Induction with floor, ceiling $n\le2^k\implies a_n\le3\cdot k2^k+4\cdot2^k-1$ for $a_n=a_{\lfloor\frac{n}2\rfloor}+a_{\lceil\frac{n}2\rceil}+3n+1$ Via induction I need to prove an expression is true. the expression is:
$n \leq 2^k \longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot 
2^k-1$
for all $n,k \in \mathbb{Z^+}$
$a_n$ is a recursive function where
$a_1 = 3$
$a_n = a_{\lfloor \frac{n}{2} \rfloor} + a_{\lceil \frac{n}{2} \rceil} + 3n+1$
I am stuck at the point where I need to prove that $P(n+1)$ is true
$a_{\lfloor \frac{n+1}{2} \rfloor} + a_{\lceil \frac{n+1}{2} \rceil} + 3(n+1) + 1 \leq 3 \cdot k  2^k + 4 \cdot 2^k -1  $ 
It is the fact that I don't know how to rewrite the floor and ceiling functions to something else. 
Can someone give me some ideas how to proceed with this?

Hint: Your induction should be for $k$, not for $n$
$P(k): n \leq 2^k \Longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot 
2^k-1$
For $k=1$ it is trivial
Suppose $P(k)$ is true and prove that $P(k+1)$ is true
$P(k+1): n \leq 2^{k+1} \Longrightarrow a_n \leq 3 \cdot (k+1)2^{k+1} + 4 \cdot 
2^{k+1}-1$
But if $n\le 2^{k+1}$ then both $\lfloor n/2\rfloor\le 2^k$ and $\lceil n/2\rceil\le 2^k$. Next, write the recurrence and apply the induction hypothesis.
$a_n = a_{\lfloor \frac{n}{2} \rfloor} + a_{\lceil \frac{n}{2} \rceil} + 3n+1 \le 2\cdot(3 \cdot k2^k + 4 \cdot 
2^k-1)+3n+1\le 3 \cdot (k+1)2^{k+1} + 4 \cdot 
2^{k+1}-1$
After simplifications, the last inequality is equivalent with $n\le 2^{k+1}$, which is true by assumption.
