Are there bounded nonconvergent sequences satisfying this recurrence relation? In this question, we ask about convergent sequences $(p_n)$ satisfying 
$p_{i} = 2p_{i+6} - {1 \over 2}p_{i+1} - {1 \over 4}p_{i+3} - {1 \over 8}p_{i+7} - {1 \over 16}p_{i+15} - {1 \over 32}p_{i+31} - \dots$.
It turned out the solutions form a linear space of dimension $7$, engendered by $(p_n) = (\lambda_i^n)$ where $\lambda_1, \ldots ,\lambda_7 = 1$ are the $7$ zeroes of the "characteristic power series" of the recurrence.
But there might be other solutions if we don't assume that $(p_n)$ is convergent.
Let $\ell^1 = \{ \sum a_n z^n \mid \sum |a_n| < \infty \} \subset \Bbb C[[z]]$. With the $1$-norm on the coefficients and the Cauchy product, this is a Banach algebra (it is also known as the Wiener algebra). Its dual is $\ell^\infty$ (our space of bounded sequences), so letting $f(z) = 1 +\frac 12 z + \frac 14 z^3-2z^6+\frac 18 z^7\dots$ and $g(z)= \prod_{i \le 7}(z-\lambda_i)$, my question is equivalent to :
"if $\forall k \ge 0, \langle fz^k, p \rangle = 0 $, do we have $ \forall k \ge 0, \langle gz^k,p \rangle = 0$ ?" or,
"is $g$ in the closure of the principal ideal $(f) \subset \ell ^1$ ?"

Now we can somewhat simplify this, because if $|\lambda| < 1$, then multiplication by $(z-\lambda)$ is a homeomorphism between $\ell^1$ and $\{f \in \ell^1 \mid f(\lambda)=  0\} = \ker \rho_\lambda$ (the evaluation-at-$\lambda$ morphism).
So after letting $\hat f = f / \prod_{i\le 6}(z-\lambda_i)$ (if I remember correctly its first coefficient is negative and then all other coefficients seem to be positive), this becomes :
"is $1-z$ in the closure of the principal ideal $(\hat f) \subset \ell^1$ ?"
and because the closure of $(1-z)$ is $\{f \in \ell^1 \mid f(1)= 0\} = \ker \rho_1$, this is equivalent to "is $\ker \rho_1$ the closure of $(\hat f)$ ?"
I'm afraid I really don't know much about the closed ideals contained in $\ker \rho_1$ so I don't know where to take this (as far as I can tell, we don't know much about the structure of closed ideals of the Wiener algebra. There goes my hope of an easy answer)
Due to $\hat f$'s nice form, one can compute the sequence $(d_n) = d(1-z, \Bbb C_n[z] \hat f)$ pretty easily, except that you need more and more precision on the $\lambda_i$ as $n$ gets larger. It seems to be very slowly converging to $0$ (a $O(1/n)$ maybe ?)
But all in all, I really don't know if the limit is $0$ or not, I think the numerical evidence is pretty weak.

