General term of this interesting sequence A sequence of positive integer is defined as follows

  
*
  
*The first term is $1$.
  
*The next two terms are the next two even numbers $2$, $4$.
  
*The next three terms are the next three odd numbers $5$, $7$, $9$.
  
*The next $n$ terms are the next $n$ even numbers if $n$ is even or the next $n$ odd numbers if $n$ is odd.
  

What is the general term $a_n?$
Please, proofs of all these formulas would be nice
 A: The general term is
$$\tag1a_n=2n-\left\lceil \sqrt{2 n}-\frac12\right\rceil.$$
Why?
We have $a_{n+1}=a_{n}+2$ unless there is an integer $\ge\frac12 +\sqrt{2n}$ and $<\frac12+\sqrt{2(n+1)}$.
We need $a_{n+1}=a_n+1$ iff $n$ is one of the numbers $1, 3, 6, 10, \ldots$, i.e. a number of the form $k \choose 2$. This is accounted for by the ceiling/sqrt term.
Proof:
That $a_{n}=a_{n-1}+1$ iff $n-1=\frac{k(k+1)}2$ because of the well-known summation $1+2+3+\cdots+k=\frac{k(k+1)}2$ should be clear.
For which $\nu$ does there exist a $k$ such that $\nu-1=\frac{k(k+1)}2$? This is a quadratic in $k$ with solutions 
$$\tag2k_{1,2}=\frac{-1\pm\sqrt{1+8(\nu-1)}}2=\frac{-1\pm\sqrt{8\nu-7}}2.$$
Thus the $\nu$ with $a_{\nu}=a_{\nu-1}+1$ are characterized by the fact that $8\nu-7$ is a square (of an odd number).
In total we have $a_n = 2n-m_n$ where $m_n$ is the number of $\nu\le n$ for which $(2)$ has a positive integer solution $k$. But since there is one $\nu$ (namely $\frac{k(k+1)}2$) for each $k=1, 2, \ldots$, we find that 
$m_n=\left\lfloor\frac{-1+\sqrt{8n-7}}2\right\rfloor$.
This proves a different formula, but it can be simplified to $(1)$ by observing that $\sqrt {2n}-\frac12$ is never an integer (that would make $2n$ the square of an odd integer). Therefore $\lfloor\sqrt{2n}+\frac12\rfloor$ may be used instead of $\lceil\sqrt{2n}-\frac12\rceil$ in $(1)$ (or just round $\sqrt{2n}$ to nearest integer). We have
$$\left\lfloor\frac{-1+\sqrt{8n-7}}2\right\rfloor\le\left\lfloor \sqrt{2 n}-\frac12\right\rfloor $$
and inequality can only hold if there is a natural number $r$ such that 
$$\frac{-1+\sqrt{8n-7}}2<s<\sqrt{2 n}-\frac12$$
i.e. 
$$\sqrt{8n-7}<2s+1<2\sqrt{2 n}$$
$$8n-7<(2s+1)^2<8 n.$$
The latter is impossible because odd squares are $\equiv 1\pmod 8$.
Therefore, rounding $\left\lfloor\frac{-1+\sqrt{8n-7}}2\right\rfloor$ may be replaced with e.g. $\left\lfloor \sqrt{2 n}-\frac12\right\rfloor$ or (with adjustment of the complete formula by a constant) with $\left\lceil \sqrt{2 n}-\frac12\right\rceil$.
A: Your sequence is the following one:
http://oeis.org/A001614
A: http://oeis.org/A001614 has a few formulas:
$$
a_n = 2n - \left \lfloor \frac{1+ \sqrt{8n-7}}{2} \right\rfloor
$$
A: Not a general term, but an interesting reformulation of your recursion. Given a(1)=1
$$
a(n):=\begin{cases}
a(n-1)+1& \text{if} \,\, a(n-1) \text{ is a square}\\
a(n-1)+2 &\text{otherwise}
\end{cases}
$$
