Solve $\frac{1}{2^\theta}\sum_{k=0}^{\theta} {\theta\choose k} \delta(k)=\theta$ for $\delta$ The following arises in unbiased estimation of a parameter for the binomial distribution, but that information is not needed for solving the question. I tried solving this by manipulating the sum to get a recursion, but failed to do so.

Find an explicit function $\delta:\{0,1,\dots\}\to\{1,2,\dots\}$ such that the following holds for all $\theta\in\{1,2,\dots\}$:
$$
 \frac{1}{2^\theta}\sum_{k=0}^{\theta} {\theta\choose k} \delta(k)=\theta
$$

EDIT
Note that the usual unbiased estimator for this binomial parameter, i.e. the function $\delta(x)=2x$ is not a solution since $\delta$ may not map to $0$. Considering $\theta = 1$ leads to $\delta(0)=\delta(1)=1$, whence considering $\theta=2$ gives that 
$$
\delta(0) + 2 \delta(1) + \delta(2) = 8 \iff \delta(2)=5,
$$
and so on one may continue to find $\delta(k)$ for every $k$, but I cannot find the pattern and solve to get an explicit function.
 A: Let's introduce the following Pascal inversion formula:
$$
{b}_{n}=\sum_{k=0}^{n}{n\choose k}a_k\\
a_n={(-1)}^{p}\sum_{k=0}^{n}(-1)^k
{n\choose k}a_k
$$
So:
$$
\sum_{k=0}^{\theta}{\theta\choose k}\delta{(k)} = {2}^{\theta}\theta
$$
Using the latter, we get:
$$
\delta{(\theta)}={(-1)}^{\theta}\sum_{k=0}^{\theta}(-1)^k
{\theta\choose k}k2^k
$$
Let's evaluate the latter sum:
We define $y$ such that:
$$
y=\sum_{k=0}^{\theta}x^k
{\theta \choose k}={(1+x)}^{\theta}\\
y'=\sum_{k=0}^{\theta}k{x}^{k-1}
{\theta \choose k}=\theta {(1+x)}^{\theta-1}\\
\sum_{k=0}^{\theta}k{x}^{k}
{\theta \choose k}=xy'=x\theta {(1+x)}^{\theta-1}
$$
Now let $x=-2$ and you'll get:
$$
\sum_{k=0}^{\theta}k{(-2)}^{k}
{\theta\choose k}=2\theta{(-1)}^{\theta}
$$
So:
$$
\delta{(k)}=2k
$$
A: Oussama Boussif,
in a truly nice use
of Pascals's inversion formula,
showed that
$\delta(k)
= 2k
$ is a solution.
However,
the OP said that
the solution must have
$\delta(k)
\in \{1, 2, ...\}
$.
To satisfy this,
use
$\sum_{k=0}^{\theta} (-1)^k \binom{\theta}{k}
= 0
$.
Therefore,
for any real $a$,
$\delta(k)
=2k+a(-1)^k
$
is a solution.
Choosing
$a = 1$
gives
$\delta(k)
=2k+(-1)^k
= 1, 1, 5, 5, 9, 9, ... 
$.
If the results have to be distinct,
I would have to think some more.
