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E Math · Number Patterns · Enrichment

How to find the general formula for a number pattern like 4, 5, 8, 14, 26

Someone shows you the start of a number pattern, like 2, 5, 8, 11, 14, and asks you for a rule that gives any number in it. Here is a way to find that rule using almost nothing but subtraction. Write your numbers along the bottom. Work out the gap between each pair and write it in the row above. Then do the same thing again to the row you just made. Sooner or later a whole row turns into the same number over and over, and that is your signal to stop. The circled numbers make a staircase going up from the bottom left, and those are the numbers that go straight into your formula.

Read this bit first

This is purely enrichment

This is purely enrichment. I wrote it down because it turns stare at the numbers until you spot something into something you can actually do, step by step. It is also rather satisfying once it clicks.

Parts of it are examinable, though, so read this carefully. The first pair of examples below, where the 1st differences are already all the same, is ordinary syllabus work. You have almost certainly met it as a pattern that goes up (or down) by the same amount every time. The second pair, where the 2nd differences settle down, can be asked as well, but a question will hand you the intermediate steps to get there rather than leaving you to produce the whole formula on your own. What is enrichment is the general recipe: one method that covers both of those, and then keeps working for the 3rd row, the 4th row and beyond.

It works on one family of patterns. The method works when the rule behind the pattern is built out of n, n × n, n × n × n and so on. Your teacher will call those polynomials later. That covers most number pattern questions you are likely to meet, which is why it is worth knowing.

It cannot do patterns that grow by multiplying. If a pattern doubles or triples each time, like 2, 4, 8, 16, this method will never get you a formula. No matter how many rows you climb, none of them will ever go constant. There is a proper worked non-example of that further down.

It cannot do patterns whose signs keep flipping. If a pattern swings from positive to negative and back again every term, like 1, −8, 27, −64, the difference rows grow instead of shrinking and nothing ever settles — even when the numbers underneath the signs are ones the method handles easily. That one is worked through below as well.

It cannot do patterns that look backwards, either. If the rule for a term is something like add the two terms before it, the ladder will not settle for that one either, and for a rather different reason. That one is worked through at the end. Knowing when a tool does not work matters just as much as knowing how to use it.

How to read the pictures

Three steps, every time

  1. Build the rows. Write the pattern along the bottom. Take each pair of neighbours, subtract, and write the answer above them, in between. Then do the same to the row you have just made.
  2. Stop when a row is all the same. The moment every number in a row is identical, you are done climbing. However many rows you climbed, that is how many pieces your formula will have.
  3. Read the circled staircase. The first number of every row is circled, including the bottom one. Copy them into the formula in that order, with no swapping and no rearranging.

Before you look at the examples

Every question gives you its own set of differences

Eight patterns are worked through below, two at each level. They are there to illustrate the method, and nothing in them is worth memorising, because your question will not be one of them. A different pattern builds different difference rows, so it circles a different staircase: a different starting number, different numbers on the way up, sometimes negative ones, and sometimes a different number of rows before anything goes constant.

That is why each level gets two examples rather than one. Read a pair side by side and you will see two ladders with almost no numbers in common, built by the same three steps in the same order. The method is what carries from one question to the next. The numbers never do.

Example 1: the 1st differences are all the same

The pattern: 2, 5, 8, 11, 14. Every jump along it is the same size, +3, so the very first row you build is already all the same. One climb and you are finished.

Circled numbers, bottom to top: the first number 2 1st differences 3

Slide the picture sideways to see the whole ladder →

The formula

Tn = 2 + 3(n−1)

Read the staircase from the bottom: 2, then 3. The 2 is where the pattern starts. The 3 is how much you add each time.

Check it: put n = 4 into the formula and you get 2 + 3 × 3 = 11, which is exactly the 4th number. ✓

Example 2: the 1st differences again, new numbers

The pattern: 40, 33, 26, 19, 12. Same level as Example 1, and not one number in common with it. This pattern goes down rather than up, so the row you build is constant at −7. The method does not treat that as a special case: a negative goes into the formula exactly the way it comes off the ladder.

Circled numbers, bottom to top: the first number 40 1st differences −7

Slide the picture sideways to see the whole ladder →

The formula

Tn = 40 + (−7)(n−1)

Read the staircase from the bottom: 40, then −7. Same two slots as Example 1, filled with different numbers. If you would rather write it as 40 − 7(n−1), that is the same formula tidied up.

Check it: put n = 4 into the formula and you get 40 − 7 × 3 = 19, which is exactly the 4th number. ✓

Example 3: the 2nd differences are all the same

The pattern: −6, −2, 5, 15, 28, 44. The jumps keep getting bigger (+4, +7, +10 and so on), but they grow by the same amount every time. So the second row you build settles down.

