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Pascal's Triangle and Cube Numbers

To help explain where cube numbers can be found in Pascal's triangle, I will first briefly explain how the square numbers are formed. The third diagonal in of Pascal's triangle is 1,3,6,10,15,21... If we add together each of these numbers with its previous number, we get 0+1=1, 1+3=4, 3+6=9, 6+10=16... , which are the square numbers. The way cube numbers can be formed from Pascal's triangle is similar, but a little more complex. Whilst the square numbers could be found in the third diagonal in, for the cube numbers, we must look at the fourth diagonal. The first few rows of Pascal's triangle are shown below, with these numbers in bold:

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

This sequence is the tetrahedral numbers, whose differences give the triangle numbers 1,3,6,10,15,21 (the sums of whole numbers e.g. 21 = 1+2+3+4+5). However, if you try adding up consecutive pairs in the sequence 1,4,10,20,35,56, you do not get the cube numbers. To see how to get this sequence, we will have to look at the formula for tetrahedral numbers, which is (n)(n+1)(n+2)/6. If you expand this, it you get (n^3 + 3n^2 + 2n)/6. Basically, we are trying to make n^3, so a good starting point is that here we have a n^3/6 term, so we are likely to need to add together six tetrahedral numbers to make n^3, not 2. Have a go at trying to find the cube numbers from this information. If you're still stuck, then look at the next paragraph.

List the tetrahedral numbers with two zeros first: 0,0,1,4,10,20,35,56... 

Then, add three consecutive numbers at a time, but multiply the middle one by 4:

0 + 0 x 4 + 1 = 1 = 1^3

0 + 1 x 4 + 4 = 8 = 2^3

1 + 4 x 4 + 10 = 27 = 3^3

4 + 10 x 4 + 20 = 64 = 4^3

10 + 20 x 4 + 35 = 125 = 5^3

This pattern does in fact, always continue. If you want to see why this is the case, then try exanding and simplifying (n(n+1)(n+2))/6 + 4(n-1)(n)(n+1)/6 + ((n-2)(n-1)n)/6, which are the formulas for the nth, (n-1)th and (n-2)th tetrahedral numbers, and you should end up with n^3. Otherwise, as I expect is the case (and I don't blame you), just enjoy the this interesting result and test it out on your friends and family to find out if they can spot this hidden link between Pascal's triangle and cube numbers!

Pascal's Triangle and Pascal's Tetrahedron

I have begun by showing the first 4 layers of Pascal's tetrahedron below:

Layer 0:

1

Layer 1:

1

1 1

Layer 2:

1

2 2

1 2 1

Layer 3:

1

3 3

3 6 3

1 3 3 1

Layer 4:

1

4 4

6 12 6

4 12 12 4

1 4 6 4 1

In layer 3, the final row of the layer is 1,3,3,1, row three of Pascal's triangle, the final row of layer 4 the 4th, and so on for all the layers listed above. In fact, if you write out each of the layers shown above in a centered equilateral triangle, you will notice that every edge of each triangular layer is that layer's corresponding row in Pascal's triangle.

This pattern does, in fact, always continue. Without delving too deeply into rigorous mathematical proofs, you should if you think about it be able to see that on the edges of layers, you only ever add together two numbers from the layer above, and so really it's just like Pascal's triangle (where you add two numbers from the row above) repeated three times at various angles.

However, the links run even deeper than this (in more ways than one). Its not just about the edges of the pyramid, there are links deep down inside the very core of the tetraheron.

To understand this, we are going to use layer 4 as an example, but this time, we will not look at an edge but as some of the rows running through the middle of the layer. For example, is there anything interesting about the second to last row, which goes 4,12,12,4? In fact there is (otherwise I wouldn't be asking). For the moment, let's just forget about the numbers themselves, and think only about the ratio between them. This gives a 1:3:3:1 ratio, as the middle two numbers are thrice the outside two numbers. This ratio happens to be the third row of Pascal's triangle. Is that just a coincidence?

Next, let's investigate the third last row in layer 4, which goes 6,12,6. This time, the ratio is 1:2:1, the second row. Something is definitely going on here. Below is the whole of layer 4, split into rows, with their ratios and where they can be found in Pascal's triangle:

Layer 4:

1

4 4 - ratio 1:1 (row 1)

6 12 6 - ratio 1:2:1 (row 2)

4 12 12 4 - ratio 1:3:3:1 (row 3)

1 4 6 4 1 - ratio 1:4:6:4:1 (row 4)

So, amazingly, every single row in layer 4 is in the ratio of the row in Pascal's triangle which has the same number of numbers in it! However, just when you thought it couldn't get any more exciting, look at what we have to multiply the ratios by to get the actual numbers in Pascal's tetrahedron again:

1 (1) x 1

4 4 - (1,1) x 4

6 12 6 - (1,2,1) x 6

4 12 12 4 - (1,3,3,1) x 4

1 4 6 4 1 - (1,4,6,4,1) x 1

They are the 4th row of Pascal's triangle! Only now do we truly see the extent of the links between these two patterns of numbers. Every single number in the pyramid is simply two numbers from Pascal's triangle multiplied together. Not only is this in my opinion a beautiful discovery which is an excellent demonstration of the interconnected nature of mathematics, it makes what seemed like the much more complex Pascal's tetrahedron easy to work with. In fact, it is these links that have helped mathematicians to modify the formula for Pascal's triangle to one which applies to Pascal's tetrahedron and even to suit higher dimensions, so it is certainly a very powerful discovery!

