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?
 A: 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: $$\begin {array}&32&&21&&14&&11\\&11&&7&&3\\&&4&&4 \end {array}$$
A: 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 $3$
A: 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.
A: There are many possible answers. One possible pattern is subtraction of primes leading to a finite end, yet an infinite beginning.
...
81-19=62
62-17=45 45 <-- preceding number of sub-sequence
45-13=32 32 <-- start of given sub-sequence
32-11=21 21
21- 7=14 14 <-- end of given sub-sequence
14- 5= 9  9 <-- next number of sub-sequence
9- 3= 6
6- 2= 4 <- end of sequence
