<s>[INST] Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins I encountered the following problem in Herber Wilf's book Generatingfunctionology: 

Prove that, in country that has $1$-cent, $2$-cent, and $3$-cent coins
  only, the number of ways of changing $n$ cents is exactly the integer
  nearest to $ \frac{(n+3)^2}{12} $.

Of course, the solution provided was one that makes use of a generating function. However, it is unbelievably tedious. Is there a way to do this without making use of a generation function?

For reference, here is the solution given in the book. 

If $f(n)$ is that number, then
   $$ \begin {gather*} 
\sum_{n\ge 0} f(n) x^n =
 \frac {1}{(1-x)(1-x^2)(1-x^3)}\qquad\qquad\qquad\qquad \\
= \frac{1}{6(1-x)^3} + \frac{1 }{4(1-x)^2} + 
\frac {17}{72(1-x)} + \frac {1}{8(1+x)} + \frac {1}{9(1-\omega x)} +
 \frac {1}{9(1-\overline{\omega}x)}.
 \end {gather*} $$ 
  If we expand each of
   the fractions on the right, we find the formula $$ f(n) = \frac {1}{6}
 \dbinom {n+2}{2} + \frac {1}{4} (n+1) + \frac {17}{72} + \frac
 {(-1)^n}{8} + \frac {2}{9} \cos \left( \frac {2n \pi}{3} \right), $$
   which can be rewritten as 
  $$ f(n) = \frac {(n+3)^2}{12} + \frac {-7 + 9 (-1)^n 
+ 16 \cos \left( \frac {2n\pi}{3} \right)}{72}. $$
   The second
   fraction cannot exceed $ \frac {32}{72} < \frac {1}{2} $ in absolute
   value, so $f(n)$ is the unique integer who distance from 
  $ \frac {(n+3)^2}{12} $ is less than $\frac{1}{2}$, as required. $\Box$

 [/INST] It is possible to shorten a bit the argument by noticing that
$$\operatorname{Res}\left(\frac{1}{(1-z)(1-z^2)(1-z^3)},z=1\right)=-\frac{17}{72},$$
hence the contribute given by the simple poles $-1,\omega,\omega^2$ lying on the unit circle cannot exceed $\frac{17}{72}$, and is exactly equal to $\frac{17}{72}$ when $6\mid n$. The pole in $z=1$ is a triple pole, hence $f(n)$ grows like a second-degree polynomial. By interpolating the values of $f(n)$ on $n=0,n=6,n=12$ we can derive our formula.
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