Circled numbers, bottom to top: the first number −6 1st differences 4 2nd differences 3

Slide the picture sideways to see the whole ladder →

The formula

Tn = −6 + 4(n−1) + 3 1 × 2 (n−1)(n−2)

Staircase: −6, 4, 3. Those three numbers are exactly what you can see in the formula, in that same order. The colours match the rows they came from.

Check it: put n = 4 into the formula and you get −6 + 4 × 3 + (3 ÷ 2) × 3 × 2 = −6 + 12 + 9 = 15, which is exactly the 4th number. ✓

Example 4: the 2nd differences again, new numbers

The pattern: 10, 7, 6, 7, 10, 15. This one falls, flattens out, then climbs again, which looks nothing like Example 3 — and yet it settles on the second row just the same. Its 1st differences run −3, −1, +1, +3, +5, and its 2nd differences are all +2.

Circled numbers, bottom to top: the first number 10 1st differences −3 2nd differences 2

Slide the picture sideways to see the whole ladder →

The formula

Tn = 10 + (−3)(n−1) + 2 1 × 2 (n−1)(n−2)

Staircase: 10, −3, 2. Set that beside Example 3's −6, 4, 3 and there is not a number shared between them — but the formula they slot into is identical, bracket for bracket.

Check it: put n = 5 into the formula and you get 10 − 3 × 4 + (2 ÷ 2) × 4 × 3 = 10 − 12 + 12 = 10, which is exactly the 5th number. ✓

Example 5: the 3rd differences are all the same

The pattern: −20, −19, −16, −8, 8, 35. This time you climb three rows before the numbers finally settle at +3. There is nothing new to learn. Just one more row, and one more piece in the formula.

Circled numbers, bottom to top: the first number −20 1st differences 1 2nd differences 2 3rd differences 3

Slide the picture sideways to see the whole ladder →

The formula

Tn = −20 + 1(n−1) + 2 1 × 2 (n−1)(n−2) + 3 1 × 2 × 3 (n−1)(n−2)(n−3)

Staircase: −20, 1, 2, 3. Copy them in just as they are. The brackets and the dividing take care of themselves.

Check it: put n = 5 into the formula and you get −20 + 1 × 4 + (2 ÷ 2) × 4 × 3 + (3 ÷ 6) × 4 × 3 × 2 = −20 + 4 + 12 + 12 = 8, which is exactly the 5th number. ✓

Example 6: the 3rd differences again, new numbers

The pattern: 1, 8, 27, 64, 125, 216. You may recognise the cube numbers. Three climbs again, but not one rung of the staircase matches Example 5's, and the row that goes constant sits at +6 rather than +3. You do not have to spot that these are cubes. The ladder gets there whether you notice or not.

Circled numbers, bottom to top: the first number 1 1st differences 7 2nd differences 12 3rd differences 6

Slide the picture sideways to see the whole ladder →

The formula

Tn = 1 + 7(n−1) + 12 1 × 2 (n−1)(n−2) + 6 1 × 2 × 3 (n−1)(n−2)(n−3)

Staircase: 1, 7, 12, 6. Four circled numbers, four pieces, exactly as in Example 5 — and multiply this one out and it collapses all the way down to n × n × n, which is where the pattern came from. You never had to guess that.

Check it: put n = 4 into the formula and you get 1 + 7 × 3 + (12 ÷ 2) × 3 × 2 + (6 ÷ 6) × 3 × 2 × 1 = 1 + 21 + 36 + 6 = 64, which is exactly the 4th number. ✓

Example 7: the 4th differences are all the same

The pattern: 4, 5, 8, 14, 26, 49. This one takes four climbs before a row is finally steady at +2. Same recipe, one step further. You can see that it would keep going for as long as you need it to.

Circled numbers, bottom to top: the first number 4 1st differences 1 2nd differences 2 3rd differences 1 4th differences 2

Slide the picture sideways to see the whole ladder →

The formula

Tn = 4 + 1(n−1) + 2 1 × 2 (n−1)(n−2) + 1 1 × 2 × 3 (n−1)(n−2)(n−3) + 2 1 × 2 × 3 × 4 (n−1)(n−2)(n−3)(n−4)

Staircase: 4, 1, 2, 1, 2. Five circled numbers, five pieces in the formula.

Check it: put n = 5 into the formula and you get 4 + 1 × 4 + (2 ÷ 2) × 4 × 3 + (1 ÷ 6) × 4 × 3 × 2 + (2 ÷ 24) × 4 × 3 × 2 × 1 = 4 + 4 + 12 + 4 + 2 = 26, which is exactly the 5th number. ✓

Example 8: the 4th differences again, new numbers

The pattern: 1, 4, 8, 15, 28, 51, 89. Four climbs again, and an entirely different staircase from Example 7. I have written out one extra term here, so the top row shows three matching +1s rather than two — which is what you want to see before you trust a row (more on that below).

