16.5.1. Recurrence Condition Derivation

The trivial recurrence condition for $s(n)$ is the following: $s(n) = s(n-1) + s(n-2) + s(n-3)$. This is because the second last value in the sequence, namely $k$, could be 1. If that is the case, then find all possible sequences starting and ending with 1 and append it with the value $n$.

This chunk introduces a basic recurrence relationship for a function denoted as $s(n)$. It suggests that to find valid sequences, we can build on smaller sequences: the number of valid sequences ending in $n$ can be recursively determined by considering valid sequences that end in 1, 2, and possibly others. Each smaller sequence can just have the last number replaced by $n$, thus creating a new valid sequence.

Imagine it like a series of building blocks. If the last block of your tower can either be red or blue, and you know how many ways there are to arrange the blocks if the last block is red, you can simply add those arrangements to the ones possible if the last block is blue, and so on.

There are two categories:

1. If $k = n-1$, then all sequences starting with 1 and ending with $n$ are possible by appending $n$ to the sequences ending with $n-1$. There are $s(n-1)$ such sequences.
2. If $k$ can be 1, 2, or any number up to $n-2$, then any valid sequence that ends with those numbers can also have $n$ appended at the end.

This chunk breaks down the concept of sequence generation further into two distinct categories based on the permissible last term. In the first category, valid sequences that end in $n-1$ can simply be taken and have $n$ added in order to produce valid sequences. In the second category, we can also append $n$ to sequences ending in any legitimate earlier number, thus expanding our possibilities for the count of valid sequences.

Think of writing numbers in increasing order. You have the sequence 1, 2, 3. If you can add a 4, you can extend every sequence by attaching a 4 at the end. If your limit goes below 4, you can still use numbers 1 or 2 to achieve new valid sequences.

To obtain a more compact representation of the recurrence equation, we need to derive an alternate recurrence that depends only on the previous term. Consider valid sequences where the second last value is $n-1$ or lesser. Category 1 captures sequences with the second last equal to $n-1$ and has $s(n-1)$ contributions. Category 2 encompasses those ending with 1 or 2 or lower, which can all be recast to show dependence only on $s(n-2)$.

Here, we transition towards achieving a simpler recurrence equation of degree 1, illustrating that we can capture the total number of arrangements with fewer dependencies. By assessing what sequences qualify to be appended with $n$, we can determine an effective methodology that allows us to represent the dependencies accurately with just the last two terms. The core idea is to emphasize how fewer categories enhance simplicity.

Imagine you are folding a piece of paper in half several times. Each fold can only be marked on the last fold you just made, so rather than tracking every single fold back to the beginning, you find a way to only note the last two. Thus, it greatly simplifies tracking changes over time.

For the recurrence condition to hold, we need specific initial conditions: $s(1) = 1$, and $s(2) = 1$. The reasoning is simple: with 1 or 2, the only valid sequences are single elements, which validate that these base cases lead us correctly into recursively generating further sequences.

This chunk clarifies that to effectively apply our recurrence relation, we need to initiate it correctly with established starting values. This prevents any errors in calculating subsequent terms. Without these specific starting conditions, recursive calculations can lead to incorrect or undefined results.

Think of planting a tree. You need a strong seed (the initial condition) before your tree can grow and branch out in subsequent years. If you don't plant the seed correctly, your tree won't grow properly.

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