recurrence relations - Show the recursive sequence is increasing

How do I show that the recursive sequence

$$a_n = a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1, \quad n\geq 2, \phantom{x} a_1 = 3$$

is an increasing sequence?

1. attempt:
If I can show that $a_{n+1}-a_n>0$, I would be able to show it is increasing.

$$\begin{align*} a_{n+1} - a_n & = a_{\left\lfloor \frac{n+1}{2} \right\rfloor} + a_{\left\lceil \frac{n+1}{2} \right\rceil} +3(n+1)+1 - ( a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1) \\ & = a_{\left\lfloor \frac{n+1}{2} \right\rfloor} + a_{\left\lceil \frac{n+1}{2} \right\rceil}+3 - a_{\lfloor n/2 \rfloor} - a_{\lceil n/2 \rceil} \end{align*}$$

I can't put $a$ together because the indexes are different (because of the ceils and floors).

2. attempt:
Proof by induction

Base case: $n=2$ then $a_2 = a_1 + a_1 + 3\cdot 2 + 1 = 3+3+7=13$ so $a_1

Testing with more gives: $a_1 < a_2 < a_3 =26 < a_4 = 39<...$

Assume $a_nNow I want to show that $a_{n+1} < a_{n+2}$

$$a_{n+2} = a_{\lfloor (n+2)/2 \rfloor} + a_{\lceil (n+2)/2 \rceil} +3(n+2)+1$$

Again I get stuck since I don't know how to handle the ceils and floors.

$\phantom{x}$

How do I go about showing the sequence is increasing? Maybe I'm making it harder than it actually is - is there by any chance an easier way?

Answer

For an even index we have
$$a_{2n}=2a_n+6n+1$$
and for an odd one,
$$a_{2n+1}=a_n+a_{n+1}+6n+4$$
so if we assume (using induction) that $a_{n+1}\ge a_n$ then

$$a_{2n+2}=2a_{n+1}+3(2n+2)+1=2a_{n+1}+6n+7>a_n+a_{n+1}+6n+4=a_{2n+1}$$
and
$$a_{2n+1}=a_n+a_{n+1}+6n+4>2a_n+6n+1=a_{2n}$$

Remark: Perhaps one of two more base cases are needed.