Pascal's Tetrahedron

We are trying to create a triangular pyramid of numbers. Specifically, this should be a triangular based pyramid, not a square based pyramid like those in Egypt (there's nothing to stop you exploring a square based Pascal's pyramid, however, which is bound to have many interesting patterns, properties and links to the triangular version waiting to be discovered). At the very tip of the pyramid, we start with the number 1. Instead of looking down rows as in Pascal's triangle, we are interested in the layers of this pyramid, and each layer should be a triangle of numbers. Whilst in Pascal's triangle, each number is the sum of the two above, in Pascal's tetrahedron is the sum on numbers on the layer above.

It's easy to get confused at first when writing out the layers of Pascal's tetrahedron and thinking about what is supposed to add up to make what. Most people start off OK by writing down the first couple of layers like this:

Layer 0:

1

Layer 1:

1

1 1

Then, however, they want to add up all three numbers in layer 1 and put a 3 directly below the middle of them in layer 2. This is where they get confused as they can't make the numbers in layer 2 form a triangular shape.

What we actually need to do for layer 2 is take the sums of each of the three edges from layer 1 and also directly outwards, treating each number as a corner. Then, for layer 2 we get what is shown below:

1

2 2

1 2 1

Although in layer 2 we never added three numbers from the previous layer together, sometimes you have to. I find Pascal's pyramid very hard to visualise, so if you're finding this hard, then you're not alone! To avoid arranging and adding the numbers incorrectly, I have a couple of suggestions. Firstly, when writing out layers, centralise them rather than left justifying them. This makes is easier to see the symmetry in the layers and see triangles of numbers which you might have to add together.

Secondly, if you can, try to create some sort of model of Pascal's pyramid. This done most easily with cubes - if you have enough dice you can cover each one with paper and write your own numbers on them, and then stack them in a pyramid. This is, however, a little fiddly, so you may find it easier to draw equilateral triangles of dots of different sizes on separate laminates, tracing paper or thin tissue paper. If you use a different colour dot on each sheet, and place them on top of each other, you can see quite easily which dots from the top sheet are adjacent to any dot from the bottom sheet, telling what numbers you have to add together to calculate the numbers on the bottom sheet.

So you can check you've got the hang of it, I have listed the first the first few layers of Pascal's pyramid below:

Layer 0:

1

Layer 1:

1

1 1

Layer 2:

1

2 2

1 2 1

Layer 3:

1

3 3

3 6 3

1 3 3 1

Layer 4:

1

4 4

6 12 6

4 12 12 4

1 4 6 4 1

Layer 5:

1

5 5

10 20 10

10 30 30 10

5 20 30 20 5

1 5 10 10 5 1

Once you are confident at how Pascal's tetrahedron works, there is no end of fun to be had. The first and most obvious question is exactly how it links with Pascal's triangle. This can be done by thinking about how patterns from Pascal's triangle can be applied to Pascal's tetrahedron, and from there making comparisons between the two. Try to discover some new patterns and properties in the more complex world of Pascal's tetrahedron for yourself!

Properties of Pascal's Pyramid

One famous property of Pascal's triangle is that the sums of the rows are the doubling numbers. Rather than looking at the sums of rows in Pascal's pyramid, we can see if we get any similar patterns when we look at the sums of layers. This has been done for layers 0 to 4 below:

Layer 0:

1

Total = 1

Layer 1:

1

1 1

Total = 1 + 1 + 1 = 3

Layer 2:

1

2 2

1 2 1

Total = 1 + 2 + 2 + 1 + 2 + 1 = 9

Layer 3:

1

3 3

3 6 3

1 3 3 1

Total = 1 + 3 + 3 + 3 + 6 + 3 + 1 + 3 + 3 + 1 = 27

Layer 4:

1

4 4

6 12 6

4 12 12 4

1 4 6 4 1

Total = 1 + 4 + 4 + 6 + 12 + 6 + 4 + 12 + 12 + 4 + 1 + 4 + 6 + 4 + 1 = 81

The sums of the layers triple each time, producing a formula for the sum of the nth layer of 3^n. In fact, this property can tell us something else about Pascal's pyramid. Things can easily get very complicated with this, so I'm not going to try too explain this too much. If we look at the average of the numbers in each row of Pascal's triangle, we get the following results for the first few rows:

1, 1, 1.33, 2, 3.2, 5.33, 9.14...

Now, if we write down what you have to multiply each term by to get to next, you get

1, 1.33, 1.5, 1.6, 1.67, 1.71, 1.75...

If you kept on going, you would get a value closer and closer to 2, so you would get the average of the numbers in the rows eventually doubling each time.

If you try this with the average of the layers in Pascal's tetrahedron, you should find you get a sequence which gets closer and closer to tripling each time. This explains why the numbers get large so much faster in Pascal's tetrahedron.

