You might be familiar with the triangular numbers: the number of objects that form equilateral triangles like this:

The sequence is 3, 6, 10, 15, 21, ... The first one, 3, is 1 + 2, the second one, 6, is 1 + 2 + 3, the third one, 10, is 1 + 2 + 3 + 4 and so on. For a triangle with a base of \(n\) blobs the number of blobs in the triangle is \(n * (n + 1) / 2\) (which can be proved by various means and was famously figured out by the mathematician Gauss while a school boy).

But what about the hollow triangular numbers? The ones where you take the middle blobs out of the triangles and just leave the border. Like this:

The pattern there is very simple. They are all multiples of three. And the formula would be that the hollow triangle with a base of size \(n\) has \(3 * (n-1)\) blobs in it (and since that's a multiple of three so is the number of blobs in a hollow triangle).

Here are three ways to prove that that formula is correct.

Because we already have a formula for the filled in triangle (see above) we can use it to figure out the formula for the hollow triangle. Look at the triangle with a base of 6 blobs.

You can see that the number of blobs in the hollow triangle with a base of six is the number of blobs in the filled in triangle of base six minus the number of blobs in the filled in triangle of base three. Or, more, generally the number of blobs in a hollow triangle of base \(n\) is

\[\begin{aligned}

&n * (n+1) / 2 - (n-3) * (n-3 + 1) / 2 \\

&= 1/2 * (n^2 + n - n^2 + 5n - 6) \\

&= (6n - 6)/2 \\

&= 3 * (n - 1) \\

\end{aligned}\]

Which is the formula above.

Another way is to follow around the edge of a triangle and observe that, for example, for the hollow triangle with a base of six the total number of blobs is \(6 + 5 + 4\) (the blue blobs, plus the red blobs, plus the yellow blobs).

And the more general formula would be that for a hollow triangle of base \(n\) the total number of blobs is \(n + (n-1) + (n - 2)\) which simplifies to \(3 * (n - 1)\).

But a real mathematician is likely to prove this in a different way using mathematical induction. That works by beginning with some starting case that can be seen to meet the required statement and a way of going from one case to another.

For example, it's pretty clear that the hollow triangle with four blobs in its base (see above) has nine blobs in total and since nine is divisible by three the statement we are trying to prove 'all hollow triangular numbers are divisible by three' is at least true for it.

Then you just need a way of working up from there. For example, if there was a way to prove that the next hollow triangular number is divisible by three if we know that the previous one was, it would be possible to know that they all were. That's because you'd start with the hollow triangle with four blobs in the base and know that the one with five blobs was divisible by three and then you'd know that the one with six blobs was divisible by three and so on... forever.

So, all you need is to examine one step in this infinite chain of reasoning.

First let's say that \(hollow_n\) is the number of blobs in the hollow triangle with a base containing \(n\) blobs. And let's assume that it is divisible by three. For mathematical induction to be applicable we just need to show that \(hollow_{n+1}\) (the next hollow triangular number) is divisible by three. So, we just need to know how to go from \(hollow_n\) to \(hollow_{n+1}\).

Take a look at the following diagram showing the hollow triangular number with base six (on the left).

You can think of going from \(hollow_6\) to \(hollow_7\) as the same adding a row of seven blobs (on the right) and removing the inside of the old bottom row (coloured in red). So, \(hollow_7 = hollow_6 + 7 - 4\) which simplifies to \(hollow_7 = hollow_6 + 3\). Since we've added three and know that \(hollow_6\) is divisible by three we know that \(hollow_7\) is also.

That can easily be generalized \(hollow_{n+1} = hollow_n + (n+1) - (n-2)\) (added on a row of \(n+1\) at the bottom and removed the interior \(n-2\) blobs leaving just the two blobs on the outside). That simplifies to \(hollow_{n+1} = hollow_n + 3\). And so if we know that \(hollow_n\) is divisible by three then so is \(hollow_{n+1}\) and thus by mathematical induction we know that all hollow triangular numbers are divisible by three.

Now that we've got the idea of mathematical induction it's possible to return to a claim a made at the start: for the filled in triangles the number of blobs is \(n * (n + 1) / 2\) when the base has \(n\) blobs. That too can be proved by induction.

