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How many distinct ways can a sequence of $n$ $1$s be partitioned into pairs or singles, in which $\{1,1\}=\{2\}$ is considered a pair and $\{1\}$ is considered a single?

For example $\{1,1,1,1\}$ can be partitioned into:

$\{2,2\}$

$\{1,2,1\}$

But

$\{2,1,1\}$ and $\{1,1,2\}$ are equivalent to $\{2,2\}$

No result containing $\{1,1,1\}$ should be enumerated since this is a triple and has not been partitioned into pairs or singles.

So for $n=4$, the answer is $2$ ways.

I think the answer to this question describes the the number of Dyck words which give unique results when exponentiating powers of $2$... as discussed in this question. Or it is at least part of the answer in respect of the fact that it factors out the identity $2^4=4^2$.

Robert Frost
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  • Did you mean to say that ${ 2,1,1}$ and ${1,1,2}$ are the same as ${1,2,1}$? Also, would ${ 1,1,1,1 }$ be a third partition for $n = 4$? – Bram28 May 08 '17 at 15:33
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    I don't understand your equivalence relation. – mercio May 08 '17 at 15:35
  • I think it is safe to assume he meant to write what Bram28 suggested... – Corey May 08 '17 at 15:37
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    but then everyone is equivalent to ${1,1,1,1}$ – mercio May 08 '17 at 15:40
  • @Bram28 no, the opposite; ${ 2,1,1}\sim{ 2,2}\sim { 1,1,2}$ but ${ 1,2,1}$ is distinct from them. – Robert Frost May 08 '17 at 16:01
  • @PaoloLeonetti I am defining an equivalence relation. I think it is ${2,1}\sim{1,2}$ – Robert Frost May 08 '17 at 16:03
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    @RobertFrost But then why not ${1,2,1} \sim {1,1,1,1} \sim {1,1,2 }$? – Bram28 May 08 '17 at 16:03
  • @PaoloLeonetti I think the interpretation in your answer is correct. – Robert Frost May 08 '17 at 16:04
  • So, for $n=3$ does this mean that ${1,1,1}$, ${2,1}$ and ${1,2}$ are all equivalent? Because if so that doesn't seem to be the prevailing interpretation below (I.e. no two consecutive 1s). – N. Shales May 08 '17 at 16:10
  • @N.Shales My intent - which is perhaps not clear in the question - is that ${1,1,1}$ has not been partitioned into pairs and singles and is therefore not a valid solution. It's a triple. – Robert Frost May 08 '17 at 16:16
  • @RobertFrost Okay I misspoke ${1,1,1}$ was just to represent the set to be partitioned but if ${1,2}$ is equivalent to ${2,1}$ then this isn't the same as an ordered sum of 1s and 2s with no consecutive 1s. If it was then ${1,2}$ and ${2,1}$ would be counted as distinct. – N. Shales May 08 '17 at 16:31
  • @N.Shales good point – Robert Frost May 08 '17 at 16:35
  • @N.Shales then I think the relation is more like ${2,{1,1}}={{1,1},{1,1}}$ – Robert Frost May 08 '17 at 16:47
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    @RobertFrost so ${{1,1},1}$ is different from ${1,{1,1}}$ then? I'm still confused about the problem statement although I do understand both answers, I just can't quite see that the problem must be interpreted this way. Despite this I cannot think of a better interpretation. – N. Shales May 08 '17 at 16:56
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    @N.Shales: What Robert wants here is certainly not an equivalence relation for the reason both mercio and Bram28 already mentioned above. But one can define a partial order where for any two sequences $x,y$ we have $x<y$ iff $x$ can be obtained by replacing some pair of adjacent "1"s in $y$ with a "2", some number of times. Then the class of sequences Robert really wants is precisely the class of sequences that each contains only "1"s and/or "2"s and is minimal under this ordering. – user21820 May 09 '17 at 14:36
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    @user21820 what may have thrown N.Shales is my statement that this problem is related to the Dyck Words corresponding to unique values of $2^{2^{\ldots}}$ because in that instance $(2^2)^2=2^{(2^2)}$ which, if this problem were equivalent would indeed suggest ${1,2}\sim{2,1}$; but that's not the case. I saw a relation between the two when I wrote the question yesterday. It's to do with the fact every pair ${2,1}$ can then be interchanged without changing the value of the tetration. – Robert Frost May 09 '17 at 14:57
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    @RobertFrost: Whatever it was, I'm just providing a rigorous way to formalize the intuitive notion of pairing; we get a partial order by taking the transitive closure of the ordering given by a reduction by a single pairing operation. So (1,1,2) is not equivalent to (2,2) but rather (1,1,2) > (2,2). Similarly (2,1,1) > (2,2). And (1,2,1),(2,2) are incomparable minimal elements. – user21820 May 09 '17 at 15:12
  • @user21820 thank you for formalising the question. This partial order formality, I feel, should be part of the problem statement to avoid any confusion. It certainly avoids the confusion that arises due to the term equivalent. – N. Shales May 09 '17 at 16:02

