16.7. Combinatorial Proof of Identity

The trivial recurrence condition for sequences is described as follows: Let $A_n$ denote the number of valid sequences ending with $n$. The recurrence can be given by: $A_n = A_{n-1} + A_{n-2} + A_{n-3}$ This is interpreted as follows: If the second last value in the sequence is 1, find all sequences starting and ending with 1 and append $n$ to those sequences. This results in valid sequences that start with 1 and end with $n$. This applies for valid sequences ending with 2 and $n-1$ as well, confirming that all discussed sequences are disjoint.

The recurrence condition for valid sequences describes how to count sequences based on the last value. Specifically, the function $A_n$ counts the number of valid sequences that end with the value $n$. The function can be expressed as the sum of the sequences ending with the previous numbers (i.e., $n-1$, $n-2$, and $n-3$). This relationship shows that if you add a valid sequence ending with 1, 2, or any other number less than $n$, plus the value $n$, you create a new sequence ending with $n$. This captures all possible valid sequences, noting that the sequences formed in this way are disjoint from one another, meaning they don’t overlap.

Imagine you are building a tower with blocks, where the last block you place can only be a specific color representing a certain value. The blocks leading up to the last must be of colors representing values less than the last block's value. Thus, if your last block is red (representing 3), the blocks below could be blue (1) or green (2), and no two blocks can be the same color in this sequence due to the disjoint nature of the sequences.

To find a more compact recurrence relation, we consider sequences where the second last term is $n - 1$ and how we can interpret the last sequence differently. The derived recurrence can be expressed as: $A_n = 2A_{n-1}$ This equation reduces complexity as now the nth term depends only on the previous term, providing a simpler calculation for the number of sequences.

In this approach, we focus on sequences where the second last term is specifically the value of $n - 1$. By changing our perspective on the last value, we recast sequences ending in $n$ as though they end with $n - 1$. This transformation shows that we can define a more efficient recurrence since all sequences with a second last item being anything less than $n - 1$ will still yield valid increasing sequences. Therefore, counting all valid sequences merely from the last position's perspective becomes simpler, and we derive a single recurrence equation that just tracks one previous term instead of several.

Think about organizing a relay race. Initially, you must consider each runner who completed before the last runner. However, if you realize that the focus of your tracking only needs to be on the person handing the baton to the last runner, you simplify your notes to just keep track of one transition instead of every runner's details leading up to the end.

Despite the new recurrence relation being of degree 1, we still need two initial conditions such that $A_1 = 1$ and $A_2 = 1$. This ensures that we reach accurate counts of strictly increasing sequences starting with 1 and ending with 2. Such initial conditions are necessary because they set the groundwork for recurrences to function accurately.

Initial conditions are crucial in recurrence relations because they provide the starting values needed for further calculations. In the context of our recurrence, both $A_1$ and $A_2$ are set to 1 because there’s precisely one way to form a strictly increasing sequence using only the values 1 and 2 (i.e., just 1 or just 1 and 2 in increasing order). Without these base cases, the recurrence wouldn’t be able to produce valid total counts beyond just the formulations, leading to incorrect results.

Consider starting a new recipe. The initial ingredients (the first step) set the tone for the recipe's progress. If you don’t have the right first ingredient (your initial condition), you cannot complete the dish effectively!

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