Recurrence relations and limits, tough. I would like a hint for the following, more specifically, what strategy or approach should I take to prove the following?
Problem: Let $P \geq 2$ be an integer. Define the recurrence
$$p_n = p_{n-1} + \left\lfloor \frac{p_{n-4}}{2} \right\rfloor$$
with initial conditions:
$$p_0 = P + \left\lfloor \frac{P}{2} \right\rfloor$$
$$p_1 = P + 2\left\lfloor \frac{P}{2} \right\rfloor$$
$$p_2 = P + 3\left\lfloor \frac{P}{2} \right\rfloor$$
$$p_3 = P + 4\left\lfloor \frac{P}{2} \right\rfloor$$
Prove that the following limit converges:
$$\lim_{n\rightarrow \infty} \frac{p_n}{z^n}$$
where $z$ is the positive real solution to the equation $x^4 - x^3 - \frac{1}{2} = 0$.
Note: I've already proven the following:
$$\lim_{n\rightarrow \infty} \frac{p_n}{p_{n-1}} = z$$
Any ideas? Not sure if this result helps. Also $\lim_{n\rightarrow \infty}p_n/z^n$ is also bounded above and below. I've attempted to show $\lim_{n\rightarrow \infty} \frac{p_n}{z^n}$ is Cauchy, but had no luck with that. I don't know what the limit converges to either.
Edit: I believe the limit should converge as $p_n$ achieves an end behaviour of the form $cz^n$ for $c \in \mathbb{R}$ (this comes from the fact that the limit of the ratios of $p_n$ converge to $z$), however I do not know how to make this rigorous.
Edit 2: Proving the limit exists is equivalent to showing
$$p_0 \cdot \prod_{n=1}^{\infty} \left( \frac{p_n/p_{n-1}}{z} \right)$$
converges.
UPDATED:
If someone could prove that $|p_n-z \cdot p_{n-1}|$ is bounded above (or converges, or diverges), then the proof is complete.
 A: Let us start with the solution of the homogeneous recurrence 
$$\phi_n = \phi_{n-1} + \frac{\phi_{n-4}}{2}$$
its characteristic equation is 
$$x^4 - x^3 - \frac{1}{2} = 0$$
this equations has $4$ solutions, two of them are complex and the other two are a negative and a positive real number. Their approximate values, as given by Mathematica, are: 
$$z_1=1.25372, \ \ z_2=-0.669107, \ \ z_3=0.207691 + 0.743573 i, \ \ z_4=0.207691 - 0.743573 i,$$
(in your answer you label $z$ the one labeled $z_1$ above). Notice that the positive real solution $z=z_1=1.25372$ is the one with the greatest magnitude among the $4$ solutions (actually, it is the only one whose magnitude exceeds $1$). 
Now, the general solution to the homogeneous recurrence is:
$$\phi_n=c_1z^n_1+c_2z^n_2+c_3z^n_3+c_4z^n_4$$ 
where $c_1, c_2, c_3, c_4$ are constants to be determined from the initial conditions posed in your question. Since $z=z_1=1.25372$ has the greatest magnitude among the roots of the characteristic equation, the above general  solution asymptotically (for $n$ large enough) tends to $c_1z^n$ i.e.
$$\phi_n\sim c_1z^n$$
Consequently, 
$$\frac{\phi_n}{z^n}\sim c_1 \ \ \textrm{ i.e. } \ \  \lim_{n\rightarrow\infty}\frac{\phi_n}{z^n}=c_1$$
where $c_1$ will be determined by the solution of the $4\times 4$ linear system of equations 
$$
\phi_i=c_1z^i_1+c_2z^i_2+c_3z^i_3+c_4z^i_4
$$
for $i=0,1,2,3$, $p_i$ given by the initial conditions posted in the question and $z_1=z,z_2,z_3,z_4$ the roots of the characteristic equation given above. 
Let me now try to justify why the convergence of $\frac{\phi_n}{z^n}$ implies also the convergence of $\frac{p_n}{z^n}$. The recurrence 
$$p_n = p_{n-1} + \left\lfloor \frac{p_{n-4}}{2} \right\rfloor = p_{n-1} + \frac{p_{n-4}}{2} - \epsilon_n$$
differs from the homogeneous, by a bounded function $0\leq\epsilon_n< 1$ of $n$. Since we are dealing with linear recurrences, increasing sequences $p_n$, $\phi_n$ and we are interested in the asymptotic behaviour of the solutions, in the limit of large $n$, the two are essentially the same. The solutions $p_n$ and $\phi_n$ differ by a $O(1)$ special solution (of the posted, non-homogeneous  recurrence): 
$$
p_n=\phi_n+O(1) \Rightarrow p_n\sim\phi_n\Rightarrow\frac{p_n}{z^n}\sim \frac{\phi_n}{z^n}\Rightarrow\lim_{n\rightarrow\infty}\frac{p_n}{z^n}=c_1
$$
 We can also see that the bigger the value of $P\geq 2$ (given in the initial conditions), the quicker  $\frac{p_n}{z^n}$ converges.
