Wikipedia:Reference desk/Archives/Mathematics/2014 October 11

= October 11 =

polygon disjoint sets
Let a polygon with m vertices labeled as 1,2,3,..,m. How many subsets of {1,2,..m} having two or more elements exists with the property that they should not include any two consecutive vertices of polygon. — Preceding unsigned comment added by 182.187.95.175 (talk) 21:29, 11 October 2014 (UTC)
 * Let's count this for subsets of size k, then sum over k. We'll break these into two categories: those containing 1, and those omitting it.
 * Let's start with those containing one. Then there are k "gaps"--one after every element.  Each of these gaps has at least 1 element in it, which leaves (m-2k) elements unaccounted for.  These remaining elements need to be distributed amongst the gaps.  Stars and bars is good for counting the number of ways to do this, giving us $$\frac{(m-k-1)!}{(m-2k)!(k-1)!}$$ different sets with k elements including 1.
 * Next, we'll consider those sets omitting 1. Here there are k+1 gaps--one after every element, and one before 1.  Each has at least 1 element, except possibly the last one.  So still (m-2k) elements unaccounted for, but this time k+1 many gaps to distribute them in.  So $$\frac{(m-k)!}{(m-2k)!k!}$$ sets with k elements and excluding 1.
 * Note that k can be at most half of m, so the total number of sets is $$\sum_{k = 2}^{m/2} \frac{(m-k-1)!}{(m-2k)!(k-1)!} + \frac{(m-k)!}{(m-2k)!k!}$$.  The interior can be simplified a bit, getting you $$\sum_{k=2}^{m/2}  \frac{m(m-k-1)!}{(m-2k)!k!}$$.--80.109.80.31 (talk) 23:19, 11 October 2014 (UTC)


 * This is A023548. I don't know why it is, but the first 20 terms all match. -- BenRG (talk) 00:42, 12 October 2014 (UTC)
 * As perhaps a clue, I understand why the terms representing the prime m have to be divisible by m in the polygon disjoint sets, if we understood why it was true in A023548... (note A023548 needs to be shifted by 3, since a square is the first one that can be disjoint...Naraht (talk) 01:05, 12 October 2014 (UTC)
 * It's well known that the Lucas numbers represent the number of cyclic 0-1 sequences that have no consecutive 1's; it's not mentioned in our article but the oeis entry mentions it. It's a small leap to get from that the numbers of subsets of the vertices of a polygon with no consecutive elements are Lucas numbers. You can also state this as the number of independent sets in a cyclic graph. The number of sets with 0 or 1 element is easy to determine so that gives a formula for the number in question. Lucas numbers can be given in terms of Fibonacci numbers so it's probably not hard to turn this formula into a proof that the sequence in question is A023548. --RDBury (talk) 13:21, 12 October 2014 (UTC)