IB Maths and GCSE Maths Resources from British International School Phuket. Theory of Knowledge (ToK). Maths explorations and investigations.  Real life maths. Maths careers. Maths videos. Maths puzzles and Maths lesson resources.

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British International School Phuket

Welcome to the British International School Phuket’s maths website.  I am currently working at BISP and so I am running my site as the school’s maths resources website for both our students and students around the world.

We are a British international school located on the tropical island of Phuket in Southern Thailand. We offer a number of scholarships each year, catering for a number of national and international standard sports stars as well as for academic excellence. You can find out more about our school here.

maths students

BISP has a very proud tradition in mathematical excellence.  Our students have  achieved the top in Thailand awards for the Cambridge Maths IGCSEs 3 years in a row.  Pictured above are our world-class maths students for this year, Minjin Kang and Natchongrat (Oy) Terdkiatkhachorn.

Please explore the site – there is a huge amount of content!  Some of the most popular includes:

A large “Flipping the classroom” videos section for IB students.  This covers the entire IB HL, SL and Studies syllabus.

A new School Code Challenge activity which allows students to practice their code breaking skills – each code hides the password needed to access the next level.

Over 200 ideas to help with students’ Maths Explorations – many with links to additional information to research.


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Plotting Stewie Griffin from Family Guy

Computer aided design gets ever more important in jobs – and with graphing software we can  create art using maths functions.  For example the above graph was created by a user, Kara Blanchard on Desmos.  You can see the original graph here, by clicking on each part of the function you can see which functions describe which parts of Stewie.  There are a total of 83 functions involved in this picture.  For example, the partial ellipse:

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when x is bounded between 3.24 and 0.9, and y is bounded as less than 1.5 generates Stewie’s left cheek.  This is what he looks like without it:

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By clicking on the various functions you can discover which ones are required to complete the full drawing.  Other artwork designed by users includes:

A minion from Despicable Me 

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A sunflower:

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And Hello Kitty:

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See if you can create some designs of your own!  This could make an interesting maths investigation for anyone thinking about a career in computer design or art – as it is a field which will grow in importance in the coming years.

You might also like to look at a similar post on using Wolfram Alpha to plot the Batman and Superman logos.  


Modeling Volcanoes – When will they erupt?

A recent post by the excellent Maths Careers website looked at how we can model volcanic eruptions mathematically. This is an important branch of mathematics – which looks to assign risk to events and these methods are very important to statisticians and insurers. Given that large-scale volcanic eruptions have the potential to end modern civilisation, it’s also useful to know how likely the next large eruption is.

The Guardian has recently run a piece on the dangers that large volcanoes pose to humans.  Iceland’s Eyjafjallajökull volcano which erupted in 2010 caused over 100,000 flights to be grounded and cost the global economy over $1 billion – and yet this was only a very minor eruption historically speaking.  For example, the Tombora eruption in Indonesia (1815) was so big that the explosion could be heard over 2000km away, and the 200 million tones of sulpher that were emitted spread across the globe, lowering global temperatures by 2 degrees Celsius.  This led to widespread famine as crops failed – and tens of thousands of deaths.


Super volcanoes

Even this destruction is insignificant when compared to the potential damage caused by a super volcano.  These volcanoes, like that underneath Yellowstone Park in America, have the potential to wipe-out millions in the initial explosion and and to send enough sulpher and ash into the air to cause a “volcanic winter” of significantly lower global temperatures.  The graphic above shows that the ash from a Yellowstone eruption could cover the ground of about half the USA. The resultant widespread disruption to global food supplies and travel would be devastating.

So, how can we predict the probability of a volcanic eruption?  The easiest model to use, if we already have an estimated probability of eruption is the Poisson distribution:

 P(X{=}k)= \frac{\lambda^k e^{-\lambda}}{k!},

 This formula calculates the probability that X equals a given value of k.  λ is the mean of the distribution.  If X represents the number of volcanic eruptions we have Pr(X ≥1) = 1 – Pr(x = 0).  This gives us a formula for working out the probability of an eruption as 1 -e.  For example, the Yellowstone super volcano erupts around every 600,000 years.  Therefore if λ is the number of eruptions every year, we have λ = 1/600,000  ≈ 0.00000167 and 1 -e also ≈ 0.00000167. This gets more interesting if we then look at the probability over a range of years. We can do this by modifying the formula for probability as 1 -e-tλ where t is the number of years for our range.

