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I've been looking for a formula determining the number of possible isomers for a given organic compound. As such formula does not exist I though I could do a little investigation on my own. I looked at alkanes:

Structural formulas of alkanes may be represented as graphs. For instance, propane:

enter image description here

can be represented as:

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(source: chemistry-reference.com)

We assume the carbon atoms to be the vertices and the bonds between them as edges. It is clear that for $n$ vertices there are $n-1$ edges. Due to the valency of carbon it is also true that $\forall$ $v \in V$ $1\leq$ deg($v$) $\leq$ $4$.

Now, in the case of butane, the adjacency matrix for a straight-chain is:

$$\begin{bmatrix} 0&1&0&0 \\ 1&0&1&0 \\ 0&1&0&1 \\ 0&0&1&0 \end{bmatrix}$$

assuming we number the vertices in order from left to right.

Observations:

  1. The main diagonal will always consist of zeros, since the carbon cannot bond to itself.
  2. By extrapolation, the total number of 1's in the matrix is $2(n-1)$, which also corresponds to the sum of the degrees of all vertices in the tree.
  3. Summing the rows or columns gives the degree of the vertex, where analogically $1$ $\leq$ $\sum_{k=1}^{n} a_k$ $\leq$ $4$
  4. An isomer exists for a distinct combination of the degrees of the vertices. For instance, the matrix :

$$\begin{bmatrix} 0&0&1&1 \\ 0&0&0&1 \\ 1&0&0&0 \\ 1&1&0&0 \end{bmatrix}$$

gives the same straight-chain butane, with only the order of the vertices changed. This is because the combination of the degrees of the vertices is the same as in the previous matrix (two $2$'s and two $1$'s). However, the matrix:

$$\begin{bmatrix} 0&0&0&1 \\ 0&0&0&1 \\ 0&0&0&1 \\ 1&1&1&0 \end{bmatrix}$$

gives isobutane, a structural isomer of butane. This is because the combination of the degrees is different (three $1$'s and one $3$)

I was wondering whether there is a way to find the total number of combinations of degrees for such matrices for larger $n$ considering all the restrictions (and also the symmetry about the main diagonal)?

I tried examining only one of the "sections" of the matrix (above or below the main diagonal), say, the upper one. But this is futile, as I need to know the total value of a row/column.

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