Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Ok, so I was in chemistry today and we where talking about isomery and alkanes. Basically an alkane is a compound made only of carbons and hydrogens that has only simple bonds and only carbon-hydrogen and carbon-carbon bonds. Since the carbons always have the 4 bonds then every alkane is determined uniquely by a spanning tree between the carbons which has no vertex with degree above 4. However I just want to find the different number of alkanes.

I tried using Caley's formula which states that the number of spanning trees on n vertices (carbons) is $n^{n-2}$ however trees which are automorphic are giving us the same compound. Also, no vertex can have degree greater than 4, so this doesn't work so well.

My question is, what is the number of different isomeric alkanes of $n$ carbons or in other words: the highest cardinality a set of trees of $n$ vertices can have such that none of them are isomorphic and no vertex has degree greater than 4. I think there might not be an easy way to solve this, if this is the case can you give me a reasonable bound? My teacher said the growth is exponential. Is this correct?

share|cite|improve this question
A related family of chemicals: – Will Jagy Feb 27 '13 at 1:35
up vote 2 down vote accepted

Look at OEIS sequences A000602 (ignoring stereoisomers) and A000628 (taking stereoisomers into account), and references given there.

share|cite|improve this answer
Ok, I meant without stereosimores, I haven't seen that in class yet. – The Kind Cat Feb 26 '13 at 22:41
I take it this means there is no easy way? – The Kind Cat Feb 26 '13 at 22:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.