How find the closed $a_{n}$ form 
let sequence $\{a_{n}\}$,such $a_{1}=0,a_{2}=2,a_{3}=5$,and for $n\in N^{+}$,such
  $$\begin{cases}
a_{2n}=2n+2a_{n}\\
a_{2n+1}=2n+1+a_{n}+a_{n+1}
\end{cases}$$
  How can I find the closed form of $\{a_{n}\}$?

My try:
we have

$$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
  so if $n=2k$,then
  $$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
  $$a_{2n}-a_{2n-1}=a_{n}-a_{n-1}+1$$
  $\cdots\cdots\cdots$
  $$a_{3}-a_{2}=a_{2}-a_{1}+1$$
  so add all this equation,we have
  $$a_{2n+1}-a_{2}=a_{n+1}-a_{1}+(2n-1)$$
  then
  $$a_{2n+1}=a_{n+1}+2n+1$$
  My idea is true? and How solve this problem,and this This problem background  is  china comption today

 A: By calculating the first few dozen values by hand, one notices the pattern that
$$
a_{n+1} = a_n + (k+2) \quad\text{for all } 2^k \le n < 2^{k+1}.
$$
This is quickly verified by induction on $k$ from the original recursion; for example,
$$
a_{2n+1}-a_{2n} = (2n+1+a_n+a_{n+1}) - (2n+2a_n) = 1 + (a_{n+1}-a_n).
$$
From here, it is not hard to prove by induction that $a_{2^k} = k2^k$ for all $k\ge0$; combining these "landmark" values with the new pattern above will provide a closed form for every $a_n$.
A: Take care, if I had one more line, you'll see that we cannot sum like you did : 
$$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
$$a_{2n}-a_{2n-1}=a_{n}-a_{n-1}+1$$
$$a_{2n-1}-a_{2n-2}=a_{n}-a_{n-1}+1$$ and not 
$$a_{2n-1}-a_{2n-2}= a_{n-1}-a_{n-2}+1$$
$\cdots\cdots\cdots$
$$a_{3}-a_{2}=a_{2}-a_{1}+1$$
And the equation you end up with is $a_{2n+1}=2n+1+a_{n}+a_{n+1}$, which unfortunately is something you already knew.
A: Here is a closed form formula for $a_n$ 
$$a_n = (\lfloor \log_2{n}\rfloor + 2)n - 2^{\lfloor \log_2{n}\rfloor +1}$$
To get this simply notice that $$a_n = n + a_{\lfloor \frac{n}{2}\rfloor}+ a_{\lceil\frac{n}{2}\rceil}$$
Then follow Greg Martin's sugestions.