So the probability of a Yellowstone eruption in the next 1000 years is 1 -e-0.00167 ≈ 0.00166, and the probability in the next 10,000 years is 1 -e-0.0167 ≈ 0.0164. So we have approximately a 2% chance of this eruption in the next 10,000 years.


A far smaller volcano, like Katla in Iceland has erupted 16 times in the past 1100 years – giving a average eruption every ≈ 70 years. This gives λ = 1/70 ≈ 0.014. So we can expect this to erupt in the next 10 years with probability 1 -e-0.14 ≈ 0.0139. And in the next 30 years with probability 1 -e-0.42 ≈ 0.34.

The models for volcanic eruptions can get a lot more complicated – especially as we often don’t know the accurate data to give us an estimate for the λ.  λ can be estimated using a technique called Maximum Likelihood Estimation – which you can read about here.

If you enjoyed this post you might also like:

Black Swans and Civilisation Collapse. How effective is maths at guiding government policies?


Can you find a sequence of consecutive integers that add up to 1000?

This puzzle is based on the excellent book A First Step to Mathematical Olympiad Problems – which is full of problems that could be extended to become exploration ideas.

Step 1 – arithmetic formula

Our first step is to write out what we want:

a + (a+1) + (a+2) + … (a +n) = 1000

next we notice that the LHS is an arithmetic series with first term a, last term a+n and n+1 terms.  Therefore we can use the sum of an arithmetic sequence formula:

Sn = 0.5n(u1 + un)

Sn = 0.5(n+1)(a + a+n) = 1000

Sn = (n+1)(2a+n) = 2000

Step 2 – logic

However, we currently have 2 unknowns, n and a, and only 1 equation – so we can’t solve this straight away.  However we do know that both a and n are integers – and n can be taken as positive.

The next step is to see that one of the brackets (n+1)(2a+n) must be odd and the other even (if  n is odd then 2a + n is odd.  If n is even then n+1 is odd).   Therefore we can look at the odd factors of 2000:

Step 3 – prime factorisation

Using prime factorisation: 2000 = 24 x 5³

Therefore any odd factors must solely come from the prime factor combinations of 5 – i.e 5, 25 and 125.

Step 4 – trial and error

So we now know that either (n+1) or (2a+n) must be 5, 25, 125.  And therefore the other bracket must be 400, 80 or 16 (as 5 x 400 = 2000 etc).  Next we can equate the (n+1) bracket to one of these 6 values, find the value of n and hence find a.  For example:

If one bracket is 5 then the other bracket is 400.

So if (n+1) = 5 and (2a+n) = 400 then n = 4 and a = 198.

This means that the sequence: 198+199+200+201+202 = 1000.

If (n+1) = 400 and (2a+n) = 5 then n = 399 and a = -197.

This means the sequence: -197 + -196+ -195 … + 201 + 202 = 1000.

We follow this same method for brackets 25, 80 and 125,16.  This gives the following other sequences:

28+29+30+…+51+52 = 1000

-54+-53+-52+…+69+70 = 1000

-27+-26+-25+…+51+52 = 1000

55+56+57+…+69+70 = 1000

So with a mixture of mathematical formulae, prime factorisation, logic and trial and error we have our solutions.  A good example of how mathematics is often solved in reality!

Mandelbrot and Julia Sets – Pictures of Infinity

The above video is of a Mandelbrot zoom.  This is a infinitely large picture – which contains fractal patterns no matter how far you enlarge it.  To put this video in perspective, it would be like starting with a piece of A4 paper and enlarging it to the size of the universe – and even at this magnification you would still see new patterns emerging.

