Uncertain Sequences

We’ve often pointed out that pattern or sequence problems, when nothing is given but a list of numbers, are not really math, in the sense that there is no one correct answer. They are psychology questions: What would a math teacher think is an interesting sequence to ask about? Mathematically, any number could come next, and infinitely many formulas or patterns could be found that produce the given terms. Let’s look at some questions we’ve had that took this to an extreme.

Three terms, and an unsatisfying answer

First, from 1997:

Alternating Sequence

Dear Dr. Math,

I need help. I have a problem that neither my friends nor I can get.  I thought you could help.  We have to find a pattern and find the next three numbers for this: 0, 8, 27, _, _, _.  I hope you can find the answer.
			
Sincerely,
Shoushou

Three terms are not enough to determine a sequence, unless something very obvious is going on. If this had been 1, 8, 27, it would have been obvious: Even though there are many options, ranging from the repeating sequence

1, 8, 27, 1, 8, 27, … ,

to the quadratic sequence \(a_n = 6n^2 – 11x + 6\), which produces

1, 8, 27, 58, 101, 156, … ,

any reasonable person would guess that the intended sequence is \(a_n = n^3\).

But what of the sequence that was given? It isn’t at all obvious, and therefore is really impossible to know. Doctor Wallace answered, suggesting one idea to start with:

For this series, the first thing to notice is that all of the numbers are perfect cubes:  0 = 0^3, 8 = 2^3, and 27 = 3^3.

If we list the numbers that are the cube roots of these numbers, we get this series:

           0, 2, 3...

Now, all you have to do is find a pattern among these three.  The one I see is an alternating series.  There may be others.  Let me know how your work goes on this problem.

Since the cubes are so obvious, his idea is that the sequence might be the cubes of this sequence. But what is this sequence? He suggests what may be the simplest idea, an alternating sequence (that is, alternating terms follow different rules). We could instead use \(\displaystyle -\frac{x^2-7x+6}{2}\), which generates

0, 2, 3, 3, 2, 0, …,

so that the cubes would be

0, 8, 27, 27, 8, 0, …;

but that certainly isn’t the first thing that comes to mind.

Shoushou wrote back:

Thanks for giving me that great hint.   I'm not sure if I have it right, but I think it might be +2, +1, +2, +1.  This doesn't quite satisfy me, but it's the best I can come up with for now.

His suggestion is that the sequence is formed by starting with 0 and alternately adding 2, then 1, then repeating, so that the sequence would continue

0, 2, 3, 5, 6, 8, … ,

and the sequence of cubes would be

0, 8, 27, 125, 216, 512, … .

Doctor Wallace agreed:

I think you're right.  That's what I came up with, too.  And I feel the same way - I wasn't quite satisfied.  Somehow, it didn't feel just right.  I think it was because there were only 3 numbers given, and it's hard to come to just one pattern with only three numbers. But it seems that, since the number 1 was missing, +2 +1 +2 +1 ... is the only thing that makes sense.

Then the next two numbers in the pattern would be 5^3 and 6^3.  What did your teacher have in mind for the answer?  If you can come up with another pattern that fits the series, please write and let me know.

We never heard what the teacher intended; I certainly hope that any answer that made some sort of sense was accepted.

I’ve mentioned a couple other possibilities. If we find a quadratic function that gives the required values, it’s \(\displaystyle a_n = \frac{11x^2-17x+6}{2}\), and continues

0, 8, 27, 57, 98, 150, … .

Or, we could continue decreasing the amount added, so that we next add 0, then -1. The base sequence is then 0, 2, 3, 3, 2, 0 — which is my quadratic sequence above!

To find other possibilities, we could go to OEIS, the Online Encyclopedia of Integer Sequences, which catalogs sequences that people have found interesting. The site recommends entering about 6 terms, which will filter out many alternatives. Entering “0, 8, 27” gives me sequences like “Sum of cubes of primes dividing n,” or “Numbers that are the sum of cubes of distinct primes,” to list only the most comprehensible of those that start with the 0. And it doesn’t include any of those I’ve mentioned.

But my guess is that the zero was a typo.

Four terms, still insufficient

Here’s another from 2001:

Finding a Pattern

My daughter received this in a homework assignment, and I don't believe there is enough specific information to logically give the next four numbers in the sequence: 2, 8, 7, 28.

