First of all, note that $f(n)$ = $\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \le n$ . We let $f(0) = 1$ for convenience.
From here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \in \mathbb{Z}$
We induct on $a$ . It is trivially true for $a = 0$ and $a = 1$ . From here, we show that, if the only numbers $n \le 2^{a-1} - 1$ where $f(n)$ is odd are of the form described above, then the only numbers $n \le 2^{a} -1$ that are odd are of that form. We first consider all numbers $b$ , such that $2^{a-1} \le b \le 2^{a} - 2$ , going from the lower bound to the upper bound (a mini induction, you might say). We know that $f(b) = \sum_{i=0}^{a-1} f(b-2^{i})$ . For a number in this summation to be odd, $b - 2^i = 2^m -1 \rightarrow b = 2^i + 2^m - 1$ . However, we know that $b > 2^{a-1}$ , so $m$ must be equal to $a-1$ , or else $b$ cannot be in that interval. Now, from this, we know that $i < a-1$ , as $b<2^{a} - 1$ . Therefore, $i$ and $m$ are distinct, and thus $f(b - 2^i)$ and $f(b- 2^{a-1})$ are odd; since there are just two odd numbers, the ending sum for any $b$ is even. Finally, considering $2^{a} - 1$ , the only odd number is $f(2^{a} - 1 - 2^{a-1})$ , so the ending sum is odd. $\Box$
The smallest $n$ greater than $2013$ expressible as $2^d - 1, d \in \mathbb{N}$ is $2^{11} -1 = \boxed{2047}$

Let $f(n)$ denote the number of ways to express $n$ as a sum of powers of 2, considering the order of summation. To solve this problem, we must determine the smallest integer $n > 2013$ for which $f(n)$ is odd.

Firstly, we understand that $f(n)$ is related to binary representations of numbers. Each integer $n$ can be uniquely represented in base 2, meaning $n = \sum a_i 2^i$, where $a_i$ are binary digits (0 or 1). The task is to count all partitions utilizing available powers of 2, ordered permutations included.

A key insight into solving this problem is:
- $f(n) \equiv 1 \pmod{2}$ if and only if $n$ is a Mersenne number, which is of the form $2^k - 1$.

To determine this, consider complete sets of powers of 2. The set of all subsets of $\{ 2^0, 2^1, \ldots, 2^{k-1} \}$, when considered with repetition and order, represents all sums $f(n)$ for $n = 1$ to $2^k - 1$.

For $n = 2^k - 1$, the binary representation contains all 1s, i.e., $(111...1)_2$ with $k$ ones. Only for $n$ of this form, the number of ordered sums equates to an odd number due to properties of binomial coefficients (odd summation through subsets).

Our strategy is then to find the smallest $k$ such that $n = 2^k - 1 > 2013$.

Calculating $2^k - 1$:
$$\begin{align*} 2^{10} - 1 &= 1023, \\ 2^{11} - 1 &= 2047, \\ 2^{12} - 1 &= 4095. \end{align*}$$

For $n = 2047$, $2^{11} - 1$ is greater than 2013 and $f(n)$ is odd since 2047 is a Mersenne number. Thus, this satisfies the given condition.

Therefore, the smallest $n > 2013$ for which $f(n)$ is odd is:
$$\boxed{2047}$$