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To understand how to make the Mandelbrot set, we first need to understand Julia sets.  Julia sets are formed by the iterative process:

Zn+1 = Zn2 + c

Here Z is a complex number (of the form a + bi) and c is a constant that we choose.  So, for example if we choose Z1 = 1+i and c = 1 then:

Z2 = Z12 + 1
Z2 =(1+i)2 + 1
Z2 = 2i + 1

We then repeat this process:
Z3 = Z22 + 1
Z3 = (2i+1)2 + 1
Z3 = 4i-3

and so on – what we are looking for is whether this iterated Z value will diverge to infinity (i.e get larger and larger) or if it will remain bounded. If diverges to infinity we colour the initial point 1+i as red on a complex axis. If it remains bounded we will colour it in black. In this case our initial point 1 + i will diverge to infinity and so it will be coloured in red.

Next we do this for every single point in the complex plane – each time seeing what happens when we iterate it many times. Each time we colour it in as red if it diverges and black if it remains bounded. Once we have done that we will have a picture which represents what happens to every point in the complex plane. This then is our Julia set.

For example the Julia set for c = 1 looks like this:

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This is because every single complex number when iterated by  Zn+1 = Zn2 + 1 will diverge to infinity (get infinitely big).

Not very interesting so far, but different values of c provide some amazing patterns.

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This above pattern is generated by c = 0.376 – 0.1566i.

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and this pattern is for c = 0.376 – 0.1566i.

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and this one is c = -0.78 + 0.1i.

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This last one for c = 0.4 + 0.1i looks different to the others – this one has patterns but they are not connected together as in the other examples.

Mandelbrot Set

This brings us on to how to calculate the Mandelbrot set.  We calculate every possible Julia set for all complex numbers c, and then for every Julia set which is connected then we colour the c value in black, and every value of c which the Julia set is disconnected we colour the c value in red.  We then have a new plot in the complex plane of c values.  This gives us the Mandelbrot set shown below:

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Don’t worry if this seem a bit complicated – it is!  You can play around making your own Julia sets by choosing a c value at this online generator.  You might also like towatch the Numberphile video on the same topic:

 If you enjoyed this post you might also like Dan Pearcy’s post on this topic which explains how Geogebra can be used to generate these sets.  Also PlusMaths have a number of posts on this amazing subject

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Reaction times – How fast are you?

Go to the Human Benchmark site and test your reaction times.  You have five attempts to press the mouse as soon as you see the screen turn green.  You can then see how your reaction times compare with people around the world.  According to the site over there have been over 15 million clicks – with a median reaction time of 251 milliseconds and a mean reaction time of 262 milliseconds.

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We can see how this data looks plotted on a chart.  As we can see this is quite a good approximation of a bell curve – but with a longer tail to the right (some people have much longer reaction times than we would expect from a pure normal distribution).  In a true normal distribution we would have the mean and the median the same.  Nevertheless this is close enough to model our data using a normal distribution.

From the data we can take the mean time as 255 milliseconds, and a standard deviation of around 35 (just by looking at the points where around 68% are within 1s.d)

So, with X ∼ N(255, 35²) we can then see how we compare with people around the world.  Reaction times significantly faster than average would suggest an ability to do well in sports such as baseball or cricket where batters need to react to the ball in a fraction of a second.

I just tried this, and got an average of 272.  I can work out what percentage of the population I’m faster than by doing the normal distribution calculation – which gives 31% of people slower than this.  Trying it again gives an average of 261 – this time 43% of people would be slower than this.

Have a go yourselves and see how you get on!

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Are you Psychic?

There have been people claiming to have paranormal powers for thousands of years.  However, scientifically we can say that as yet we still have no convincing proof that any paranormal abilities exist.  We can show this using some mathematical tests – such as the binomial or normal distribution.

ESP Test 

You can test your ESP powers on this site (our probabilities will be a little different than their ones).  You have the chance to try and predict what card the computer has chosen.  After repeating this trial 25 times you can find out if you possess psychic powers.  As we are working with discrete data and have a fixed probability of guessing (0.2) then we can use a binomial distribution.  Say I got 6 correct, do I have psychic powers?