Four terms might be enough to support a very simple sequence; for example, if the difference from one term to the next was the same, we’d have three identical numbers and a reasonably strong case for an arithmetic progression. But the differences here are 6, -1, 21, which reveals nothing.

I agreed with Ray, but I had the same idea Doctor Wallace had above, and with a little more justification:

I agree, there really is not enough information here. I can guess what they probably want, however; most likely they have had other examples where they alternated two simple operations to get successive terms, and you are expected to assume that this pattern is similar. 

If so, then we are first multiplying by 4 (2*4 = 8), then subtracting 1 (8-1 = 7), then multiplying by 4 again (7*4 = 28), so you would continue in the same way: 27, 108, 107, 428.

The evidence we have is that multiplication by 4 appears twice, and subtraction of 1 in between is a simple thing to do. So I felt justified in using the word “probably”. My sequence, then, is

2, 8, 7, 28, 27, 108, 107, 428, … .

But there isn’t a lot of evidence for the subtraction step, as we see only one instance of it; and some other alternation is quite reasonable:

But another perfectly valid pattern would be "for odd terms, add 5 each time; for even terms, add 20." That would give 12, 48, 17, 68.

That is, the sequence could arise from merging the arithmetic progressions 2, 7, 12, 17, … (common difference 5) and 8, 28, 48, 68, … (common difference 20). This is a different kind of alternating sequence:

2, 8, 7, 28, 12, 48, 17, 68, …

On the other hand, if we’re merging sequences, we have only two terms of each, so they could be just about anything. What if the even terms were a geometric progression where we multiply each term by 28/8 = 3.5 to get the next: 8, 28, 98, 343, … ? And, interestingly, the odd terms would have the same common ratio, 7/2 = 3.5, and we’d get 2, 7, 24.5, 85.75, … . So our answer would be

2, 8, 7, 28, 24.5, 98, 85.75, 68, … .

Not very pretty, but just as logical!

Who is to say whether any of these is “correct”? All yield the same first four terms, so a teacher would have to give credit to all of them — unless there is some information we haven’t been given, such as that the class has only learned about merging arithmetic progressions, or all the examples given had alternated addition and multiplication.

I continued:

If a problem merely says "give the next four numbers" or "find the pattern in this sequence," there are infinitely many possible answers, since the word "pattern" has no precise definition; it's really a matter of guessing what pattern they had in mind, which is not math but psychology or ESP. To make this a valid problem, they should say something at least as clear as "This sequence was formed by a pattern similar to those you saw in this chapter. Make a reasonable guess as to what the pattern is, and show how it continues." Or, I suppose, they could say "Find a pattern in this sequence, explain how it works, and use that pattern to predict the next four numbers. There may be more than one correct answer." 

But to imply that students can determine _the_ correct answer by looking at four numbers is a misleading lesson in what math is all about. It's not a guessing game.

Four terms, and a thoughtful student

Here is another with four terms, from 2002:

The Perils of Predicting Patterns

What is the next number in the pattern 1,3,6,10 ___?  

If the pattern is to add 2, 3, 4, and then 2, 3, 4 again and again, it should be 12.  But if the pattern is to add 2, 3, 4, 5, and so on, then it should be 15.  Which is correct?

To my mind, it’s “obvious” what they intended, because this is a well-known arithmetic progression called the triangular numbers (1, 1+2, 1+2+3, 1+2+3+4, …). If so, then Su’s second guess is “correct”. But Su has the makings of a mathematician, and sees that there are other possibilities, one being a mere repetition of the pattern of differences. Doctor Greenie had an excellent answer:

There can't be a single "correct answer" for any question like this, since no matter what list of numbers you give me, I can find a formula that will fit them to any following number.

Usually what is desired is the "simplest" answer, and unfortunately, different people's definition of what it means to be "simple" varies.

Obviously, adding 2, then 3, then 4, then 2, then 3, then 4, then 2, and so on, is one way to extend the sequence; and adding 2, then 3, then 4, then 5, then 6 is another.

To my mind, the second is slightly simpler but only because the pattern presented so far does not give us any reason to think that we should go back to adding 2 to obtain the next number.