The starting case is a base of two blobs. Just by looking at it we can see that it has three blobs in the triangle and \(2 * (2 + 1) / 2 = 3\). So the formula works for that case.

Then consider going from a triangle with a base of \(n\) blobs to one with \(n+1\). All that happens there is that \(n+1\) blobs get added (the new bottom row). So, if the triangle with a base of \(n\) blobs has \(n * (n + 1) / 2\) in total, the triangle with base \(n +1\) has \(n * (n + 1) / 2 + (n + 1)\) which can be simplified as follows:

\[\begin{aligned}

&n * (n + 1) / 2 + n + 1 \\

&= 1/2(n * (n +1) + 2(n+1)) \\

&= 1/2(n^2 + n + 2n + 2) \\

&= 1/2(n^2 + 3n + 2) \\

&= 1/2(n+1)(n+2) \\

&= (n+1) * ((n+1)+1)/2

\end{aligned}\]

Which is just what was expected.

PS I spotted all this staring at the duvet cover the other night. But then again there was that flight I took back in 2008.

PPS In the comments reader tz points out another way to see that the number of blobs in a hollow triangle is a multiple of three. There are three corners and three equal length sides (when ignoring the corners).

The sequence is 3, 6, 10, 15, 21, ... The first one, 3, is 1 + 2, the second one, 6, is 1 + 2 + 3, the third one, 10, is 1 + 2 + 3 + 4 and so on. For a triangle with a base of \(n\) blobs the number of blobs in the triangle is \(n * (n + 1) / 2\) (which can be proved by various means and was famously figured out by the mathematician Gauss while a school boy).

But what about the hollow triangular numbers? The ones where you take the middle blobs out of the triangles and just leave the border. Like this:

The pattern there is very simple. They are all multiples of three. And the formula would be that the hollow triangle with a base of size \(n\) has \(3 * (n-1)\) blobs in it (and since that's a multiple of three so is the number of blobs in a hollow triangle).

Here are three ways to prove that that formula is correct.

**Subtracting the middle**Because we already have a formula for the filled in triangle (see above) we can use it to figure out the formula for the hollow triangle. Look at the triangle with a base of 6 blobs.

You can see that the number of blobs in the hollow triangle with a base of six is the number of blobs in the filled in triangle of base six minus the number of blobs in the filled in triangle of base three. Or, more, generally the number of blobs in a hollow triangle of base \(n\) is

\[\begin{aligned}

&n * (n+1) / 2 - (n-3) * (n-3 + 1) / 2 \\

&= 1/2 * (n^2 + n - n^2 + 5n - 6) \\

&= (6n - 6)/2 \\

&= 3 * (n - 1) \\

\end{aligned}\]

Which is the formula above.

**Counting around the outside**Another way is to follow around the edge of a triangle and observe that, for example, for the hollow triangle with a base of six the total number of blobs is \(6 + 5 + 4\) (the blue blobs, plus the red blobs, plus the yellow blobs).

And the more general formula would be that for a hollow triangle of base \(n\) the total number of blobs is \(n + (n-1) + (n - 2)\) which simplifies to \(3 * (n - 1)\).

**Mathematical induction**But a real mathematician is likely to prove this in a different way using mathematical induction. That works by beginning with some starting case that can be seen to meet the required statement and a way of going from one case to another.

For example, it's pretty clear that the hollow triangle with four blobs in its base (see above) has nine blobs in total and since nine is divisible by three the statement we are trying to prove 'all hollow triangular numbers are divisible by three' is at least true for it.

Then you just need a way of working up from there. For example, if there was a way to prove that the next hollow triangular number is divisible by three if we know that the previous one was, it would be possible to know that they all were. That's because you'd start with the hollow triangle with four blobs in the base and know that the one with five blobs was divisible by three and then you'd know that the one with six blobs was divisible by three and so on... forever.

So, all you need is to examine one step in this infinite chain of reasoning.

First let's say that \(hollow_n\) is the number of blobs in the hollow triangle with a base containing \(n\) blobs. And let's assume that it is divisible by three. For mathematical induction to be applicable we just need to show that \(hollow_{n+1}\) (the next hollow triangular number) is divisible by three. So, we just need to know how to go from \(hollow_n\) to \(hollow_{n+1}\).