We have the Binomial model B(25, 0.2), 25 trials and 0.2 probability of success.  So we want to find the probability that I could achieve 6 or more by luck.

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The probability of getting exactly 6 right is 0.16.  Working out the probability of getting 6 or more correct would take a bit longer by hand (though could be simplified by doing 1 – P(x ≤ 5).  Doing this, or using a calculator we find the probability is 0.38.  Therefore we would expect someone to get 6 or more correct just by guessing 38% of the time.

So, using this model, when would we have evidence for potential ESP ability?  Well, a minimum bar for our percentages would probably be 5%.  So how many do you need to get correct before there is less than a 5% of that happening by chance?

Using our calculator we can do trial and error to see that the probability of getting 9 or more correct by guessing is only 4.7%.  So, someone getting 9 correct might be showing some signs of ESP.  If we asked for a higher % threshold (such as 1%) we would want to see someone get 11 correct.

Now, in the video above, one of the Numberphile mathematicians manages to toss 10 heads in a row.  Again, we can ask ourselves if this is evidence of some extraordinary ability.  We can calculate this probability as 0.510 = 0.001. This means that such an event would only happen 0.1% of the time. But, we’re only seeing a very small part of the total video. Here’s the full version:

Suddenly the feat looks less mathematically impressive (though still an impressive endurance feat!)

You can also test your psychic abilities with this video here.

Medical Data Mining

It’s worth watching the video above, where Derren Brown manages to flip 10 heads in a row.  With Derren being a professional magician, you might expect some magic or sleight of hand – but no, it’s all filmed with a continuous camera, and no tricks.  So, how does he achieve something which should only occur with probability (0.5)10 ≈ 0.001, or 1 time in every thousand?  Understanding this trick is essential to understanding the dangers of accepting data presented to you without being aware of how it was generated.

At 7 minutes in Derren reveals the trick – it’s very easy, but also a very persuasive way to convince people something unusual is happening.  The trick is that Derren has spent the best part of an entire day tossing coins – and only showed the sequence in which he achieved 10 heads in a row.  Suddenly with this new information the result looks much less remarkable.

Scientific tests are normally performed to a 5% confidence interval – that is, if there is a less than 5% chance of something happening by chance then we regard the data as evidence to reject the null hypothesis and to accept the alternate hypothesis.   In the case of the coin toss, we would if we didn’t know better, reject the null hypothesis that this is a fair coin and conjecture that Derren is somehow affecting the results.

Selectively presenting results from trials is called data mining – and it’s a very powerful way to manipulate data.  Unfortunately it is also a widespread technique in the pharmaceutical industry when they release data on new drugs.  Trials which show a positive effect are published, those which show no effect (or negative effects) are not.  This is a massive problem – and one which has huge implications for people’s health.  After all, we are prescribed drugs based on scientific trials which attest to their efficiency.  If this data is being mined to skew results in the drug company’s favour then we may end up taking drugs that don’t work – or even make us worse.

Dr Ben Goldacre has written extensively on this topic – and an extract from his article “The Drugs Don’t Work” is well worth a read:

The Drugs Don’t Work

Reboxetine is a drug I have prescribed. Other drugs had done nothing for my patient, so we wanted to try something new. I’d read the trial data before I wrote the prescription, and found only well-designed, fair tests, with overwhelmingly positive results. Reboxetine was better than a placebo, and as good as any other antidepressant in head-to-head comparisons. It’s approved for use by the Medicines and Healthcare products Regulatory Agency (the MHRA), which governs all drugs in the UK. Millions of doses are prescribed every year, around the world. Reboxetine was clearly a safe and effective treatment. The patient and I discussed the evidence briefly, and agreed it was the right treatment to try next. I signed a prescription.

But we had both been misled. In October 2010, a group of researchers was finally able to bring together all the data that had ever been collected on reboxetine, both from trials that were published and from those that had never appeared in academic papers. When all this trial data was put together, it produced a shocking picture. Seven trials had been conducted comparing reboxetine against a placebo. Only one, conducted in 254 patients, had a neat, positive result, and that one was published in an academic journal, for doctors and researchers to read. But six more trials were conducted, in almost 10 times as many patients. All of them showed that reboxetine was no better than a dummy sugar pill. None of these trials was published. I had no idea they existed.

