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This is a nice investigation idea from Nrich. The above screen capture is from their Picture Story puzzle. We have successive cubes – a 1x1x1 cube, a 2x2x2 cube etc.
The cubes are then rearranged to give the following shape. The puzzle is then to use this information to discover a mathematical relationship. This was my first attempt at this:
13 = 12
23 = (1+2)2 – 12
33 = (1+2+3)2 – (1+2)2
43 = (1+2+3+4)2 – (1+2+3)2
n3 = (1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2
This is not an especially attractive relationship – but nevertheless we have discovered a mathematical relationship using the geometrical figures above. Next let’s see why the RHS is the same as the LHS.
(1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2
= ([1+2+3+4+…+ (n-1)] + n)2 – (1+2+3+…+ (n-1))2
= (1+2+3+…+ (n-1))2 + n2 + 2n(1+2+3+4+…+ (n-1)) – (1+2+3+…+ (n-1))2
= n2 + 2n(1+2+3+4+…+ (n-1))
next we notice that 1+2+3+4+…+ (n-1) is the sum of an arithmetic sequence first term 1, common difference 1 so we have:
1+2+3+4+…+ (n-1) = (n-1)/2 (1 + (n-1) )
1+2+3+4+…+ (n-1) = (n-1)/2 + (n-1)2/2
1+2+3+4+…+ (n-1) = (n-1)/2 + (n2 – 2n + 1)/2
2n(1+2+3+4+…+ (n-1)) = 2n ( (n-1)/2 + (n2 – 2n + 1)/2 )
2n(1+2+3+4+…+ (n-1)) = n2 -n + n3 – 2n2 + n
n2 + 2n(1+2+3+4+…+ (n-1)) = n2 + n2 -n + n3 – 2n2 + n
n2 + 2n(1+2+3+4+…+ (n-1)) = n3
and we have shown that the RHS does indeed simplify to the LHS – as we would expect.
An alternative relationship
13 = 12
13+23 = (1+2)2
13+23+33 = (1+2+3)2
13+23+33+…n3 = (1+2+3+…+n)2
This looks a bit nicer – and this is a well known relationship between cubes and squares. Could we prove this using induction? Well we can show it’s true for n =1. Then we can assume true for n=k:
13+23+33+…k3 = (1+2+3+…+k)2
Then we want to show true for n = k+1
13+23+33+… k3 + (k+1)3= (1+2+3+…+k + (k+1))2
13+23+33+… k3 + (k+1)3
= (1+2+3+…+k)2 + (k+1)3
(1+2+3+…+k + (k+1))2
= ([1+2+3+…+k] + (k+1) )2
= [1+2+3+…+k]2 + (k+1)2 + 2(k+1)[1+2+3+…+k]
= [1+2+3+…+k]2 + (k+1)2 + 2(k+1)(k/2 (1+k)) (sum of a geometric formula)
= [1+2+3+…+k]2 + (k+1)2 + 2(k+1)(k/2 (1+k))
= [1+2+3+…+k]2 + k3+ 3k2 + 3k + 1
= (1+2+3+…+k)2 + (k+1)3
Therefore we have shown that the LHS = RHS and using our induction steps have shown it’s true for all n. (Write this more formally for a real proof question in IB!)
So there we go – a couple of different mathematical relationships derived from a simple geometric pattern – and been able to prove the second one (the first one would proceed in a similar manner). This sort of free-style pattern investigation where you see what maths you can find in a pattern could make an interesting maths IA topic.
This is a great puzzle which the Guardian ran last week:
Fill in the equations below using any of the basic mathematical operations, +, –, x, ÷, and as many brackets as you like, so that they make arithmetical sense.
10 9 8 7 6 5 4 3 2 1 = 2017
There are many different ways of solving this – see if you can find the simplest way!
Scroll down to see some possible answers.
I had a play around with this and this is my effort:
(10+(9 x 8 x 7) -(6 + 5) ) x 4 + 3 + (2 x 1) = 2017
An even nicer answer was provided on the Guardian – which doesn’t even need brackets:
10 x 9 x 8 x 7 / 6 / 5 x 4 x 3 + 2 – 1 = 2017
Any other solutions?
The Watson Selection Task – a logical puzzle
The Watson Selection Task is a logical problem designed to show how bad we are at making logical decisions. Watson first used it in 1968 – and found that only 10% of the population would get the correct answer. Indeed around 65% of the population make the same error. Here is the task:
The participants were given the following instructions:
Here is a rule: “every card that has a D on one side has a 3 on the other.” Your task is to select all those cards, but only those cards, which you would have to turn over in order to discover whether or not the rule has been violated. Each card has a number on one side and a letter on the other.