2 Answers2

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I guess what you are trying to ask is:

"In how many way you can express $n\ge 1$ as ordered sum of $1$ and $2$ so that there are no two consecutive ones?"

Solution: Suppose that $n$ is even, i.e., $n=2k$. Then we can place at most $k$ times the number $2$. So, given $i \in \{0,\ldots,k\}$, we miss to place (if possible) $n-2i$ times the number $1$ in the $i+1$ possible positions. Hence the total number is $$ 1+\binom{k}{2}+\binom{k-1}{4}+\cdots=\sum_{j\ge 0}\binom{k+1-j}{2j}. $$

With a similar argument, if $n$ is odd, i.e., $n=2k+1$, we obtain $$ (k+1)+\binom{k}{3}+\binom{k-1}{5}+\cdots=\sum_{j\ge 0}{\binom{k+1-j}{2j+1}}. $$

Paolo Leonetti
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    This is my interpretation of the question too. – TonyK May 08 '17 at 15:48
  • This is the correct interpretation. Sorry if it was vague. The set ${1,1,1}$ would be a triple and therefore not partitioned into pairs and singles and should not be enumerated. – Robert Frost May 08 '17 at 16:19
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A composition of $n$ can either be a composition of $n-2$ followed by a $2$ or a composition of $n-3$ followed by $2,1$. This gives the recurrence $f(n)=f(n-2)+f(n-3)$ with $f(1)=1, f(2)=1, f(3)=2.$ We find it, offset, in OEIS A000931 and it begins $1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329, 4410, 5842, 7739, 10252, 13581, 17991, 23833, 31572, 41824, 55405, 73396, 97229, 128801, 170625$

Ross Millikan
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  • Does this agree with Paolo's interpretation of the question? I've clarified that no sequence of three $1$'s is permitted. Sorry if it was vague. – Robert Frost May 08 '17 at 16:22
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    Yes, it agrees. In particular, you can find explicitely the $n$-term of the sequence as a convex combination of the $n$-th powers of the roots of the polynomial $x^3-x-1$. – Paolo Leonetti May 08 '17 at 16:29
  • @PaoloLeonetti we're looking at the Padovan sequence aren't we? – Robert Frost May 08 '17 at 16:34
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    This is using the same interpretation as @Paolo Leonetti. In fact, I found it first by essentially his approach, then computed some terms and looked it up in OEIS. That gave me the recurrence and I recognized how it arose. Yes, this is the Padovan sequence. There are many combinatorial interpretations in the OEIS entry. I didn't see this one. – Ross Millikan May 08 '17 at 16:39
  • @RossMillikan I like both answers... but can only accept 1 :( – Robert Frost May 08 '17 at 19:16
  • No worries. I think the two answers complement each other. I have plenty of points not to be concerned. – Ross Millikan May 08 '17 at 19:34