It got worse. The trials comparing reboxetine against other drugs showed exactly the same picture: three small studies, 507 patients in total, showed that reboxetine was just as good as any other drug. They were all published. But 1,657 patients’ worth of data was left unpublished, and this unpublished data showed that patients on reboxetine did worse than those on other drugs. If all this wasn’t bad enough, there was also the side-effects data. The drug looked fine in the trials that appeared in the academic literature; but when we saw the unpublished studies, it turned out that patients were more likely to have side-effects, more likely to drop out of taking the drug and more likely to withdraw from the trial because of side-effects, if they were taking reboxetine rather than one of its competitors.

The whole article is a fantastic (and worrying) account of regulatory failure.  At the heart of this problem lies a social and political misunderstanding of statistics which is being manipulated by drug companies for profit.  A proper regulatory framework would ensure that all trials were registered in advance and their data recorded.  Instead what happens is trials are commissioned by drugs companies, published if they are favourable and quietly buried if they are not.  This data mining would be mathematically rejected in an IB exploration coursework, yet these statistics still governs what pills doctors do and don’t prescribe.

When presented data therefore, your first question should be, “Where did this come from?” shortly followed by, “What about the data you’re not showing me?”  Lies, damn lies and statistics indeed!

If you enjoyed this post, you might also like:

How contagious is Ebola? – how we can use differential equations to model the spread of the disease.


Tetrahedral Numbers – Stacking Cannonballs

This is one of those deceptively simple topics which actually contains a lot of mathematics – and it involves how spheres can be stacked, and how they can be stacked most efficiently.  Starting off with the basics we can explore the sequence:

1, 4, 10, 20, 35, 56….

These are the total number of cannons in a stack as the stack gets higher.  From the diagram we can see that this sequence is in fact a sum of the triangular numbers:

S1 = 1

S2 1+3

S3 1+3+6

S4 1+3+6+10

So we can sum the first n triangular numbers to get the general term of the tetrahedral numbers. Now, the general term of the triangular numbers is 0.5n2 + 0.5n therefore we can think of tetrahedral numbers as the summation:

\bf \sum_{k=1}^{n}0.5k+0.5k^2 = \sum_{k=1}^{n}0.5k+\sum_{k=1}^{n}0.5k^2

But we have known results for the 2 summations on the right hand side:

\bf \sum_{k=1}^{n}0.5k =\frac{n(n+1)}{4}


\bf \huge \sum_{k=1}^{n}0.5k^2 = \frac{n(n+1)(2n+1)}{12}

and when we add these two together (with a bit of algebraic manipulation!) we get:

\bf S_n= \frac{n(n+1)(n+2)}{6}

This is the general formula for the total number of cannonballs in a stack n rows high. We can notice that this is also the same as the binomial coefficient:

\bf S_n={n+2\choose3}


Therefore we also can find the tetrahedral numbers in Pascals’ triangle (4th diagonal column above).

The classic maths puzzle (called the cannonball problem), which asks which tetrahedral number is also a square number was proved in 1878. It turns out there are only 3 possible answers. The first square number (1) is also a tetrahedral number, as is the second square number (4), as is the 140th square number (19,600).

We can also look at something called the generating function of the sequence. This is a polynomial whose coefficients give the sequence terms. In this case the generating function is:

\bf \frac{x}{(x-1)^4} = x + 4x^2 + 10x^3 + 20x^4 ...


Having looked at some of the basic ideas behind the maths of stacking spheres we can look at a much more complicated mathematical problem. This is called Kepler’s Conjecture – and was posed 400 years ago. Kepler was a 17th century mathematician who in 1611 conjectured that there was no way to pack spheres to make better use of the given space than the stack above. The spheres pictured above fill about 74% of the given space. This was thought to be intuitively true – but unproven. It was chosen by Hilbert in the 18th century as one of his famous 23 unsolved problems. Despite much mathematical efforts it was only finally proved in 1998.