Give yourself a couple of minutes to work out what you think the answer is – and then highlight the space below where the answer is written in white text.
The correct answer is to pick the D card and the 7 card
This result is normally quite unexpected – but it highlights one of the logical fallacies that we often fall into:
A implies B does not mean that B implies A
All cats have 4 legs (Cats = A, legs = B, A implies B)
All 4 legged animals are cats (B implies A)
We can see that here we would make a logical error if we concluded that all 4 legged animals were cats.
In the logic puzzle we need to turn over only 2 cards, D and 7. This is surprising because most people will also say that you need to turn over card with a 3. First we need to be clear about what we are trying to do: We want to find evidence that the rule we are given is false.
If we turn over the D and find a number other than 3, we have evidence that the rule is false – therefore we need to turn over D.
If we turn over the 7 and find a D on the other side, we have evidence that the rule is false – therefore we need to turn over the 7.
But what about the 3? If we turn over the 3 and find a D then we have no evidence that the rule is false (which is what we are looking for). If we turn over the 3 and find another letter then this also gives us no evidence that the rule is false. After all our rule says that all Ds have 3s on the other side, but it doesn’t say that all 3s have Ds on the other side.
Are mathematicians better at this puzzle than historians?
Given the importance of logical thought in mathematics, people have done studies to see if undergraduate students in maths perform better than humanities students on this task. Here are the results:
You can see that there is a significant difference between the groups. Maths students correctly guessed the answer D7 29% of the time, but only 8% of history students did. The maths university lecturers performed best – getting the answer right 43% of the time.
Making different mistakes
You can also analyse the mistakes that students made- by only looking at the proportions of incorrect selections. Here again are significant differences which show that the groups are thinking about the problem in different ways. DK7 was chosen by around 1/5 of both maths students and lecturers, but by hardly any history students.
You can read about these results in much more depth in the following research paper Mathematicians and the Selection Task – where they also use Chi Squared testing for significance levels.
This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).
How to Avoid The Troll: A Puzzle
In these situations it’s best to look at the extreme case first so you get some idea of the problem. If you are feeling particularly pessimistic you could assume that the troll is always going to be there. Therefore you would head to the top of the barrier each time. This situation is represented below:
The Pessimistic Solution:
Another basic strategy would be the optimistic strategy. Basically head in a straight line hoping that the troll is not there. If it’s not, then the journey is only 2km. If it is then you have to make a lengthy detour. This situation is shown below:
The Optimistic Solution:
The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.
The question is now, is there a better strategy than either of these? An obvious possibility is heading for the point halfway along where the barrier might be. This would make a triangle of base 1 and height 1/2. This has a hypotenuse of root (5/4). In the best case scenario we would then have a total distance of 2 x root (5/4). In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2. We find the expected value by multiply both by 0.5 and adding. This gives 2.63 (2 dp). But can we do any better? Yes – by using some algebra and then optimising to find a minimum.
The Optimisation Solution:
To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum. Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us. Wolfram Alpha is incredibly powerful -and also very easy to use. Here is what I entered:
and here is the output:
So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.
So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint. Not a huge difference – but nevertheless we’ll save ourselves a few seconds!
This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across? Maybe the troll appears only 1/3 of the time? Could you even generalise the results for when the troll is y distance away or appears z percent of the time?
This is a classic puzzle which is discussed in some more detail by the excellent Wired article. The puzzle is best represented by the picture below. We have a hunter who whilst in the jungle stumbles across a monkey on a tree branch. However he knows that the monkey, being clever, will drop from the branch as soon as he hears the shot being fired. The question is therefore, at what angle should the hunter aim so that he still hits the monkey?
(picture from the Wired article – originally from a UCLA physics textbook)
The surprising conclusion is that counter to what you would expect, you should actually still aim at the monkey on the branch – and in this way your bullet’s trajectory will still hit the monkey as it falls. You can see a video of this experiment at the top of the page.
You can use tracking software (such as the free software tracker ) to show this working graphically. Tracker provides a video demo with the falling monkey experiment:
As you can see from the still frame, we have the gun in the bottom left corner, lined up with the origin, the red trace representing the bullet and the blue trace representing the falling monkey.
We can then generate a graph to represent this data. The red line is the height of the bullet with respect to time. The faint blue line (with yellow dots) is the height of the monkey with respect to time. We can see clearly that the red line can be modeled as a quadratic. The blue line should in theory also be a quadratic (see below):
but in our model, the blue line is so flat as to be better modeled as a linear approximation – which is shown in pink. Now we can use regression technology to find the equation of both of these lines, to show not only that they do intersect, but also the time of that intersection.
