16.7.2. RHS Expression Explanation

So let $f(n)$ denote my function which is the number of valid sequences ending with $m$.

In this section, we introduce a function, denoted as $f(n)$, which counts how many valid sequences can be formed that end with the number $m$. The function helps in understanding how sequences can be constructed under certain rules or conditions, such as strict increase or valid endpoints.

Imagine you are building different towers with blocks, where each tower can only end with a specific colored block (representing $m$). The function $f(n)$ helps you figure out how many different ways you can construct these towers based on specific rules.

So a trivial recurrence condition for $f(n)$ is the following. I can say that $f(n) = f(n-1) + f(n-2) + f(n-3)$.

Here, we derive a basic recurrence relation for the function $f(n)$. This relation says that the number of valid sequences for a number $n$ depends on summing the valid sequences of the previous numbers $n-1$, $n-2$, and $n-3$. The reason for this is that a valid sequence can directly depend on the last number chosen from the previous sequences.

Think of this like a game where each time you score a point, the total score you can achieve depends on the scores from the previous three rounds. Each round can contribute differently, just like each previous sequence contributes to forming a longer valid sequence.

Now there can be 2 categories: Category 1 where the second last value is $m-1$ and Category 2 where it can be 1 or 2 or $m-2$.

We define two specific categories of sequence endings based on what the second to last number can be. Category 1 includes those where the second last value is $m-1$, which relates to sequences that are just one step lower than the final number. Category 2 includes sequences that can have a second last number of 1, 2, or up to $m-2$.

Imagine these categories are like stepping stones in a river. In Category 1, you can only step on the stone right next to your last stone, while in Category 2, you have more options of stones to step onto, which allows for greater diversity in the paths you can take.

And hence we can say that our alternate recurrence equation will be $f(n) = 2 \cdot f(n-1)$.

In the end, we derive a more compact and effective recurrence relation: $f(n) = 2 \cdot f(n-1)$. This equation states that the number of valid sequences for number $n$ is just double the valid sequences for the previous number, simplifying our calculations significantly by reducing dependencies.

Picture deciding between two routes at every intersection. If there are two possible routes you could take at each step, your choices effectively double as you proceed, just like the relation saays that each valid sequence builds upon the previous one.

It turns out that even though this equation is of degree 1, we need 2 initial conditions: $f(1) = 1$ and $f(2) = 1$.

To use the recurrence relation effectively, we need to establish clear initial conditions. For this function, establishing that both $f(1) = 1$ and $f(2) = 1$ informs our subsequent calculations about the sequences since these base cases anchor the growth of the sequence.

Think of these initial conditions like the foundation of a building. Just as you need a strong base to build up, you need defined initial values for your function's calculations to ensure everything else stacks correctly and logically.

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