Introduction

Warmup: The Table Seating Problem

You work for a company that provides seating for various functions. The tables you use can seat four people comfortably. When placed end to end, the tables can accommodate a group by pushing the tables end to end allowing for seating as indicated in the diagram below.

One Table

Seats four

Two Tables

Seat six

Three Tables

Seat eight

When you arrive at a function you like to imagine how many people could be seated if all of the tables were set up end to end (e.g. as if to make a very long banquet table).

Given a number of tables, what is the maximum number of people that can be seated comfortably if the tables are set up end to end?

Solutions:

Counting pairs across each table and a pair on the ends:

1 table seats 2 across (top and bottom) from one another at 1 table and 2 on the end (one on each end, left and right).

2 tables seat 2 across (top and bottom) from one another at 2 tables and 2 on the end (one on each end, left and right).

3 tables seat 2 across (top and bottom) from one another at 3 tables and 2 on the end (one on each end, left and right).

N tables seat 2 across (top and bottom) from one another at N tables and 2 on the end (one on each end, left and right).

Doubling across each table and adding a pair on the ends:

1 table seats 1 on top and 1 on the bottom at at 1 table and 2 on the end (one on each end, left and right).

2 tables seat 2 on top and 2 on the bottom at at 2 tables and 2 on the end (one on each end, left and right).

3 tables seat 3 on top and 3 on the bottom at at 1 tables and 2 on the end (one on each end, left and right).

N tables seat N on top and N on the bottom at N tables and 2 on the end (one on each end, left and right).

Counting around the corners on the left and top and on the right and bottom:

1 table seats 2 on the left and top and 2 on the right and bottom at at 1 table.

2 tables seat 3 on the left and top and 3 on the right and bottom at at 2 tables.

3 tables seat 4 on the left and top and 4 on the right and bottom at at 3 tables.

N tables seat N + 1 on the left and top and N + 1 on the right and bottom at at N tables.

Add two more seats for every additional table:

If we look at the seating for the first several tables we start to see a pattern emerge:

1 table seats 4
2 tables seat 4 + 2 = 6
3 tables seat 6 + 2 = 8
4 tables seat 8 + 2 = 10

Every time a table is added, 2 more seats are added to the number of seats prior. We recognize this as an arithmetic sequence starting at 4 and adding a common difference of 2.

Connecting Solutions:

It should be clear that out first three solutions are equivalent to one another. However, what about the sequence description? Recall that a sequence is a special, real-valued function that is defined on the natural numbers. To help us quickly identify a sequence, we often use subscript notation, an, to refer to the terms of a sequence rather than the traditional function notation a(n). In our last example, we defined the (n + 1)th term of the sequence recursively as the sum of the previous (nth) term and 2. Luckily, because the number of seats that are added every time a new table is introduced is constant, our sequence is arithmetic, which has a closed form formula. Given the first term a1 and common difference d, the closed form description of our sequence is a(n) = a1 + d(n – 1). The first term of the sequence, when we only have one table, is a1 = 4 seats and the common difference is d = 2. By substituting, we see that a(n) = a1 + d(n – 1) = 4 + 2(n – 1). Below, we can see algebraically that all our solutions are equivalent.

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