We have the linear approximation as y = -18.5t + 14.5
and the quadratic approximation as y = -56t2+39t +0.1
So the 2 graphs will indeed intersect when -18.5t + 14.6 = -56t2+39t +0.1
which will be around 0.45 seconds after the gun is fired.
(A more humane version, also from Wired – where we can throw the monkey a banana)
The next question is can we prove this using some algebra? Of course! The key point is that the force of gravity will affect both the bullet and the falling monkey equally (it will not be affected by the different weights of the two – see the previous post here about throwing cannonballs from the Leaning Tower of Pisa). So even thought the bullet deviates from the straight line path lined up in the gun sights, the distance the bullet deviates will be exactly the same distance that the monkey falls. So they still collide. Mathematically we have:
The vertical height of the bullet given by:
y = V0t – 0.5gt2
Where V0 is the initial vertical speed, t is the time, g is the gravitational force (9.8)
The vertical height of the monkey is given by:
y = h – 0.5gt2
where h is the initial vertical height of the monkey.
Therefore these will intersect when:
V0t – 0.5gt2 = h – 0.5gt2
V0t = h
V0/h = t
And for any given non-zero value of V0 we will have a t value – which represents the time of collision.
Well done – you have successfully shot the monkey!
If you like this you might also like:
Throwing cannonballs off the Leaning tower of Pisa – why weight doesn’t affect falling velocity
War Maths – how cannon operators used projectile motion to win wars
How to Win at Rock, Paper, Scissors
You might think that winning at rock, paper, scissors was purely a matter of chance – after all mathematically each outcome has the same probability. We can express the likelihood of winning in terms of a game theory grid:
It is clear that in theory you would expect to win, draw and lose with probability 1/3. However you can actually exploit human psychology to give yourself a significant edge at this game. Below is a report of a Chinese study on the psychology of game players:
Zhijian and co carried out their experiments with 360 students recruited from Zhejiang University and divided into 60 groups of six players. In each group, the players played 300 rounds of Rock-Paper-Scissors against each other with their actions carefully recorded.
As an incentive, the winners were paid in local currency in proportion to the number of their victories. To test how this incentive influenced the strategy, Zhijian and co varied the payout for different groups. If a loss is worth nothing and a tie worth 1, the winning payout varied from 1.1 to 100.
The results reveal a surprising pattern of behavior. On average, the players in all the groups chose each action about a third of the time, which is exactly as expected if their choices were random.
But a closer inspection of their behavior reveals something else. Zhijian and co say that players who win tend to stick with the same action while those who lose switch to the next action in a clockwise direction (where R → P → S is clockwise).
So, for example if person A chooses Rock and person B chooses Paper, then person B wins. Human nature therefore seems to mean that person B is more likely to stick to a winning strategy and choose Paper again, whilst person A is more likely to copy that previous winning behaviour and also choose Paper. A draw.
So you can exploit this by always moving anticlockwise i.e R → S → P. To look at our example again, person A chooses Rock and person B chooses Paper, then person B wins. This time person A follows his previous pattern and still chooses Paper, but person B exploits this new knowledge to choose Rock. Player B wins.
You can play against a Wolfram Alpha AI player here. This program will track your win percentage, and will also adapt its behavior to exploit any non-random behavior that you exhibit. Even though you may not be conscious of your biases, they probably will still be there – and the designers of this simulator are confident that the program will be beating you after about 50 games. Have a go!
There are some additional tips for winning at rock paper scissors – if you are in a single game competition then choose paper. This is because men are most likely to choose rock, and scissors are the least popular choice. Also you should try some reverse psychology and announce what you will throw. Most opponents will not believe you and modify their throw as a result.
Rock, Paper, Scissors, Lizard, Spock
You can of course make the game as complicated as you wish – the version above was popularised (though not invented by) The Big Bang Theory. The grid below shows the possible outcomes for this game:
And of course, why stop there? Below is a 15 throw version of the game
If you’ve honed your strategy then maybe you could compete in the a professional rock, paper, scissors tournament – here you can watch the final of a $50,000 Las Vegas competition.
If you liked this post you might also like:
Game Theory and Tic Tac Toe – Tic Tac Toe has already been solved using Game Theory – this topic also brings in an introduction to Group Theory.
Does it Pay to be Nice? Game Theory and Evolution. How understanding mathematics helps us understand human behaviour
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!
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
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:
But we have known results for the 2 summations on the right hand side:
and when we add these two together (with a bit of algebraic manipulation!) we get:
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:
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:
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.
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:
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
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
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
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:
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.
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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.