Circled numbers, bottom to top: the first number 1 1st differences 3 2nd differences 1 3rd differences 2 4th differences 1

Slide the picture sideways to see the whole ladder →

The formula

Tn = 1 + 3(n−1) + 1 1 × 2 (n−1)(n−2) + 2 1 × 2 × 3 (n−1)(n−2)(n−3) + 1 1 × 2 × 3 × 4 (n−1)(n−2)(n−3)(n−4)

Staircase: 1, 3, 1, 2, 1. Five circled numbers, five pieces, just like Example 7 — and going down the two staircases side by side, not one of the five positions holds the same number.

Check it: put n = 6 into the formula and you get 1 + 3 × 5 + (1 ÷ 2) × 5 × 4 + (2 ÷ 6) × 5 × 4 × 3 + (1 ÷ 24) × 5 × 4 × 3 × 2 = 1 + 15 + 10 + 20 + 5 = 51, which is exactly the 6th number. ✓

Spot the pattern in the formulas

Every row up adds one more bracket

Put the eight formulas side by side and the rhythm is hard to miss. The numbers change from one example to the next — that is the whole point of doing two of each — but the brackets never do. The piece that comes from the 1st differences carries (n−1). The piece from the 2nd differences carries (n−1)(n−2), divided by 1 × 2. The 3rd carries (n−1)(n−2)(n−3) divided by 1 × 2 × 3, and the 4th carries (n−1)(n−2)(n−3)(n−4) divided by 1 × 2 × 3 × 4. Every time you climb one more row, you multiply by one more bracket and divide by one more counting number. Nothing else changes.

A check that always works. Put n = 1 into your formula. Every piece after the first one contains (n−1), which turns into 0, so the whole lot disappears and you are left with the first number of your pattern. If you do not get it back, something has gone wrong somewhere, and it is usually a slip in one of the difference rows.

Where the method stops

When the pattern grows by multiplying

Take the powers of two: 2, 4, 8, 16, 32. Now start climbing. The 1st differences are 2, 4, 8, 16, which is the same pattern all over again. Climb once more and you get 2, 4, 8. Then 2, 4. Every row is just the row below it repeated, so no row ever turns constant. That is what multiplying does to a pattern, and it is exactly why this method cannot find a formula for it. The same goes for any pattern that grows by multiplying instead of adding.

There is a trap hiding in that example. Because I only wrote down five numbers, the ladder ran out of room and finished on a single 2. If you are not paying attention, one lonely number looks like a row that has gone constant. Take it at face value, and the staircase reads 2, 2, 2, 2, 2, which gives you this:

Tn = 2 + 2(n−1) + 2 1 × 2 (n−1)(n−2) + 2 1 × 2 × 3 (n−1)(n−2)(n−3) + 2 1 × 2 × 3 × 4 (n−1)(n−2)(n−3)(n−4)

That formula gives all five of your numbers back perfectly. It really is a proper formula, and it really does fit everything you fed it. But ask it for the 6th number and it hands you 2 + 10 + 20 + 20 + 10 = 62, when 26 is 64. It was never the real rule. It was only a formula that happened to pass through five points.

So here is the honest test. A row has only gone constant if you can see the same number repeated several times over. One number on its own is not a constant row. It just means you have run out of paper. Write out enough of the pattern that the bottom row repeats itself two or three times before you trust it.

Where the method stops

When the signs flip from one term to the next

Here is 1, −8, 27, −64, 125, −216. You have already met most of that on this page. Rub the minus signs out and you are back at Example 6 — the cube numbers, 1, 8, 27, 64, 125, 216 — which the ladder polished off in three rows and handed you a formula for. All I have done is flip the sign of every second term. Nothing else about the numbers has changed. So put it on the ladder and see what that one small change costs you.

Slide the picture sideways to see the whole ladder →

Hold that up against Example 6. There the 3rd row was +6, +6, +6 and the work was over. Here the 3rd row is −170, +406, −810, and the 4th is +576, −1216. Those numbers are not settling down. They are getting louder, and they are still changing sign at every step, which is what the rows underneath were doing too. Nothing in this picture is heading anywhere near a constant row.

Subtracting is supposed to wear a pattern down. That is the engine of the whole method: take two neighbours, subtract, and what you are left with is smaller than what you started with, so after a few rows there is nothing left to shrink. It does that as long as neighbouring terms are pulling the same way. When two of them sit on opposite sides of zero, the subtraction quietly stops being a subtraction: going from −8 to 27 means taking away a negative, and taking away a negative adds, so the gap is 35 — that is 8 + 27. From 27 to −64 the gap is 91, which is 27 + 64.

