I currently have an open question about counting the possible ways of summing numbers. I am still exploring all the ideas provided - those within my level of understanding. This is a question involving looking into one avenue and possible solution. The question involves finding the possible ways of creating a number $n\in \mathbb{I}$ only by adding 1, 2, or 3. For example, for the number 4:
$$1+1+1+1, 1+1+2, 2+2, 3+1$$
There are four possible combinations. My question involves breaking down the number $n$ using as many 3's as possible. For instance, the number 12 can be written $3+3+3+3.$ Any number greater than or equal to 3 can be broken into
$$3+3+\cdot \cdot \cdot+3$$ or $$3+3+\cdot \cdot \cdot+3+2$$ or $$3+3+\cdot \cdot \cdot+3+1$$
For the first case, it would be a number $n$ that when divided by three yields an integer. Suppose $n=12$, then $n$ can be written as $3+3+3+3$. However, each of these 3's can be written as $3$, $2+1$, or $1+1+1$. That is, there are three possible ways of summing to 3.
How can I calculate the possible combinations? I've thought for the case that $n=12$, there are four 3's that can be either $1+1$, or $1+2$, or $1+1+1$. Can any given number be broken down into a number of 3's, then the possible combinations of $3, 2+1, 1+1+1$ calculated? I would also have to take into account the cases where there would be a remainder of 1 or 2 after dividing by 3.
Does this make any sense?
This is not an answer to the problem itself; rather, it’s some ideas that you might find useful in attacking the problem.
Consider the set $P(n)$ of partitions of $n$ into parts no larger than $3$. They come in three flavors: those with smallest part $1$, those with smallest part $2$, and those with smallest part $3$. For $k=1,2,3$ let $P_k(n)$ be the set of $p\in P(n)$ with smallest part $k$.
Each $p\in P_1(n)$ is of the form $1+q$, where $q\in P(n-1)$; conversely, each $q\in P(n-1)$ gives rise to a partition $1+q\in P_1(n)$. Thus, $|P_1(n)|=|P(n-1)|=p(n-1,3)$.
Each $p\in P_2(n)$ is of the form $2+q$, where $q\in P_2(n-2)\cup P_3(n-2)$; conversely, each $q\in P_2(n-2)\cup P_3(n-2)$ gives rise to a partition $2+q\in P_2(n)$. Since $P_2(n-2)$ and $P_3(n-2)$ are disjoint, $|P_2(n)|=|P_2(n-2)|+|P_3(n-2)|$.
$P_3(n)$ has one member, $\underbrace{3+3+\ldots+3}_{n/3}$, if $n$ is a multiple of $3$; otherwise, $P_3(n)=\varnothing$.
You could try using these ideas as a starting point to develop recurrences that would at least allow you to calculate $|P_k(n)|$ (and hence $p(n,3)$ fairly efficiently for $n$ of reasonable size. Once you’ve done that, the data themselves may reveal further regularities that are worth pursuing.