# Trivial sequence: $32,21,14,\dots$

My little brother (third grade) asked me for help with this math problem on his homework, which was:

Find the next number in the sequence $32,21,14,\dots$

I was not able to see a trivial solution, so I ran a linear regression (for an equation for $a_n$) which turned up an equation with an $r^2$ of .$98$, which (considering I had three points) was not suitable, so I ran a quadratic regression and got the polynomial $$2n^2-17n+47$$ which yielded an $r^2$ of 1, and so was a perfect fit. Therefore, I said the answer was $11$.

My question is, considering that the question assumes no knowledge of algebra, or exponents, is there a simple (recursive, maybe?) formula for the next number in the sequence?

• Another possibility is $7$. Just multiply the leading digit by $7$. This is something doable by a $3$-grader. – achille hui May 1 '14 at 1:02
• I really like that! If you put it as an answer, I would up-vote. However, if you see my comment on Lost's post, they are expected to do operations on the numbers, not on the individual digits. – Juan Sebastian Lozano May 1 '14 at 1:03
• I always thought these kind of problems were silly because there is no right or wrong answer. If you look hard enough you can make all sorts of contrived patterns. I guess these kind of things can be "fun" but they could also be frustrating, especially for a kid. – Seth May 1 '14 at 1:04
• @seth I think these problems are supposed to have a disclaimer with a modified version Occam's Razor, saying "among competing hypotheses, the one with the fewest operations should be selected." :P – Juan Sebastian Lozano May 1 '14 at 1:09
• An alternate answer is $9$. one take the two thirds of previous number. i.e. $a_{n+1} = \lfloor 2/3 * a_n \rfloor$. This one now operates on numbers instead of just digits. Of course, everyone knows the ultimate answer should be $42$ but no one knows what is the real question. – achille hui May 1 '14 at 1:16

Note that you have not proved that the pattern is what you say. The next term could be anything and you could find a cubic that went through the four points. There are also many other rules that could form these numbers. Accepting the quadratic, I think that formula is fine. Another way to indicate it is with a difference table, where each term is the difference of the two above it: $$\begin {array}&32&&21&&14&&11\\&11&&7&&3\\&&4&&4 \end {array}$$

• This is a really nice way to explain it to a third-grader, "it's the difference of the differences"; no need to know calculus! (yet) – NothingsImpossible May 1 '14 at 2:59

I see a pattern of subtracting the number you get by adding the digits together, doubling the resulting sum, then adding one.

e.g. $3+2 = 5 \times 2 = 10 + 1 = 11$

$32-11 = 21$ and so on

In which case, the next term would be $3$

• That is a nice pattern, but to be expressed in operations it would probably be $$a_n = a_{n-1} - \left(\left(\lfloor \frac{a_{n-1}}{10} \rfloor + \left(a_{n-1} - 10 \cdot \lfloor \frac{a_{n-1}}{10} \rfloor \right) \right) \cdot 2 + 1 \right)$$ (or something similar) which is not using elementary school- level operations. – Juan Sebastian Lozano May 1 '14 at 0:53
• You don't need to know that formula in order to be able to compute this; it seems pretty reasonable to me for the third grade. – Lost May 1 '14 at 0:58
• Fair point, but my point is that they are expected to do operations on numbers, not work with the digits. However, for curiosity's sake, I like this pattern. – Juan Sebastian Lozano May 1 '14 at 1:01
• Ah alright then, that explains things. I'll try to think of something else. – Lost May 1 '14 at 1:02

A humble attempt:

To me doesn't seem trivial at all, but a possible pattern that comes to my mind is the following:

Let's say a number $a_1$ in the sequence is given, to get $a_2$ take $a_2=a_1+0.5$, to get $a_3$ take $a_3=a_2-0.75$. I believe this is a pattern that could be given to a third grade, even if they can't find a formal expression for the sequence.

Now, looks like we have something like $a_n=a_{n-1}+(-1)^{n}0.25n$ . If the first number is 1.5, follows 2.0, then 0.25, and then 1.0; notice anything curious?, those numbers can be written as $3/2,2/1,1/4,1/1,\dots$. So maybe this could be thinked as the nth term of the sequence being the fractional form of the number given by the rule above.

I don't know, maybe the way I'm thinking this is kinda dumb, but is the only thing that comes to my mind right now. But I insist, to me it doesn't look trivial at all.

Edit: I post this here because I added a couple of things for clarification and no longer fits in the comment section.