Solution:

Note that these $a_{k}$'s are in fact the Fibonacci numbers. Denote by $f(n)$ the number of distinct ways to express a number as a sum of $a_{k}$. Note that $2017=1597+377+34+8+1=a_{15}+a_{12}+a_{8}+a_{5}+a_{1}$.

We prove the following lemma:
$$a_{1}+a_{2}+\cdots+a_{k}=a_{k+2}-2$$
This follows simply from induction. It is true for $k=1$; adding $a_{k+1}$ to both sides and using the fact that $a_{k+1}+a_{k+2}=a_{k+3}$ gives the result.

Now, denote by $f(n)$ the number of ways to express $n$ as a sum of distinct $a_{k}$'s; we are looking for $f(2017)$. Now, note that any such sum must contain either $1597$ or $987$. If the sum does not contain $1597$, it must certainly contain $987$; otherwise, from the lemma, the sum would be at most $1+2+\cdots+610=1595$. Moreover, if the sum contains $987$ (but not $1597$), it must also contain $610$; otherwise, it will be at most $987+(1+2+\cdots+377)=987+985=1972$.

Hence, $f(2017)=2 f(420)$.

By a similar argument, any sum of $420$ must contain either $377$ or $233$. However, this time, it is perfectly possible for this sum to contain $233$ but not $144$, since $233+(1+2+3+\cdots+89)=464>420$. We thus have two cases to deal with.

Case 1: If the sum contains $377$, then we have to compute $f(43)$. Now, note that as in the argument from earlier, any sum adding up to $43$ contains either $34$ or $13+21$. Hence $f(43)=2 f(9)$. Repeating this argument, we get $f(9)=2 f(1)=2$. This gives us $f(43)=4$.

Case 2: The sum does not contain $377$. In this case, the sum must contain $233$. We then have to compute $f(187)$. Any sum adding up to $187$ must contain either $144$ or $89$; moreover, if it contains $89$, it must contain $55$ as well. Hence, $f(187)=2 f(43)=8$ by our previous computation.

Thus, $f(420)=3 f(43)=12$, and $f(2017)=24$.