Background:
This is a chemistry inspired problem. Alkanes are simply chains of carbon with hydrogens attached to each carbon (the general formula is $C_n H_{2n+2}$).
Alkenes are chains of carbon similar to alkanes, although they have a double bond and have the general formula $C_n H_{2n}$
Alkanes can be cracked to produce alkanes and alkenes. Cracking involves breaking a bond betweek two carbon atoms. In each "crack," only one bond in the chain can be broken. For each "crack," one alkene and one alkane is produced.
Puzzle:
Given that the smallest possible alkene contains chain length two, and given that the alkane produced from cracking a larger alkane can also be cracked, how many ways can an alkane of chain length $n$ be cracked so that the orginal chain is cracked at least one time.
Steps towards a solution:
As far as I am concerned, the hydrogen is not relevant; so we are trying to work out how to split a chain so that one part of it can be split again. My best idea so far was to use an algorithm almost identical to the change making algorithm to work out how many combinations of $1, 2, 3... n$ length chains there are that create a $n$ length chain
Edit:
Another way I think this can be thought about is how many ways can a chain of length nn be broken so that all but 1 or less constituent chains has a length of two or greater. I believe that only one methane molecule could be produced as producing an methane molecule is the final possible crack in any sequence seeing as the other part must be an alkene so it is impossible to crack further.
In its current form, the question asks how many distinct ways there are to divide the number $n$ into one or more values that are $\ge 2$, with possibly one exceptional value of $1$. Here is the answer to that specific question:
Let $f(n)$ be the numbers of ways to partition $n$ into values $\ge 2$, where the order of the values within a given partition does not matter (only the number of occurrences of each value). Then the generating function $\sum_{n\ge 0} f(n) x^n$ is given by the formal product
$$\prod_{k=2}^\infty \frac{1}{1-x^k} = \prod_{k=2}^\infty (1+x^k+x^{2k}+x^{3k}+\cdots).$$
Comparing this to the very similar g.f. for the standard partition function we see that $$\sum_{n\ge 0} f(n) x^n = (1-x) \sum_{n\ge 0} p(n) x^n = \sum_{n\ge 0} (p(n) - p(n-1))x^n,$$
so that $f(n) = p(n)-p(n-1)$ for $n\ge 1$. Finally the number of partitions of $n$ with exactly one part of size $1$ is the same as $f(n-1)$ from the remaining chain of length $n-1$. Therefore the answer to the question posed at the top of this answer is $f(n) + f(n-1) = p(n) - p(n-2)$, valid for $n\ge 2$. If we wish to exclude the trivial partition ($n$ itself) then subtract $1$ from this.
For example, when $n=7$, $p(7) = 15$, $p(6) = 11$ and $p(5) = 7$, so we expect there to be $4$ partitions with all parts $\ge 2$ and $4$ partitions with exactly one part $=1$. These are indeed given by: