Let's say that you set $$f(1)=f(3)=0,\,f(2)=f(5)=1,\ f(4)=2\\f(n)=f(n-2)+f(n-4)+f(n-5)\quad\text{if }n\ge6$$
Then $f(7)=f(2)+f(5)=2$, which corresponds to the stamps $2+5$ and $5+2$. That's fine if you actually want to distinguish between ordered sums (like if you were interested in the order in which stamps were placed on an envelope to add up to $7$ cents. But it probably isn't.
The problem can't actually be solved with a recurrence relation, not even a non-linear one. $^{\color{blue}{\text{[citation needed]}}}$ What you need to do is to check out the last chapter of your combinatorics textbook, which will talk about generating functions. Here is a thread that talks about the canonical problem of this genre, which is counting the $293$ ways to make change for a dollar from coins with values of $1,5,10,25,50,$ and $100$ cents. In the case of your problem, the solution is the coefficient of $x^{20}$ in the polynomial $$\frac1{(1-x^2)(1-x^4)(1-x^5)}$$
This sequence is OEIS A025802, and the values from $f(0)$ through $f(61)$ are
1, 0, 1, 0, 2, 1, 2, 1, 3, 2, 4, 2, 5, 3, 6, 4, 7, 5, 8, 6, 10, 7, 11, 8, 13, 10, 14, 11, 16, 13, 18, 14, 20, 16, 22, 18, 24, 20, 26, 22, 29, 24, 31, 26, 34, 29, 36, 31, 39, 34, 42, 36, 45, 39, 48, 42, 51, 45, 54, 48, 58, 51
So f(20)=10. If $(x,y,z)$ represents buying $x$ 2-cent stamps, $y$ 4-cent stamps, and $z$ 5-cent stamps, the ten different arrangements are
(10,0,0), (8,1,0), (6,2,0), (4,3,0), (2,4,0), (0,5,0)
(5,0,2), (3,1,2), (1,2,2)
(0,0,4)
Best Answer
For each integer $n$, let $p(n)$ be the number of nonnegative integer triples $(x,y,z)$ such that $$3x+4y+5z=n$$ Using a generating function approach as in
$\qquad$What's the number of natural solutions of $x_1 + 2x_2 + 3x_3 = n$?
we get the recursion $$ p(n)= {\small{\begin{cases} \text{if}\;n<0,\;\text{then}\\[3.5pt] \qquad 0\\[2.5pt] \text{else if}\;n=0,\;\text{then}\\[3.5pt] \qquad 1\\[.6pt] \text{else}\\[.4pt] \qquad p(n-3)+p(n-4)+p(n-5)-p(n-7)-p(n-8)-p(n-9)+p(n-12)\\ \end{cases} }} $$ Next we show that $p$ is non-decreasing for $n\ge 1$ . . .
For each positive integer $n$, let \begin{align*} a(n)&=p(n+60)-p(n)\\[4pt] b(n)&=a(n+1)-a(n)\\[4pt] c(n)&=b(n+1)-b(n)\\[4pt] \end{align*} Applying the recursion as many times as needed (using Maple), we can reduce each of $a(n),b(n),c(n)$ to an integer linear combination of $p(n-1),...p(n-12)$, yielding \begin{align*} a(n)&={\large{\langle}}{37, 38, 39, 3, -34, -72, -74, -39, -3, 34, 35, 36}{\large{\rangle}}\cdot \vec{P}\\[4pt] b(n)&={\large{\langle}}{ 1, 1, 1, 0, -1, -2, -2, -1, 0, 1, 1, 1}{\large{\rangle}}\cdot \vec{P}\\[4pt] c(n)&={\large{\langle}}{0,0,0,0,0,0,0,0,0,0,0,0}{\large{\rangle}}\cdot \vec{P}\\[4pt] \end{align*} for all $n\ge 1$, where $\vec{P}={\large{\langle}}{p(n-1),...,p(n-12)}{\large{\rangle}}$.
By direct evaluation, we get $b(1)=1$.
Since $c(n)=0$ for all $n\ge 1$, we get $b(n)=b(1)=1$ for all $n\ge 1$.
Our goal is to prove $p(n)\le p(n+1)$ for all $n\ge 1$.
Proceed by strong induction on $n$.
By direct evaluation, we get $p(1) \le p(2) \le p(3) \le \cdots \le p(61)$.
Next suppose that for some $n\ge 61$, we have $p(k)\le p(k+1)$ for all positive integers $k < n$. \begin{align*} \text{Then}\;\;&b(n-60)=1\\[4pt] \implies\;&a(n-59)-a(n-60)=1\\[4pt] \implies\;&\bigl(p(n+1)-p(n-59)\bigr)-\bigl(p(n)-p(n-60)\bigr)=1\\[4pt] \implies\;&p(n+1)-p(n)=\bigl(p(n-59)-p(n-60)\bigr)+1\\[4pt] \implies\;&p(n+1)-p(n)\ge 1\\[4pt] \implies\;&p(n) < p(n+1)\\[4pt] \implies\;&p(n)\le p(n+1)\\[4pt] \end{align*} which completes the induction.
Hence, $p$ is non-decreasing for $n\ge 1$, as was to be shown.