If the answer is yes, one can ask the same question in a "finer" Banach algebra $B = \{f \in \ell^1 \mid f' \in \ell^\infty\}$ with the norm $N(f) = |f|_1 + |f'|_\infty$ (both norms are the norms on the coefficients sequence). Everything I've said above is still valid in $B$ (so in particular, $\hat f \in B$), and in case this is also true this should entail a stronger result.
 A: It turns out the more general fact that I hoped was true is in fact true :
Let $\mathcal L^1 = \{ \sum_{n,m \ge 0} a_{n,m}z^n t^m \in \Bbb C[[z,t]] \mid \sum_{n,m \ge 0} |a_{n,m}| < \infty \}$, equipped with the $1$-norm.
$\mathcal L^1$ is again a Banach algebra, and if $0 < a < 1$, we have that $1-z$ is in the closure of the principal ideal $(1-az-(1-a)t)$.
Write $(1-z)/(1-az-(1-a)t) = \sum_{n \ge 0} r_n t^n$ where $r_n \in \Bbb C[[z]]$.  
To show this, we take $Q_n = \sum_{k=0}^{n-1} r_k t^k \in \Bbb C[[z]][t]$ to be its truncation modulo $t^n$, and we look at the remainder $(1-z)-(1-az-(1-a)t)Q_n$.
By construction of $Q_n$, this is just $(1-a)r_{n-1}t^n$, so the goal is to show that $|r_n|_1 \to 0$.
Observe that $1/(1 - az - (1-a)t) = \sum_{k \ge 0} (az + (1-a)t)^k = \sum_{k,l \ge 0} a^k(1-a)^l \binom{k+l}{k}z^kt^l$, is a power series with only positive coefficients.
Evaluating at $z=1$ gives $1/((1-a)(1-t)) = \sum_{l \ge 0} 1/(1-a) t^l$ which shows that if you sum each coefficient of $t^l$ separately, the sums converge to $1/(1-a)$ and so after multiplication by $(1-z)$, $r_n \in \ell^1$ and $|r_n|_1 \le 2/(1-a)$.
More precisely we have $r_n = (1-a)^n(1-z)\sum_{k \ge 0}a^k\binom{k+n}{k}z^k = (1-a)^n\sum_{k \ge 0}a^{k-1}(a\binom{k+n}{k}-\binom{k+n-1}{k-1})
\\ = (1-a)^n\sum_{k \ge 0}a^{k-1}(\frac {a(k+n)}k - 1) \binom {k+n-1}{k-1}$
So its coefficients start positive, then switch sign and get negative when $a(k+n) \le k$, or when $k \ge an/(1-a)$. Then, $|r_n|_1$ is made of two telescopic summations and we get  
$|r_n|_1 = 2 \max \{a^k(1-a)^n \binom {k+n}{k}\}$, where the maximum is achieved for $k = an/(1-a)$ (approximately). Then Stirling's formula gives (after everything simplifies) $|r_n|_1 \sim 2/\sqrt{2\pi an}$, which shows $|r_n|_1 \to 0$ and that $1-z \in \overline{(1-az-(1-a)t)}$

Now, if $g \in \ell^1$ and $|g|_1 \le 1$, the evaluation morphism $\phi : \mathcal L^1 \to \ell^1$ given by $t \mapsto g$ is well-defined, continuous 
($|\phi| = 1$), and is a ring morphism. Therefore $\phi((1-z) - Q_n(1-az-(1-a)t)) = (1-z) - \phi(Q_n) (1-az-(1-a)g)$ and $|(1-z) - \phi(Q_n) (1-az-(1-a)g)|_1 \le |(1-z) - Q_n(1-az-(1-a)t)| \to 0$ too.
This shows that $1-z$ is in the closure of the principal ideal $(1-az-(1-a)g)$.
Going back to the original question, we have to show that $\hat f$ can be written in this form ($f$ itself cannot)
$\hat f = f / \prod_{i=1}^6 (z - \lambda_i) = f \times \prod_{i=1}^6 (\sum_{k \ge 0} (\lambda_i^k z^{-k)})) \times z^{-6}$
Multiplying an element of $\Bbb C[[z]]$ by an element of $\Bbb C[[1/z]]$ is usually of a pretty bad taste, but we are fine here because both sequences of coefficients converge to $0$ exponentially (and the six roots are defined so that the result is actually in $\Bbb C[[z]]$, i.e. all the coefficients for negative powers of $z$ cancel between themselves).
Now, a computation shows that $\sum \lambda_i^k$ is always positive, (the largest root is very close to $1$ so eventually it stays positive, and you only have to check finitely many coefficients) ; all the coefficients in $f$ from exponent $7$ are negative, and after doing the product and dividing by $z^6$, all the coefficients from exponent $1$ are negative.
Since $\hat f(1) = 0$, the constant coefficient is positive, and a small computation shows that the coefficient of $z$ is nonzero, so we are in a situation where we can apply the above result with $\hat f/\hat f(0)$ and $a = - \hat f'(0)/\hat f(0)$.
