Infinite recurrence relation which depends on subsequent sequence values I'm trying to solve a problem and I have it reduced to solving the following recurrence relation which goes "backwards" as the value of $p_i$ depends on subsequent values.
$p_i\in [0,1]$
$p_i = 0 (i < 7) $
$p_i = {1\over 2} p_{i-7+1} + {1\over 4} p_{i-7+2} + {1\over 8} p_{i-7+4} + {1\over 16} p_{i-7+8} + {1\over 32} p_{i-7+16} + ... (i>=7)$
which I rewrote as 
$p_{i+1} = 2p_{i+7} - {1 \over 2}p_{i+2} - {1 \over 4}p_{i+4} - {1 \over 8}p_{i+8} - {1 \over 16}p_{i+16} - {1 \over 32}p_{i+32}...$ 
Note that we know
$\lim_{i \to \infty } p_i = 1$
Is there any way to solve this?  I thought about using characteristic equations as used to find closed form for say, fibonacci numbers, but it doesn't seem to work here as it's an infinite series.  I tried reducing it via some sort of telescopic operation but I got nowhere.  Could it be solved with generating functions?  Or something else?
 A: Let $f(x)= \sum_{i=0} p_i x^i$ and $g(x) = x^7 - \sum_{i \ge 0} x^{2^i} 2^{-1-i}$.
$f$ has radius of convergence exactly $1$ and $g$ has radius of convergence $2$. If I am not mistaken, your relations says that $f(1/x)g(x)$, that converges for $1<|x|<2$, is actually a power series in $x$.
Hence it has radius of converge at least $2$,
And this shows that $f$ has a meromorphic continuation on the whole Riemann sphere, so it is a rational fraction.
Moreover, it has no poles of modulus less than $1$, and so its poles have to be inverses of the zeros of $g$ of modulus less than $1$.
This should complete the answer above, and if the solution is unique you should be able to show it from there.

Edit : Uniqueness was in fact easy :
Suppose you have 2 solutions $p$ and $q$. Since $p-q$ satisfies the recurrence relations, is zero on the first six terms and converges to zero, it has to be zero :
Suppose it has an element with maximal modulus. Since it can't be one of the first six terms, the recurrence relation says that it is a weighted average of some other terms. Then all those terms have to be equal, and so the sequence can't converge to zero, contradiction.