If you like this post you might also like:

The Poincare Conjecture – the search for a solution to one of mathematics greatest problems.


Hailstone Numbers

This is a post inspired by the article on the same topic by the ever brilliant Plus Maths. Hailstone numbers are created by the following rules:

if n is even: divide by 2

if n is odd: times by 3 and add 1

We can then generate a sequence from any starting number.  For example, starting with 10:

10, 5, 16, 8, 4, 2, 1, 4, 2, 1…

we can see that this sequence loops into an infinitely repeating 4,2,1 sequence.  Trying another number, say 58:

58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1…

and we see the same loop of 4,2,1.

In fact we can use the generator in the Plus Maths article to check any numbers we can think of, and we still get the pattern 4,2,1 looping.  The question is, does every number end in this loop?  Well, we don’t know.  Every number mathematicians have checked do indeed lead to this loop, but that is not a proof.  Perhaps there is a counter-example, we just haven’t found it yet.

Hailstone numbers are called as such because they fall, reach one (the ground) before bouncing up again.  The proper mathematical name for this investigation is the Collatz conjecture. This was made in 1937 by a German mathematian, Lothar Collatz.

One way to investigate this conjecture is to look at the length of time it takes a number to reach the number 1.  Some numbers take longer than others.  If we could find a number that didn’t reach 1 even in an infinite length of time then the Collatz conjecture would be false.


The following graphic from wikipedia shows how different numbers (x axis) take a different number of iterations (y axis) to reach 1.  We can see that some numbers take much longer than others to reach one.  For example, the number 73 has the following pattern:

73, 220, 110, 55, 166, 83, 250, 125, 376, 188, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1…

so investigating what it is about certain numbers that leads to long chains is one possible approach to solving the conjecture. This conjecture has been checked by computers up to a staggering 5.8 x 1018 numbers. That would suggest that the conjecture could be true – but doesn’t prove it is. Despite looking deceptively simple, Paul Erdos – one of the great 20th century mathematicians stated in the 1980s that “mathematics is not yet ready for such problems” – and it has remained unsolved over the past few decades.  Maybe you could be the one to crack this problem!

If you liked this post you might also like:

Friendly Numbers, Solitary Numbers, Perfect Numbers – a look at some other number sequence problems.


Stellar Numbers Investigation

This is an old IB internal assessment question and so can not be used for the new IB exploration – however it does give a good example of the sort of pattern investigation that is possible.

The task starts off with the fairly straightforward problem of trying to find the nth term formula for the triangular numbers:


Method 1

There are a number of ways to do this, probably the easiest is to notice that the second differences are always constant (+1 each time).  Therefore we have a quadratic sequence in the form an2 + bn + c

We can now substitute the known values when n = 1, 2, 3 into this to find 3 equations:

a(1) + b(1) + c = 1

a(2)2 + b(2) + c = 3

a(3)2 + b(3) + c = 6

this gives us:

a + b + c = 1

4a + 2b + c = 3

9a + 3b + c = 6

We can then eliminate using simultaneous equations to find a, b, c.  In fact our job is made easier by knowing that if the second difference is a constant, then the a in our formula will be half that value.  Therefore as our second difference was 1, the value of a will be 1/2.  We then find that b = 1/2 and c = 0.  So the formula for the triangular numbers is:

0.5n2 + 0.5n

Method 2

We can also derive this formula by breaking down triangular numbers into the following series:





Therefore we have the sum of an arithmetic series, with first term 1, common difference 1 and last term n, and so we can use the sum of an arithmetic series formula:

Sn = 0.5n(a1 + an)

Sn = 0.5n(1 + n) = 0.5n2 + 0.5n

Once this is done, we are asked to find the nth term for the 6-stellar numbers (with 6 vertices) below:


which give the pattern 1, 13, 37, 73

Method 1

Once again we can use the method for quadratic sequences.  The second difference is 12, giving us an2 + bn + c with a = 12/2 = 6. Substituting values gives us:

