Sequences - Finding a Rule

To find a missing number in a Sequence, first you must have a Rule

Quick Definition of Sequence

Read Sequences and Series for a more in-depth discussion, but put simply:

A Sequence is a set of things (usually numbers) that are in order.

Each number in the sequence is called a term (or sometimes "element" or "member"):

Finding Missing Numbers

To find a missing number, first find a Rule behind the Sequence.

Sometimes you can just look at the numbers and see a pattern:

Example: 1, 4, 9, 16, ?

Answer: they are Squares (12=1, 22=4, 32=9, 42=16, ...)

Rule: xn = n2

Sequence: 1, 4, 9, 16, 25, 36, 49, ...

Did you see how we wrote that rule using "x" and "n" ?

xn means "term number n", so term 3 would be written x3

And we also used "n" in the formula, so the formula for term 3 is 32 = 9. This could be written

x3 = 32 = 9

Once we have a Rule we can use it to find any term. For example, the 25th term can be found by "plugging in" 25 wherever n is.

x25 = 252 = 625

How about another example:

Example: 3, 5, 8, 13, 21, ?

After 3 and 5 all the rest are the sum of the two numbers before, that is 3 + 5 = 8, 5 + 8 = 13 and so on (it is actually part of the Fibonacci Sequence):

Rule: xn = xn-1 + xn-2

Sequence: 3, 5, 8, 13, 21, 34, 55, 89, ...

Now what does xn-1 mean? It just means "the previous term" because the term number (n) is 1 less (n-1).

So, if n was 6, then xn = x6 (the 6th term) and xn-1 = x6-1 = x5 (the 5th term)

So, let's apply that Rule to the 6th term:

x6 = x6-1 + x6-2

x6 = x5 + x4

We already know the 4th term is 13, and the 5th is 21, so the answer is:

x6 = 21 + 13 = 34

Pretty simple ... just put numbers instead of "n"

Many Rules

One of the troubles with finding "the next number" in a sequence is that mathematics is so powerful you can find more than one Rule that works.

What is the next number in the sequence 1, 2, 4, 7, ?

Here are three solutions (there can be more!):


Solution 1: Add 1, then add 2, 3, 4, ...

So, 1+1=2, 2+2=4, 4+3=7, 7+4=11, etc...

Rule: xn = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, ...

(That rule looks a bit complicated, but it works)

 

Solution 2: After 1 and 2, add the two previous numbers, plus 1:

Rule: xn = xn-1 + xn-2 + 1

Sequence: 1, 2, 4, 7, 12, 20, 33, ...

 

Solution 3: After 1, 2 and 4, add the three previous numbers

Rule: xn = xn-1 + xn-2 + xn-3

Sequence: 1, 2, 4, 7, 13, 24, 44, ...

So, we had three perfectly reasonable solutions, and they created totally different sequences.

Which is right? They are all right.

And there will be other solutions.
race  

Hey, it may be a list of the winners' numbers ... so the next number could be ... anything!

Simplest Rule

When in doubt choose the simplest rule that makes sense, but also mention that there are other solutions.

Finding Differences

Sometimes it helps to find the differences between each pair of numbers ... this can often reveal an underlying pattern.

Here is a simple case:

The differences are always 2, so we can guess that "2n" is part of the answer.

Let us try 2n:

n: 1 2 3 4 5
Terms (xn): 7 9 11 13 15
2n: 2 4 6 8 10
Wrong by: 5 5 5 5 5

The last row shows that we are always wrong by 5, so just add 5 and we are done:

Rule: xn = 2n + 5

OK, you could have worked out "2n+5" by just playing around with the numbers a bit, but we want a systematic way to do it, for when the sequences get more complicated.

Second Differences

In the sequence {1, 2, 4, 7, 11, 16, 22, ...} we need to find the differences ...

... and then find the differences of those (called second differences), like this:

The second differences in this case are 1.

With second differences you multiply by "n2 / 2".

In our case the difference is 1, so let us try n2 / 2:

n: 1 2 3 4 5
Terms (xn): 1 2 4 7 11
           
n2: 1 4 9 16 25
n2 / 2: 0.5 2 4.5 8 12.5
Wrong by: 0.5 0 -0.5 -1 -1.5

We are close, but seem to be drifting by 0.5, so let us try: n2 / 2 - n/2

n2 / 2 - n/2: 0 1 3 6 10
Wrong by: 1 1 1 1 1

Wrong by 1 now, so let us add 1:

n2 / 2 - n/2 + 1: 1 2 4 7 11
Wrong by: 0 0 0 0 0

The formula n2 / 2 - n/2 + 1 can be simplified to n(n-1)/2 + 1

So, by "trial-and-error" we were able to discover a rule that works:

Rule: xn = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, ...

Other Types of Sequences

As well as the sequences mentioned on Sequences and Series:

Look out for

In truth there are too many types of sequences to mention here, but if there is one you would like me to add just let me know.