In previous blogs, we’ve looked at some interesting integer sequences. We’ve looked at the Fibonacci sequence, the Look-and-say sequence, Golomb’s sequence and Van Eck’s sequence. Today, we’re looking at the Lazy Caterer’s sequence.
You can find this sequence on the Online Encyclopedia of Integer Sequences here.
How to generate the sequence
The name of this sequence gives an idea of how it is generated. Let’s say we have a pancake, and we want to cut it into as many different slices as we can. This sequence will tell you how many different pieces you can cut the pancake into with a given number of straight-line cuts.
The first number in the sequence is the number of pieces it can be cut into with 0 cuts. This is uninterestingly 1, as we cannot cut the pancake up at all. So the first term in the sequence is 1.
The second term will be the maximum number of pieces it can be cut into with 1 cut. This is also rather uninteresting, as the answer is clearly 2.
Now it gets more interesting. The third term is the maximum number of pieces it can be cut into with 2 cuts. On the pancake below, I’ve cut it twice and it’s ended up with three pieces.
But this isn’t the maximum number of pieces I can create. If I cut it like this, I can get four pieces (each section is coloured in with a different colour).
So the third term in the sequence is 4.
The trick to achieving the largest number of cuts is to make sure your cuts touch as many of the other cuts as possible. Let’s do the next one.
The fourth term in the sequence in the maximum number of pieces it can be cut into with three cuts.
We’ll start with our answer for the third term, as this is already optimised for two cuts and we just have to add a third cut.
We want to cut as many of the other cuts as we can. There are two cuts already, so we want to cut both of them with our next cut. We can do this a number of different ways, here’s one of them.
That’s 7 pieces. So the fourth term is 7.
Hopefully you can see what’s happening here. Let’s do one more.
Again, let’s start with our previous pancake. There are already three cuts, so we want to draw a line that cuts all three of them. Below, I have done that.
I’m running out of colours now, but that’s 11 pieces. So the fifth term is 11.
If you think about it, each new number in the sequence will be the previous number + the previous number of cuts. Because we are cutting each previous cut, which means we are creating that many additional sections.
Nth term formula for the Lazy Caterer’s sequence
Here are the first few terms of the Lazy Caterer’s Sequence:
1, 2, 4, 7, 11, 16, 22, 29
You will notice that this is a quadratic sequence (as the difference of the differences is the same). Check out this blog for help on how to determine sequence types.
Because this is a quadratic sequence, we can find the nth term of it (if you are studying higher tier maths, check out my guide to finding the nth term of a quadratic sequence). Here is the formula for the maximum number of sections (s) we can make with n cuts:
So, for example, if I wanted to know how many sections I can make with 30 cuts, I can work this out by substituting n = 30 into the formula.
s = (302 + 30 + 2) ÷ 2 = 466
So we can make 466 sections in 30 cuts. I don’t really fancy drawing that one. This is the 31st term in the Lazy Caterer’s Sequence (remember that the first term is the number of sections with 0 cuts).
These numbers are more formally known as the “Central Polygonal Numbers”, but I think the Lazy Caterer’s Sequence is a much better name.
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