In other words, in line with my thoughts above, in the absence of evidence that a change is to be made (such as going back and repeating), it is more likely that the author intended just to continue the existing pattern. But that doesn’t make either answer more correct than the other.

He continued:

These sorts of patterns are used in intelligence tests, and the "correct" answer is "whatever very intelligent people think the correct answer is".  That's not much help, is it?

I remember a wonderful example shown to me once that illustrated how silly this sort of question is.  Here it is:

  What comes next in this sequence?

  33, 23, 14, 9, ___

The answer is "Christopher Street".  The reason is that the numbers are the exits of the 6th Avenue subway in New York City.

(Looking at a current subway map, it looks like things have changed a little, but the idea is clear!)

Four terms, a proof, and a guess

A 1996 question provided a chance for a different approach to the issue:

Predicting the Next Number

When given a series of numbers and asked to predict the next number, what is the formula for doing so?  Example:  2,5,12,23, ?

This question appears on psychological exams, federal employment exams and many others.  Is there a mathematical way of determining the next number in this series?

As we’ve been saying, there is no truly mathematical way to do it; but Doctor Jerry provided a very mathematical way to show that it can’t really be done. He supposed that we knew a polynomial formula for the sequence, and showed that we can find another polynomial that will give the very same four terms, AND whatever fifth term we choose! This is the idea Doctor Greenie mentioned in his first sentence above.

First, if the first several terms of a sequence are given, there is no method for determining the general term. Suppose I'm given the numbers a, b, c, d and asked to determine the fifth and sixth terms of the sequence.  I'll show that any number of different solutions is possible.

I start by determining a polynomial 

   p(x) = Ax^3 + Bx^2 + Cx + D 

such that p(1) = a, p(2) = b, p(3) = c, and p(4) = d.  Then consider

   f(n) = p(n) + (n-1)(n-2)(n-3)(n-4) or
   g(n) = p(n) + (n-1)(n-2)(n-3)(n-4)(n-5).

Notice that both f and g determine sequences whose first four terms are a, b, c, and d.  Remaining terms are wildly different.  This idea could be elaborated.

His function f is identical to p for n = 1, 2, 3, 4, because the added product is zero in all four cases. But for n = 5, he is adding 24 to the value of p(5), giving a new polynomial with a different fifth term; and we could adjust that to make it any number we want. And his function g will match p for the first 5 terms, but differ in the sixth.

So even if we required a really “mathematical” answer in the form of a polynomial, there are infinitely many “correct” answers. You’d have to specify that the function must have the lowest possible degree, to make only one answer correct.

But, of course, that is not what problems like this are looking for:

You can, however, try to guess what was most likely in the mind of the person who made up the question.  For the sequences
  2,4,6,8,...
  1,4,9,16,...
I suppose most persons would say 10,12  or 25,36.

The two examples he gives are “skip counting” (an arithmetic progression) and a list of perfect squares, both of which most of us would recognize and suppose that anyone writing a test would be more likely to think of than of other possibilities.

For the sequence you gave, 2, 5, 12, 23, I noticed that 5 - 2 = 3, 12 - 5 = 7, and 23 - 12 = 11.  Since 3, 7, and 11 can be viewed as the odd numbers, leaving every other one out, one could argue that the next two terms are 23 + 15 and 38 + 19.  Other, more or less natural answers are possible.  However, one has no choice but to accept whatever the text makers decree is the correct answer!

Here, he observes that the differences between successive terms are 3, 7, 11, which increase by 4 each time; so the natural thing to do is to continue with 15, 19.

The problem (like all the others in this post) didn’t ask for a formula, just for the next term, so that is all that is needed. The differences imply (as we’ll see when we get to the Method of Finite Differences) that the sequence has a second-degree polynomial (quadratic) formula, which turns out to be \(2x^2 – 3x + 3\), which generates the terms

2, 5, 12, 23, 38, 57, … ,

just as Doctor Jerry found by adding 15 and 19.

If we take that as \(p(x)\), then the \(f(x)\) above is $$f(n) = p(n) + (n-1)(n-2)(n-3)(n-4) =// 2x^2 – 3x + 3 + n^4-10n^3+35n^2-50n+24 = n^4-10n^3+37n^2-53n+27.$$ The first six terms of this are

2, 5, 12, 23, 62, 177.

As intended, the first four terms agree, but the next terms are different.

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