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.