One crossing does no harm. Example 5 starts at −20 and climbs up through zero without any trouble at all, because it only does it once. In this pattern every single neighbouring pair straddles zero, so every subtraction the whole way along is an addition in disguise, and each row comes out bigger than the row it was built from rather than smaller.

And the flipping is not worn down either. Read along the 1st differences: −9, +35, −91, +189, −341. Still alternating. So the row above is built out of neighbours on opposite sides of zero in exactly the same way, and it grows for exactly the same reason — and so does the row after that, and the one after that. The ladder is not running out of steam. It is picking up speed.

This is really the multiplying pattern in disguise. The rule behind these numbers is the cube of n with a sign on the front that turns over every term, and turning a sign over is multiplying by −1, again and again. So there is a multiplying pattern buried in this one after all — the very thing the section above warns you about — and it has simply turned up wearing a perfectly respectable n × n × n on top of it. That is what makes it worth a section of its own: the cubes are exactly the sort of thing the ladder eats for breakfast, so the pattern looks safe right up until you start climbing.

Which brings the lonely number back. Suppose you had written down only the first five terms and stopped at 125. The ladder would have run out on a single +576, with nothing beside it to disagree with, and the staircase 1, −9, +44, −170, +576 gives you a formula that returns all five of your numbers without a murmur — and then claims the 6th term is 1576, when the pattern plainly says −216. The test from the section above catches it, and there is a faster one to go with it: if the signs are alternating, do not start the ladder at all. You already know how it ends.

Where the method stops

When each term is the sum of the two before it

Here is a pattern you may well have run into somewhere: 1, 1, 2, 3, 5, 8, 13. Nothing is hidden in it. Every term is just the two terms before it added together — 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8 — and you can carry on as long as you have the patience for it. It is a perfectly good rule, and it is famous enough to have a name of its own, though the name will not help you here. So put it on the ladder and watch what it does.

Slide the picture sideways to see the whole ladder →

Look at the first row of differences: 0, 1, 1, 2, 3, 5. Cover up the 0 and what is left is your own pattern, starting over. That is not a fluke. The gap between two neighbours is never anything new here: if a term is the two before it added up, then taking one neighbour away from the next just hands you back a term you already had, from two places further down the row.

And every row above does the same thing. Take any row in the picture — 1, 0, 1, 1, 2, or −1, 1, 0, 1 — and check it for yourself: each entry is still the two entries before it added together. The rule survives the climb, and it survives every climb after that. The ladder is not wearing the pattern down. It is copying it out again, row after row, for as long as you care to keep going.

So there is nothing to stop at. A row goes constant only when the climbing has finally worn the life out of a pattern. This one never loses any. The circles are still sitting there in the picture out of habit, and you could copy them down if you insisted — 1, 0, 1, −1, 2 — but the formula they give you says the 6th term is 11, when the pattern plainly says 8. It is the same trap as the one above: you never reached a row that told you to stop, so the staircase was never yours to read.

It is the same wall as the doubling pattern, reached down a different road. Near the top of this page I said the method works when the rule behind a pattern is built out of n, n × n and so on. This rule is not that sort of thing at all. It never tells you how to build the 10th term out of the number 10; it tells you to turn round and look at two terms you have already written down. A rule that looks backwards like that is a different animal, and the ladder cannot reach it. There is a formula for this pattern, as it happens, but it looks nothing like anything on this page, and you will not need it at this level.

None of this makes the method any less useful. It just means you should know what it is. It is a reliable way to find a rule when a rule of that kind exists, and it is a quick way to find out for certain when one does not.

Your turn

Try one yourself

5,  9,  17,  29,  45

  1. Build the difference rows until one of them goes constant.
  2. Circle the first number of every row.
  3. Read the staircase into the formula for Tn.
Show the answer

The 1st differences are 4, 8, 12, 16 and the 2nd differences are 4, 4, 4, which is constant on the second row. So the formula has two difference pieces. The staircase reads 5, 4, 4, giving

Tn = 5 + 4(n−1) + (4 ÷ 2)(n−1)(n−2) = 5 + 4(n−1) + 2(n−1)(n−2).

Check it with n = 4: 5 + 4 × 3 + 2 × 3 × 2 = 5 + 12 + 12 = 29, which is the 4th number. ✓

Notice that this one settles on the second row, the same as Examples 3 and 4, and yet all three staircases are made of different numbers. That is the ordinary state of affairs, not a coincidence.

Where this fits

Number patterns in the E Math syllabus

Number patterns turn up early in the course, and the same idea comes back later whenever a question gives you a table of values and asks for the general term. If the ladder makes sense but the algebra afterwards does not, that is the usual sticking point, and it is worth sitting down with someone and working through it. You are very welcome to bring this page to a lesson.

Everything else I have written up is on the secondary maths learning materials page.