P.S.  Regarding the estimation that the general solutions $p_n$ and $ϕ_n$ of the respective recurrences, differ by a $O(1)$ special solution of the non-homogeneous, my argument is the following: when dealing with non-homogeneous linear recurrences with constant coefficients i.e.
$$
p_n+c_1p_{n−1}+...+c_dp_{n−d}=h(n)
$$
and $h(n)=const$, then it is customary to seek for a special solution to be a constant. Since here, the non-homogeneous factor is the bounded function $0\leq\epsilon_n< 1$, I guess that it is reasonable to conjecture that the corresponding special solution is $O(1)$.  
A: Consider the vector $X_n = (p_n, p_{n-1}, p_{n-2}, p_{n-3})$.  We have
$$
\begin{eqnarray}
X_{n+1} &=& (p_{n+1},p_n, p_{n-1},p_{n-2}) \\
&=& \left(p_n+\left\lfloor \frac{1}{2}p_{n-3}\right\rfloor, p_n, p_{n-1}, p_{n-2}\right) \\ &=& (p_n+ \frac{1}{2}p_{n-3}, p_n, p_{n-1}, p_{n-2}) - (\varepsilon_n, 0, 0, 0)\\ &=&\hat{M}\cdot X_{n} + \varepsilon_n E,
\end{eqnarray}
$$
where $|\varepsilon_n| < 1$, $E=(-1,0,0,0)$, and
$$
\hat{M}=\left(
\begin{matrix}
1 & 0 & 0 & \frac{1}{2} \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0
\end{matrix}\right).
$$
Let $\lambda_{i=1,2,3,4}$ and $u_i$ be the eigenvalues and normalized eigenvectors of $\hat{M}$.  Only $\lambda_1 = z \approx 1.25372$ has magnitude greater than $1$; the other eigenvalues have magnitudes strictly less than $1$.  We can write $E=\sum_i e_i u_i$ for some fixed coefficients $e_i$.  Now, writing $X_n=z^n \sum_i c_{i,n} u_i$, we have
$$
z^{n+1}\sum_i c_{i,n+1}u_i = X_{n+1}=\hat{M}\cdot X_n + \varepsilon_n E = z^n \sum_i c_{i,n} \lambda_i u_i + \varepsilon_n \sum_i e_i u_i;
$$
or simply
$$
c_{i,n+1} = \left(\frac{\lambda_i}{z}\right) c_{i,n} + \frac{\varepsilon_n e_i}{z^{n+1}}.
$$
Because $|\varepsilon_n|$ is bounded, $\lim_{n\rightarrow \infty}c_{i,n}$ exists for each $i$ (and is zero for $i\neq 1$).  Therefore $X_n / z^n=\sum_i c_{i,n}u_i$ has a limit, as does its first component, $p_n/z^n$.
A: I don't know if you
can show that
$\frac{p_n}{z^n}
= 1
$.
If the sequence
$\frac{p_n}{p_{n-1}}
$
approaches $z$ from the same side,
each term in the product
exceeds $z$,
so the product will
always exceed
$z^n$.
What you can show
is that
$\lim \frac{p_n^{1/n}}{z}
= 1
$.
I will now give the
standard,
not original proof.
Once you have shown that
$\lim_{n\rightarrow \infty} \frac{p_n}{p_{n-1}} = z
$,
the hard part is done.
The rest is a standard
good-part/bad-part splitting
on $p_n$.
From that limit,
for any $c > 0$,
there is a $N = N(c)$
such that
$z-c
< \frac{p_n}{p_{n-1}}
< z+c
$
for
$n > N(c)
$.
Then
(this is how these proofs
usually go)
$\begin{array}\\
\frac{p_n}{p_0}
&=\prod_{k=1}^{n} \frac{p_k}{p_{k-1}}\\
&=\prod_{k=1}^{N(c)} \frac{p_k}{p_{k-1}}\prod_{k=N(c)1}^{n} \frac{p_k}{p_{k-1}}\\
&=P(c)\prod_{k=N(c)+1}^{n} \frac{p_k}{p_{k-1}}\\
&< P(c)(z+c)^{n-N(c)}
\qquad\text{(this is for an upper bound -  the lower bound proof is similar)}\\
\text{so}\\
\frac{p_n}{z^n}
&< \frac{P(c)(z+c)^{n-N(c)}}{z^n}\\
&= \frac{P(c)(1+c/z)^{n-N(c)}}{z^{N(c)}}\\
&= (1+c/z)^n\frac{P(c)}{z^{N(c)}(1+c/z)^{N(c)}}\\
&= (1+c/z)^n\frac{P(c)}{(z+c)^{N(c)}}\\
\text{so that}\\
\frac{p_n^{1/n}}{z}
&< (1+c/z)\left(\frac{P(c)}{(z+c)^{N(c)}}\right)^{1/n}\\
&= (1+c/z)R(c)^{1/n}
\qquad\text{where }R(c) = \frac{P(c)}{(z+c)^{N(c)}}\\
\end{array}
$
Therefore,
by taking $c$ small
and letting $n$ get large,
we have
$\lim \sup \frac{p_n^{1/n}}{z}
\le 1
$.
A almost identical,
cut-and-pasteable
proof will show that
$\lim \inf \frac{p_n^{1/n}}{z}
\ge 1
$,
so that
$\lim \frac{p_n^{1/n}}{z}
= 1
$.
