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Been stuck on this question for days... Would appreciate any help :)

  1. Devise a function $p(n)$ such that $p(n)$ is the number of different ways to create postage of $n$ cents using 3-, 4-, and 5-cent stamps.
  2. Prove that $p(n)$ is monotonic nondecreasing on $\mathbb N$.

I'm getting to a point where I can narrow down how to generate $n$ (For a number $n$, take the set of solutions of $n-3$ and add 3 i.e. $p(n-3)$, but I don't know how I would go about accounting for repetitions.

Parcly Taxel
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mzab123
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  • Do you consider using a $5$-cent stamp, then a $3$-cent stamp to be a different way of creating $8$-cents than using a $3$-cent stamp, then a $5$-cent stamp? – N. F. Taussig Nov 02 '18 at 11:18

2 Answers2

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This post How many distinct ways to climb stairs in 1 or 2 steps at a time? might help you get started. Think about how you can adapt this method to your problem (you can add 3,4, or 5 cents in the last step). The additional challenge is that you need to account for the permutations. For example, if n = 9, you can do 3+3+3,4+5, or 5+4. But there are only two ways: 3+3+3 and 4+5 if we do not care about the order.

D.B.
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  • Hey thanks for the help, I actually got this far in the reasoning as I have mentioned in the post but i'm at a loss about what to do next, I tried partitioning the sets of solutions and even tried playing around with the Inclusion-Exclusion Principle, but it has gotten me no where. – mzab123 Nov 02 '18 at 05:15
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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.

quasi
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