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Sequences and PatternsFigurate Numbers

Reading time: ~35 min

The name for geometric sequences is pretty confusing, because they don’t have anything to do with geometry. In fact, the name was developed hundreds of years ago, when mathematicians thought about multiplication and square roots in a much more geometric way.

However, there are many other sequences that are based on certain geometric shapes – some of which you already saw in the introduction. These sequences are often called figurate numbers, and in this section we will have a closer look at some of them.

Triangle Numbers

The triangle numbers are generated by creating triangles of progressively larger size:

1

triangle-1

3

triangle-2

6

triangle-3

10

triangle-4

15

triangle-5

21

triangle-6

You’ve already seen the recursive formula for triangle numbers: xn= .

It is no coincidence that there are always 10 pins when bowling or 15 balls when playing billiards: they are both triangle numbers!

Unfortunately, the recursive formula is not very helpful if we want to find the 100th or 5000th triangle number, without first calculating all the previous ones. But, like we did with arithmetic and geometric sequences, we can try to find an explicit formula for the triangle numbers.

Let’s start with a triangle of size ${n}, containing ${n×(n+1)/2} circles. Rather than counting them one by one, we can find a formula.

First, we make a second copy of the triangle, flipped upside down.

Now we can rearrange the two triangles, so that they fit together perfectly, without any gaps or overlaps.

The two triangles now form a rectangle with ${n} rows and ${n + 1} columns – ${n} × ${n + 1} = ${n×(n+1)} circles in total.

The rectangle is exactly twice as large as the original triangle. Therefore the ${n}th triangle number must be half of that: ${n×(n+1)/2}. In general, the nth triangle number is Tn=n×n+12.

Triangle numbers seem to pop up everywhere in mathematics, and you’ll see them again throughout this course. One particularly interesting fact is that any whole number can be written as the sum of at most three triangle numbers:

${n}

=

+

+

The fact that this works for all whole numbers was first proven in 1796 by the German mathematician Carl Friedrich Gauss – at the age of 19!

Problem Solving

What is the sum of the first 100 positive integers? In other words, what is the value of

1+2+3+4+5+…+97+98+99+100?

Rather than manually adding up everything, can you use the triangle numbers to help you? What about the sum of the first 1000 positive integers?

Square and Polygonal Numbers

Another sequence that is based on geometric shapes are the square numbers:

1, 4 +3, 9 +5, 16 +7, +9, +11, +13, +15, …

You can calculate the numbers is this sequence by squaring every whole number (12, 22, 32, …), but it turns out that there is another pattern: the differences between consecutive square numbers are the in increasing order!

The reason for this pattern becomes apparent if we actually draw a square. Every step adds one row and one column. The size of these “corners” starts at 1 and increases by 2 at every step – thereby forming the sequence of odd numbers.

This also means that the nth square number is just the sum of the first n odd numbers! For example, the sum of the first 6 odd numbers is

1+3+5+7+9+11= .

1 3 5 7 9 11 13

In addition, every square number is also the sum of two consecutive triangle numbers. For example, ${n×n} = ${n×(n+1)/2} + ${n×(n-1)/2}. Can you see how we can split every square along its diagonal, into two triangles?

x=

After triangle and square numbers, we can keep on going with larger polygons. The resulting number sequences are called polygonal numbers.

For example, if we use polygons with ${k} sides, we get the sequence of ${polygonName(k)} numbers.

Can you find recursive and explicit formulas for the nth polygonal number that has k sides? And do you notice any other interesting patterns for larger polygons?

Tetrahedral and Cubic Numbers

Of course, we also don’t have to limit ourselves to two-dimensional shapes and patterns. We could stack spheres to form small pyramids, just like how you would stack oranges in a supermarket:

1

20

35

Mathematicians often call these pyramids tetrahedra, and the resulting sequence tetrahedral numbers.

Just like we did for triangle and square numbers, let’s look at the differences between consecutive tetrahedral numbers:

1, 4 +3, 10 +6, 20 +10, 35 +15, +21, +28, …

The differences are the triangle numbers: 3, 6, 10, 15, 21, 28, …! To get the next tetrahedral number, you simply add the next triangle number. This gives us a recursive formula:

Ten = Ten−1 + Tn

For example, the 5th tetrahedral number is 20 + = 35 – the sum of the first five triangle numbers.

We can also find an explicit formula that jumps straight to the n-th tetrahedral number, without adding up all the previous triangle numbers:

Ten =

For example, the 5th tetrahedral number is 5 × 6 × 7 ÷ 6 = .

Tetrahedral numbers appear wherever objects get stacked into pyramids: oranges at a market stall, cannonballs in an old fort, or gifts in the song “The 12 Days of Christmas” – which we will count in a moment. Here is the pyramid with 6 layers, containing Te6 = 56 spheres.

Cubic Numbers

There is one more sequence hiding inside these pyramids. If we stack square layers on top of each other – or simply multiply a number by itself twice – we get the cubic numbers:

1, 8 +7, 27 +19, 64 +37, 125 +61, +91, …

Unlike arithmetic or geometric sequences, the cubic numbers have no constant difference or ratio: the differences (7, 19, 37, 61, 91, …) keep changing, and even the differences between those differences (12, 18, 24, …) keep growing!

But there is a very simple explicit formula: the n-th cubic number is

xn =

Cubic numbers hide one of the most beautiful identities in all of mathematics. Move the slider to arrange the cubes 13, 23, 33, … as coloured “corners” of a big square:

With n = 1: 1

= (1 + 2 + … + n)² = 1² = 1

The coloured corners are the cubes 13, 23, 33, … and together they always form a perfect square! The side length of this square is the triangle number 1+2+…+n, which gives us Nicomachus’s theorem:

13+23+…+n3=1+2+…+n2

For example, 1+8+27+64+125+216=441=212, and 21 is the 6th triangle number!

The 12 Days of Christmas

Tetrahedral numbers have one famous real-world appearance: the song “The 12 Days of Christmas”. On day 1, your true love gives you 1 gift. On day 2, you receive 2 turtle doves plus another partridge – that’s 3 gifts. On day 3 you get 3 + 2 + 1 = 6 gifts, and so on: the number of gifts on each day is a triangle number, and the running total is a tetrahedral number. Can you fill in the table?

DayGifts that dayTotal gifts
111
234
310
410
51535
656
72884
836
945165
1055220
11286
1278

By the end of the song you have received 364 gifts in total: Te12=12×13×14÷6. That is almost exactly one gift for every day of the year – except Christmas itself!

Problem Solving

1. How many gifts would you receive in “The n Days of Christmas”, for any value of n? Which formula do you get?

2. The running totals satisfy Ten=Ten−1+Tn. Can you combine this with the explicit formula for triangle numbers to verify the explicit formula for tetrahedral numbers?