Writing down a couple of the sums:
$$1,3,6,7,9,12,13,15,18,\dots$$
and comparing that to the sequence$$1,3,5,7,9\cdots$$
gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1\dots$$ which is a sequence you can describe in closed form in a way similar to $a_n$.
That is, you can see that $$\sum_{i=n}^k a_n = 2k-1 + b_k$$
where $b_k$ is equal to $1$ if $k$ is divisible by $3$ and $0$ otherwise.
You can express $b_n$ algebraically by taking $a_n$ and any function for which $f(1)=f(2)=0$ and $f(3)=1$, and you have $b_n=f(a_n)$.
I can't think of any "elegant" function $f$ at the moment, but a quadratic polynomial can surely do it, since we only have a restriction on three points. The quadratic polynomial that satisfies $f(1)=f(2)=0$ and $f(3)=1$ is $$f(x)=\frac12x^2-\frac32 x + 1.$$
Edit:
Thanks to BarryCipra, a nicer function (more in the spirit of your solution) for $b_k$ is
$$b_k = \left\lfloor 1 + \left\lfloor\frac k3\right\rfloor - \frac k3\right\rfloor$$
Best Answer
A first useful point to consider is that $$ \eqalign{ & \left\lfloor {{{N + 1} \over a}} \right\rfloor + \left\lfloor {{{N - 1} \over a}} \right\rfloor = \cr & = {{N + 1} \over a} + {{N - 1} \over a} - \left\{ {{{N + 1} \over a}} \right\} - \left\{ {{{N - 1} \over a}} \right\} = \cr & = {{2N} \over a} - \left( {\left\{ {{{N + 1} \over a}} \right\} + \left\{ {{{N - 1} \over a}} \right\}} \right) \cr} $$
And a second one is that $$ \eqalign{ & \left\{ x \right\} + \left\{ y \right\} = \cr & = \left\{ {x + y} \right\} + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr & = \left\{ {x + y} \right\} + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \cr} $$ where $[P]$ denotes the Iverson bracket
From these, you can get a rough estimate, considering that the fractional part is less than one, and on avg. it can be take to equal $1/2$.
Then it depends on the approximation you need to reach.
A further point is that $$ \eqalign{ & \left\lfloor {x + y} \right\rfloor = \cr & = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr & = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left[ {1 - \left\{ x \right\} \le \left\{ y \right\}} \right]\quad \Rightarrow \cr & \Rightarrow \;\quad \left\lfloor {{{N + 1} \over a}} \right\rfloor = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left\lfloor {{2 \over a}} \right\rfloor + \left[ {1 - \left\{ {{2 \over a}} \right\} \le \left\{ {{{N - 1} \over a}} \right\}} \right] \cr} $$ which for $a=1$ and for $a=2$ gives simple results, and for $3 \le a$ becomes $$ \eqalign{ & \left\lfloor {{{N + 1} \over a}} \right\rfloor \quad \left| {\;3 \le a} \right.\quad = \cr & = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {{{a - 2} \over a} \le \left\{ {{{N - 1} \over a}} \right\}} \right] = \cr & = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {a - 2 \le \left( {N - 1} \right)\bmod a} \right] = \cr & = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {a - 2 = \left( {N - 1} \right)\bmod a} \right] + \left[ {a - 1 = \left( {N - 1} \right)\bmod a} \right] = \cr & = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {0 = \left( {N + 1} \right)\bmod a} \right] + \left[ {0 = N\bmod a} \right] = \cr & = \left\lfloor {{{N - 1} \over a}} \right\rfloor + \left[ {a\backslash N} \right] + \left[ {a\backslash \left( {N + 1} \right)} \right] \cr} $$ so that the sum of the two terms differ for the number of divisors of $N$ and $N+1$.