**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 an^{2} + 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:

0.5n^{2} + 0.5n

**Method 2**

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

1

1+2

1+2+3

1+2+3+4

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:

S_{n} = 0.5n(a_{1} + a_{n})

S_{n} = 0.5n(1 + n) = 0.5n^{2} + 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 an^{2} + 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 6n^{2} – 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:

S_{1}

S_{2} = S_{1} + outside star edge

S_{3} = S_{2} + outside star edge

S_{4} = S_{3} + outside star edge

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

The outside star edge of S_{3} 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 S_{4} 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 S_{5} 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(0.5n^{2} + 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:

S_{1} = 1

S_{2} = 1 + 12(n -1)

S_{3} = 1 + 12((n-1) -1) + 12(n -1)

S_{4} = 1 + 12((n-2) -1) + 12((n-1) -1) + 12(n -1)

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

So:

S_{1} = 1

S_{2} = 1 + 12(n -1)

S_{3} = 1 + 12(n-1) + 12(n-2)

S_{4} = 1 + 12(n-1) + 12(n-2) + 12(n-3)

So:

S_{1} = 1 + 0

S_{2} = 1 + 12

S_{3} = 1 + 12+ 24

S_{4} = 1 + 12 + 24 + 36

So using the formula for the sum of an arithmetic S_{n} = 0.5n(a_{1} + a_{n}) we have

S_{n} = 1 + 0.5(n-1)(12 + 12(n-1))

S_{n} = 6n^{2} – 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 5n^{2} – 5n + 1. In fact the p-stellar numbers will have the formula pn^{2} – pn + 1.

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

S_{1}

S_{2} = S_{1} + outside star edge

S_{3} = S_{2} + outside star edge

S_{4} = S_{3} + 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(0.5n^{2} + 0.5n) – p(0.5(n-2)^{2} + 0.5(n-2)) – p

which simplifies to

2p(n-1)

and so substituting this into our recurrence formula:

S_{1} = 1

S_{2} = 1 + 2p(n-2)

S_{3} = 1 + 2p(n-2) + 2p(n-1)

S_{4} = 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 S_{n} = 0.5n(a_{1} + a_{n}) we have:

S_{n}= 1 + 0.5(n-1)(2p + 2p(n-1) )

S_{n} = pn^{2} – 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.