SDR question: What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?












0












$begingroup$


Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:



$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$



$(2)$ For every $1leq ileq n$, we have $a_i < b_i$



$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$



What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?






combinatorics







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  • $begingroup$
    What have you tried so far? How far have you gotten?
    $endgroup$
    – saulspatz
    Dec 3 '18 at 18:27
















0












$begingroup$


Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:



$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$



$(2)$ For every $1leq ileq n$, we have $a_i < b_i$



$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$



What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    What have you tried so far? How far have you gotten?
    $endgroup$
    – saulspatz
    Dec 3 '18 at 18:27














0












0








0





$begingroup$


Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:



$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$



$(2)$ For every $1leq ileq n$, we have $a_i < b_i$



$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$



What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?






combinatorics







share|cite|improve this question









$endgroup$




Consider elements of $Z_{2n}$, that is, numbers $0,1,2,dots, 2n-1$. Let
$mathcal P_n={ P_1,P_2,dots, P_m}$ be a collection of unordered sets, where each $P_k$ consists of $n$ ordered pairs $(a_i,b_i)$ with the following properties:



$(1)$ $A={ a_i| 1leq ileq n}, B={ b_i| 1leq ileq n}$, and
$Acup B={0,1,2,dots,2n-1}$



$(2)$ For every $1leq ileq n$, we have $a_i < b_i$



$(3)$ For any $1leq i,jleq n, ineq j$ we have $a_i < a_j < b_i$ if and only if $a_i < b_j < b_i$



What is the number $m$ of different sets $P_k$ in $mathcal P_n$ for a given $n$?






combinatorics




combinatorics






share|cite|improve this question













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asked Dec 3 '18 at 17:52









DummKorfDummKorf

335




335












  • $begingroup$
    What have you tried so far? How far have you gotten?
    $endgroup$
    – saulspatz
    Dec 3 '18 at 18:27


















  • $begingroup$
    What have you tried so far? How far have you gotten?
    $endgroup$
    – saulspatz
    Dec 3 '18 at 18:27
















$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27




$begingroup$
What have you tried so far? How far have you gotten?
$endgroup$
– saulspatz
Dec 3 '18 at 18:27










2 Answers
2






active

oldest

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1












$begingroup$

Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$



Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    With any $P_kin mathcal P_n$, we associate a legal sequence of $n$ open and $n$ closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
    Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$



    Footnotes:
    1. https://en.wikipedia.org/wiki/Catalan_number






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

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      votes









      1












      $begingroup$

      Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$



      Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$



        Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$



          Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.






          share|cite|improve this answer









          $endgroup$



          Let $C_n$ be the number of such sets for a given $n$. Clearly, $C_0=C_1=1.$ To establish a recurrence, we ask which number can be paired with $0.$ Suppose $(0,b)$ is one of the pairs. Then in every other pair either both numbers are greater than $b$, or both are less than $b$. That is, $b$ must be odd. Further, the pairings of the numbers from $1$ to $b-1$ and the pairings of the numbers from $b+1$ to $2n-1$ must both obey rule $(3).$



          Therefore, $$E_n=sum_{k=0}^{n-1}E_kE_{n-1-k}$$ and we recognize the recurrence for the Catalan numbers. (look at the "First Proof" on the wiki page.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 19:07









          saulspatzsaulspatz

          15.5k31331




          15.5k31331























              0












              $begingroup$

              With any $P_kin mathcal P_n$, we associate a legal sequence of $n$ open and $n$ closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
              Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$



              Footnotes:
              1. https://en.wikipedia.org/wiki/Catalan_number






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                With any $P_kin mathcal P_n$, we associate a legal sequence of $n$ open and $n$ closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
                Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$



                Footnotes:
                1. https://en.wikipedia.org/wiki/Catalan_number






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  With any $P_kin mathcal P_n$, we associate a legal sequence of $n$ open and $n$ closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
                  Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$



                  Footnotes:
                  1. https://en.wikipedia.org/wiki/Catalan_number






                  share|cite|improve this answer











                  $endgroup$



                  With any $P_kin mathcal P_n$, we associate a legal sequence of $n$ open and $n$ closed parentheses by putting an open parentheses at position $a_i$ and a closed one at position $b_i$. Conversely, given any legal sequence of length $2n$, we can pair the positions of corresponding parentheses and the set of such pairs gives us an element of $mathcal P_n$.
                  Therefore, $$mid mathcal P_nmid= C_n=frac{1}{n+1}binom{2n}{n}$$



                  Footnotes:
                  1. https://en.wikipedia.org/wiki/Catalan_number







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 28 at 14:04

























                  answered Dec 3 '18 at 19:35









                  Anubhab GhosalAnubhab Ghosal

                  1,20919




                  1,20919






























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