Take a look at the following diagram showing the hollow triangular number with base six (on the left).

You can think of going from \(hollow_6\) to \(hollow_7\) as the same adding a row of seven blobs (on the right) and removing the inside of the old bottom row (coloured in red). So, \(hollow_7 = hollow_6 + 7 - 4\) which simplifies to \(hollow_7 = hollow_6 + 3\). Since we've added three and know that \(hollow_6\) is divisible by three we know that \(hollow_7\) is also.

That can easily be generalized \(hollow_{n+1} = hollow_n + (n+1) - (n-2)\) (added on a row of \(n+1\) at the bottom and removed the interior \(n-2\) blobs leaving just the two blobs on the outside). That simplifies to \(hollow_{n+1} = hollow_n + 3\). And so if we know that \(hollow_n\) is divisible by three then so is \(hollow_{n+1}\) and thus by mathematical induction we know that all hollow triangular numbers are divisible by three.

**Back to the triangular numbers**Now that we've got the idea of mathematical induction it's possible to return to a claim a made at the start: for the filled in triangles the number of blobs is \(n * (n + 1) / 2\) when the base has \(n\) blobs. That too can be proved by induction.

The starting case is a base of two blobs. Just by looking at it we can see that it has three blobs in the triangle and \(2 * (2 + 1) / 2 = 3\). So the formula works for that case.

Then consider going from a triangle with a base of \(n\) blobs to one with \(n+1\). All that happens there is that \(n+1\) blobs get added (the new bottom row). So, if the triangle with a base of \(n\) blobs has \(n * (n + 1) / 2\) in total, the triangle with base \(n +1\) has \(n * (n + 1) / 2 + (n + 1)\) which can be simplified as follows:

\[\begin{aligned}

&n * (n + 1) / 2 + n + 1 \\

&= 1/2(n * (n +1) + 2(n+1)) \\

&= 1/2(n^2 + n + 2n + 2) \\

&= 1/2(n^2 + 3n + 2) \\

&= 1/2(n+1)(n+2) \\

&= (n+1) * ((n+1)+1)/2

\end{aligned}\]

Which is just what was expected.

PS I spotted all this staring at the duvet cover the other night. But then again there was that flight I took back in 2008.

PPS In the comments reader tz points out another way to see that the number of blobs in a hollow triangle is a multiple of three. There are three corners and three equal length sides (when ignoring the corners).

## 10 comments:

Also in each case, the length of each side is equal, but the corners are shared, so the number will be the length of one side minus one, multiplied by three. Or since the lengths are identical, the non-corners will be 3N, and 3 corners, which would also be a multiple o 3.

(Reposted to fix thinko.)

@tz I would describe it as each larger triange adds another ball to each corner. Three corners is three balls. The starting case is the "null triangle" consisting of no balls.

This generalizes to any hollow regular polygon. Balls increase as multiples of the number of sides.

This is highly related to the method of differences:

http://en.wikipedia.org/wiki/Difference_engine#Method_of_differences

That's how Babbage was going to tabulate polynomials.

IRC the difference between (X+1)^n and X^n is O(X^n-1) so you can calculate successive polynomials in O(n) time using simple adding and multiplying by (often small) constants n times.

Nice Article...

As descriptive as any one can want it!

It would have been nice if you had gone just one step further to prove how many blobs make up a pyramid of base(n) blobs.

@Andywal: sounds like a good 'exercise for the reader'

@tz This is the first solution I came up with too. Algebraically:

3 * (n - 2) + 3 = x

And after some manipulation:

(n - 2) + 1 = x/3

n - 2 = x/3 - 1

n = x/3 + 1

n - 1 = x/3

3 (n - 1) = x

Fun stuff.

=D

Ummm... the proof that they're multiples of three is obvious: to make the next bigger triangle, you add one dot to each side, thus three dots - it's impossible to make one that isn't a multiple of three!

In other words, the total number of dots = 3 x (1 vertex + 1 side). This is the same as saying 3n = a multiple of 3. I don't think you really require a

prooffor that... and it works for any n-sided polygon, if you think about it.Post a Comment