1 = 6(1)2 + b(1) + c
13 = 6(2)2 + b(2) + c

This simplifies to:

1 = 6 + b + c
13 = 24 + 2b + c

Therefore we can eliminate to find that b = -6 and c = 1.

which gives 6n2 – 6n + 1

Method 2

A more interesting method makes use of the triangular numbers.  We can first note a recurrence relationship in the stellar numbers – each subsequent pattern contains all the previous patterns inside.  In fact we can state the relationship as:


S2 = S1 + outside star edge

S3 = S2 + outside star edge

S4 = S3 + outside star edge


The outside star edge of S2 can be thought of as 6 copies of the 2nd triangular number


The outside star edge of S3 can be thought of as 6 copies of the 3rd triangular number, but where we subtract 6×1 (the first triangular number) because we double count one of the internal points six times. We also subtract 6 as we double count each vertex.


The outside star edge of S4 can be thought of as 6 copies of the 4th triangular number, but where we subtract 6 x 3 (the second triangular number) because we double count three of the internal points six times. We also subtract 6 as we double count each vertex.

The outside star edge of S5 can be thought of as 6 copies of the 5th triangular number, but where we subtract 6 x 6 (the third triangular number) because we double count six of the internal points six times. We also subtract 6 as we double count each vertex.

Therefore we can form a formula for the outside star:

6(0.5n2 + 0.5n) – 6(0.5(n-2)2 + 0.5(n-2)) – 6

which simplifies to:

12(n -1)

We can now put this into our recurrence relationship:

S1 = 1

S2 = 1 + 12(n -1)

S3 = 1 + 12((n-1) -1) + 12(n -1)

S4 = 1 + 12((n-2) -1) + 12((n-1) -1) + 12(n -1)

Note that when we substituted the nth term formula for S2 into S3 we had to shift the n value to become n-1 as we were now on the 3rd term rather than 2nd term.


S1 = 1

S2 = 1 + 12(n -1)

S3 = 1 + 12(n-1) + 12(n-2)

S4 = 1 + 12(n-1) + 12(n-2) + 12(n-3)


S1 = 1 + 0

S2 = 1 + 12

S3 = 1 + 12+ 24

S4 = 1 + 12 + 24 + 36

So using the formula for the sum of an arithmetic Sn = 0.5n(a1 + an) we have

Sn = 1 + 0.5(n-1)(12 + 12(n-1))

Sn = 6n2 – 6n + 1

Quite a bit more convoluted – but also more interesting, and also more clearly demonstrating how the sequence is generated.


Generalising for p-stellar numbers

We can then generalise to find stellar number formulae for different numbers of vertices.  For example the 5-stellar numbers pictured above have the formula 5n2 – 5n + 1.  In fact the p-stellar numbers will have the formula pn2 – pn + 1.

We can prove this by using the same recurrence relationship before:


S2 = S1 + outside star edge

S3 = S2 + outside star edge

S4 = S3 + outside star edge

and by noting that the outside star edge is found in the same way as before for a p-stellar shape – only this time we subtract p for the number of vertices counted twice.  This gives:

p(0.5n2 + 0.5n) – p(0.5(n-2)2 + 0.5(n-2)) – p

which simplifies to


and so substituting this into our recurrence formula:

S1 = 1

S2 = 1 + 2p(n-2)

S3 = 1 + 2p(n-2) + 2p(n-1)

S4 = 1 + 2p(n-3) + 2p(n-2) + 2p(n-1)

We have the same pattern as before – an arithmetic series in terms of 2p, and using Sn = 0.5n(a1 + an) we have:

Sn= 1 + 0.5(n-1)(2p + 2p(n-1) )

Sn = pn2 – pn + 1

Therefore, although our second method was slower, it allowed us to spot the pattern in the progression – and this then led very quickly to a general formula for the p-stellar numbers.

If you like this you might also like:

The Goldbach Conjecture – The Goldbach Conjecture states that every even integer greater than 2 can be expressed as the sum of 2 primes.  No one has ever managed to prove this.

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