converting geometric infinite series to another infinite series Let $\{x_n\}_{n=1}^\infty$ be a sequence satisfying the recurrence relation:
$$ x_n = a\left(1- \sum_{k=0}^{n-1}x_k\right) $$
Where $ x_0 = 1 $, and $a \in [0,1]$ is chosen so that
$$ 
\sum_{k=1}^{\infty} x_k = 1$$
Given a positive integer $d$, how do I generate a sequence $\{y_n\}$ such that
$$ \sum_{k=p}^{p+(d-1)} y_k = x_{\frac {p+(d-1)}d}$$
for example, if $a = 0.5$ and $d = 4$,
$$ \sum_{k=1}^{4} y_k = x_1 = 0.5 $$
and
$$ \displaystyle \sum_{k=5}^{8} y_k = x_2 = 0.25 $$
I originally thought that this would be related to compound interest, but doing arithmetic by hand, have not found this to be the case.
I have limited mathematical knowledge, so if the answer requires anything beyond algebra, please explain or cite references to the form you are using. If the title of the question can be made clearer, please feel free to edit.
My use case is a computer application that will calculate 
$ g(x) $ from $x = 0$, so iterative solutions work for me.
 A: Note that so long as $-1 < a < 1$, we have
$$
\sum_{x = 1}^\infty x a^x = 
\left(\sum_{x = 0}^\infty (x+1) a^x\right) - \left(\sum_{x=0}^\infty a^x\right) \\= 
\frac{1}{(1-a)^2} - \frac{1}{1-a} = 
\frac{a}{(1-a)^2}
$$
So, in order to have $\sum_{x=1}^\infty f(x) = 1$, we would need to have $a$ such that $\frac a{1-a} = 1 \implies a = \frac {3 - \sqrt 5}2$.
As for your function $g$: it seems that for a fixed positive integer $d$, you require
$$
\sum_{x = p}^{p+d-1} g(x) = f\left(\frac{p + d - 1}{d}\right)
$$
In the case of $d = 4$, this precisely means that
$$
g(1) + \cdots + g(4) = f(1)\\
g(5) + \cdots + g(8) = f(2)\\
\vdots \\
g(4n + 1) + \cdots + g(4n+4) = f(n+1)
$$
One definition of $g$ that would work is
$$
g(x) = \frac 14 f\left( 1 + \lfloor (x-1)/4 \rfloor \right)
$$
Where $\lfloor x \rfloor$ is the "floor" or "greatest integer" function.  It seems that there are infinitely many suitable solutions, and this is one of them.

As for the latest iteration of your question: let $x[k]$ denote $x_k$, because subscripts are annoying.  If we are given a sequence $\{x_k\}$ satisfying those properties, then if you define
$$
y[k] = \frac 1d x\left[ 1 + \lfloor (k-1)/d \rfloor \right]
$$
then the resulting sequence $\{y_n\}_{n=1}^\infty$ should do the job.  This works regardless of the conditions on $\{x_n\}.$
A: For the beginning, rewrite $$f(x) = xa^x=a x a^{x-1}=a\frac{d(a^x)}{da}$$ So, $$\displaystyle \sum_{x=1}^{\infty} f(x) = a \sum_{x=1}^{\infty}\frac{d(a^x)}{da}=a\frac{d}{da}\sum_{x=1}^{\infty}a^x=a\frac{d}{da}\Big(\frac{a}{1-a}\Big)=\frac{a}{(1-a)^2}$$ So, if $$\displaystyle \sum_{x=1}^{\infty} f(x) =1$$ solving $$\frac{a}{(1-a)^2}=1$$ corresponds to the quadratic $a^2-3a+1=0$ the roots of which being $$a_{\pm}=\frac{1}{2} \left(3\pm\sqrt{5}\right)$$ bur since $a [0,1]$ the only solution is $$a_{\pm}=\frac{1}{2} \left(3-\sqrt{5}\right)$$ 