We can also look at the symmetry of Pascal's tetrahedron. If you arrange each layer as an equilateral triangle, it has rotational symmetry of order 3, and reflective symmetry from each of its corners to the midpoint of the opposite side. This may sound complicated, but let's think about Pascal's triangle for a moment. in each row, every number appears twice unless it is in the very centre of the row. This is due to the symmetry through the centre of the triangle. Pascal's tetrahedron, however, due to slightly more complicated symmetry, has every number repeated three times is each layer, with the exception of a number which is sometimes found in the very centre of the triangular layer, such as the 6 in layer 3.

It is clear, therefore, that many of the properties of Pascal's triangle apply in some way to Pascal's pyramid as well, but it is interesting to think about how these patterns have evolved to suit Pascal's tetrahedron, and the reasons for these changes.

Pascal's Triangle and Square Numbers

Below are the first few rows of Pascal's triangle:

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 21 15 6 1

The numbers in bold are the third diagonal in when Pascal's triangle is drawn centrally. These are the triangle numbers, made from the sums of consecutive whole numbers (e.g. 15 = 1 + 2 + 3 + 4 + 5), and from these we can form the square numbers. All we have to do is add up consecutive numbers from these and we get the square numbers. To get the first square number, we have to add a 0 on to the front of the list:

0, 1, 3, 6, 10, 15, 21...

0 + 1 = 1 = 1^2

1 + 3 = 4 = 2^2

3+ 6 = 9 = 3^2

6 +10 = 16 = 4^2

10 +15 = 25 = 5^2

15 +21 = 36 = 6^2

Incidentally, you can also get the square numbers by taking the differences of numbers two places apart on the 4th diagonal in of Pascal's triangle. The fourth diagonal goes 1, 4, 10, 20 35... , and the differences you get are 1-0 = 1, 4-0 = 4, 10-1 = 9, 20-4 = 16, 35-10 = 25 and so on.

To understand why you get the square numbers from adding together consecutive triangle numbers, you can use a variety of methods. Firstly, if you know that the formula for the nth triangle number is (n^2 + n)/2, then the previous triangle number is n less than this, as it's the same sum of numbers but with (n-1) and not n as the last number you add. If we then add together these two numbers, we get

(n^2 + n)/2 + (n^2 + n)/2 - n

= (1/2)n^2 + n/2 + (1/2)n^2 + n/2 - n

= n^2 + n - n

= n^2

If that method was not to your liking, we can also show this result pictorially. Triangle numbers derive their name from fact that you can make them by adding up the number of dots that make different sizes of triangle, and square numbers from the number of dots that make up different sized squares. So all we need to do is make a square from two triangles of dots. If you try this out with coins or counters, or on paper, and make right-angled triangles, you should find that you can make a square from two triangles, but one has to be one counter smaller on each of its sides. Okay, that wasn't that rigorous a method for proving it, but it was a lot easier than doing a lot of algebra, wasn't it?

The Formula for Pascal's Triangle

The formula for Pascal's triangle is n!/(r!(n-r)!). This needs quite a lot of explaining. Firstly, what are n and r supposed to be in this formula? In this equation, n means the row in which the number you're trying to find the value of is. For the formula to work, we must count the row 1,1 as row 1. r is the number of numbers across your number is. In slightly better English, you count from left to right how far along the number you want to calculate is, and this gives you a value for r. The catch, however, is that we count the leftmost 1 as r=0, so you then have to subtract 1. Let's look at an example:

1 1

1 2 1

1 3 3 1

1 4 6 ? 1

So, what values would I use for n and r if I wanted to calculate the number in the position of the question mark? We are in the fourth row down, so n=4. Counting from left to right, the question mark is the fourth number along (okay, a question mark isn't a number, I know, but you get my point). However, we must remember that the first 1 in the row is r=0. Counting up from 0 gives r=3. So, in this example, we get 4!/(3!(4-3)!).

This is all very well, but we still don't know what this actually equals. However, have patience, as I am about to explain what all those "!" signs were for (no, they weren't just punctuation to show how amazing the formula is!)

To calculate n! you have to multiply together every positive whole number up to and including n itself. Perhaps a few examples will clarify this:

1! = 1

2! = 1 x 2 = 2

3! = 1 x 2 x 3 = 6

4! = 1 x 2 x 3 x 4 = 24

10! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3628800

It should now be clear how to use the formula for Pascal's triangle! I have given one example here just to you how cool it is:

Say I was walking down the street one day and suddenly needed to know (as one so often does) what the 5th number along on the 14th row of Pascal's triangle was. What do I do?

Firstly, I would have to decide what n and r are. n = 14 as I am interested in the 14th row, and r = 4, as we count along five numbers starting from 0. So, we get 14!/(5!(15-5)!)

= 87178291200/(24×3628800)=1001

So, the number I was looking for was therefore 1001.

This is definitely one of my favourite mathematical formulas. I suggest you try out a few examples yourself to get a feel for how it works. Then, you can try it on your friends and impress them with how fast you can calculate numbers in Pascal's triangle. Have fun!