_{n-k}$$
Is there a way to recover $a_{n}b_{n}$ via something akin to :
$$a_{n}b_{n}=sum_{k=0}^{n}d_{n,k}c_{k}$$

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 $$a_nstar b_n=tilde a_nstar tilde b_n=2^n$$ have the same binomial convolution, but $$a_nb_n=1neq tilde b_n=tilde a_ntilde b_n.$$ Therefore, the binomial convolution is insufficient to recover the Hadamard product.

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 $Tcolon Xto X$ as
$$
(Ta)_n=sum_{k=0}^nbinom{n}{k}a_k.
$$
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
$$
a_n=(Ta)_n-sum_{k=0}^{n-1}binom{n}{k}a_k.
$$

Given
$$
c_{,n} = ,sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {binom{n}{k} a_{,k} ,b_{,n - k} }
$$
then notice that the e.g.f.'s are in the following relation
$$
eqalign{
& C(z) = sumlimits_{0, le ,n,} {{{c_{,n} } over {n!}},z^{,n} }
= ,sumlimits_{0, le ,n,} {{1 over {n!}}sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {binom{n}{k}
a_{,k} ,b_{,n - k} ,z^{,n} } } = cr
& = ,sumlimits_{0, le ,n,} {{1 over {n!}}sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {binom{n}{k}
a_{,k} z^{,k} ,b_{,n - k} z^{,n - k} ,} } = cr
& = ,sumlimits_{0, le ,n,} {sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {{{a_{,k} } over {k!}}z^{,k}
,{{b_{,n - k} } over {left( {n - k} right)!}}z^{,n - k} ,} } = cr
& = ,left( {sumlimits_{0, le ,n,} {{{a_{,k} } over {k!}}z^{,k} } } right)left( {sumlimits_{0, le ,j,} {,{{b_{,j} } over {j!}}z^{,j} ,} } right) = cr
& = ,A(z)B(z) cr}
$$

That premised, and passing for simplicity to the ordinary g.f.
$$
C(z) = sumlimits_{0, le ,n,} {c_{,n} ,z^{,n} }
= ,sumlimits_{0, le ,n,} {sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {a_{,k} ,b_{,n - k} ,z^{,n} } } = A(z)B(z)
$$
if I understand properly your question and comment, you need to compute $D(z)$
$$
D(z) = sumlimits_{0, le ,n,} {a_{,n} b_{,n} ,z^{,n} }
$$
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
$$
D(z) = {1 over {i2pi }}ointlimits_C {A(z)B(z/v){{dv} over v}}
$$
refer to last entry of the table in this Wikipedia Article