How many subsets of $\{1,2,...,n\}$ do not contain three consecutive integers?

Question

My attempt. Let us assume $n$ is a large positive integer. ${n \choose 0},{n \choose 1},{n \choose 2}$ are the numbers of such subsets having $0,1,2$ elements respectively, which is trivial. For $3$ elements, the number of such subsets is ${n \choose 3}-(n-2)$. Starting from $4$ elements, my brain begins muddling up; I have no idea how to proceed further systematically. Any hint or idea would be appreciated.

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Remark. Thanks to the partial solutions by VIVID and Masacroso, I have completely solved this problem. Following VIVID's answer, I have posted an answer which completes the solution primarily for future reference.
I am going to give the accepted answer to VIVID who has been very dedicated to this problem seen from his number of edits. Also, most importantly, VIVID was the first person who posted the core part of the solution. Masacroso, hope you would not mind.

Last, though this problem has been completely resolved, any new approach is always welcome.

Answer 1

Partial solution: Let's denote by $S \subseteq \{1,2,\dots,n\}$ all sets that satisfy the condition. And let $a_n$ be the number of such sets.

There may be some cases:

1. $n \not \in S \implies$ there are $a_{n-1}$ possibilities for $S$ (clear)
2. $n \in S$:

• a) $n-1 \not \in S \implies$ there are $a_{n-2}$ possibilities for $S$ (why?)
• b) $n-1 \in S, \ \ n-2 \not \in S \implies$ there are $a_{n-3}$ possibilities for $S$. (why?)

Hence, we get the recurrence formula $$\boxed{a_n = a_{n-1} + a_{n-2} + a_{n-3}}$$

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Answer two (why?) parts above and it will become a full solution.

Answer 2

Sketch for the solution: you can construct a recursion to find this number. Let $x_k$ the number of subsets in $\{1,\ldots ,k\}$ that doesn't contain three consecutive numbers, then $x_{k+1}=x_k+x_{k-1}+x_{k-2}$ because

• $x_k$ is the number of subsets in $\{1,\ldots ,k+1\}$ that doesn't contain $k+1$ and dont have three consecutive numbers
• $x_{k-1}$ is the number of subsets in $\{1,\ldots ,k+1\}$ that contains $k+1$ but doesn't contain $k$ and doesn't contain three consecutive numbers
• $x_{k-2}$ stay for the subsets in $\{1,\ldots ,k+1\}$ that contains $k+1$ and $k$ but doesn't contains three consecutive numbers.

Answer 3

Here's perhaps another approach. (Well it's basically equivalent to the answers already given, but maybe the binary string representation makes the problem easier to think about. At least it does for me, but that might be because I happened to already be familiar with these "run length problems".)

You can think a subset $S$ of $\{1,\dots, n\}$ as a binary string of length $n$, where $1$ at position $j$ means $j\in S$ and $0$ means $j\notin S$. Now, the subsets we want to count correspond to binary strings without a $3$-run of $1$'s.

To solve this new problem, lets denote

$$A_n = \{\text{length } n \text{ binary strings without a run of three 1's} \}$$

Now think of a $w\in A_n$ and the number of $1$'s it has in the beginning. That can be either $0$, $1$ or $2$. Then there is a zero and the rest is a word in $A_{n-1}$, $A_{n-2}$ or $A_{n-3}$, respectively. We're assuming $n>3$; the first ones are base cases. You can see the tribonacci-recursion for the size $|A_n|$.

Answer 4

First of all, a lot of thanks to VIVID and Masacroso for their partial solutions by a recursive approach which is indeed wonderful. Below, I just want to finish what they left off (which is good so that I can think) for future reference and for reinforcing my own understanding.

Let me follow VIVID's notations.
Let $s$ denote a general element of $S$.

• For those $s$ which do not contain $n$, there are $a_{n-1}$ such $s$, which is trivial.
• For those $s$ which contain $n$ but do not contain $n-1$, there are $a_{n-2}$ such $s$. Because by 'isolating' $n$ we need only consider the selection of the rest $n-2$ integers.
• For those $s$ which contain both $n$ and $n-1$, we must not include $n-2$ in such $s$, and then by 'isolating' $n$ and $n-1$ we need only consider the selection of the rest $n-3$ integers. Thus there are $a_{n-3}$ such $s$.
  

Then, we recognize that those three sets of different $s$ are disjoint, and indeed their union is $S$. Hence the recursive relation $a_n=a_{n-1}+a_{n-2}+a_{n-3}$. Then we can form a sequence similar to the Fibonacci numbers but this time adding the preceding three numbers to get the next, i.e. $1,2,4,7,13,24,44,81,...$, where the first number indicates $a_0$ which corresponds to the empty set.

Remark. I realize that using this recursive ideology, we can further derive sequences of $a_n$ with the condition replaced by not having $r$ consecutive integers where $r$ can be any positive integer.