Problem 1671

1. Transforming the Problem:
We start by noting that the bug can move from position $x = a$ to either $x = a + 2$ or $x = \frac{a}{2}$. We need to determine the number of distinct positions the bug can reach in at most $n$ steps starting from $x = 1$.

2. Characterizing Reachable Positions:
We observe that the bug's position after any number of steps can be written in the form $\frac{p}{2^q}$, where $p$ is an odd integer and $q \geq 0$. This is because starting from $x = 1$, which is $\frac{1}{2^0}$, each step either adds 2 (which keeps the numerator odd) or divides by 2 (which increases the power of 2 in the denominator).

3. **Defining the Function $f(n, m)$**:
We define $f(n, m)$ as the minimum number of steps required to reach $\frac{2n+1}{2^m}$ from 1. We can derive the following:
- $f(n, 0) = n$ because to reach $2n+1$ from 1, we need $n$ steps of adding 2.
- For $n \geq 2^m$, $f(n, m) = f(n - 2^m, m) + 1$ because we can reach $\frac{2n+1}{2^m}$ by first reaching $\frac{2(n-2^m)+1}{2^m}$ and then adding 2.
- For $n < 2^m$, $f(n, m) = f(n, m-1) + 1$ because we can reach $\frac{2n+1}{2^m}$ by first reaching $\frac{2n+1}{2^{m-1}}$ and then dividing by 2.

4. Vengeful Pairs:
We call a pair $(n, m)$ vengeful if $n \leq 2^{m+1} - 1$. We claim that for all $x \in \mathbb{X}$, the number of vengeful pairs $(n, m)$ such that $f(n, m) = x$ is $F_{x+1}$. This can be shown by induction on $m$.

5. Inductive Proof:
We prove by induction that the number of pairs $(n, m)$ such that $f(n, m) = x$ is $F_{x+2} - 1$:
- Base case: For $x = 0$, $f(0,0) = 0$ and no other pairs satisfy $f(n, m) = 0$, so the count is 1.
- Inductive step: Assume the statement holds for $x = k$. For $x = k+1$, let $S_x$ and $S_{x+1}$ be the sets of pairs such that $f(n, m) = x$ and $f(n, m) = x+1$, respectively. We construct a function $g$ from $S_{x+1}$ to $S_x$ and show that $|S_{x+1}| = |S_x| + v(x)$, where $v(x)$ is the number of vengeful pairs with $f(n, m) = x$. By induction, $|S_{x+1}| = F_{x+2} - 1 + F_{x+1} = F_{x+3} - 1$.

6. Conclusion:
By the lemma, the number of pairs $(n, m)$ such that $f(n, m) \leq x$ is $F_2 - 1 + F_3 - 1 + \cdots + F_{x+2} - 1 = F_{x+4} - (x+4)$.

The final answer is $\boxed{ F_{n+4} - (n+4) }$