Catalan numbers and numbers of subsequence












1












$begingroup$


Given a set of numbers {$1,2,...,n$}, I want to find the number of permutations without 3-term subsequence of '1-3-2' pattern



Now this would be $A_n = n! - U_n$, where $A_n$ is the number of acceptable permutations and $U_n$ is the number of unacceptable permutations.



I have a vague idea that I should be using the Catalan numbers to consider cases of $U_n$. I could think of this as the parentheses problem where "(" is the low number, "A" is the high number, and ")" is the high number but I am unsure this works.



Would this be a valid argument? Any ideas?






combinatorics catalan-numbers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Given a set of numbers {$1,2,...,n$}, I want to find the number of permutations without 3-term subsequence of '1-3-2' pattern



    Now this would be $A_n = n! - U_n$, where $A_n$ is the number of acceptable permutations and $U_n$ is the number of unacceptable permutations.



    I have a vague idea that I should be using the Catalan numbers to consider cases of $U_n$. I could think of this as the parentheses problem where "(" is the low number, "A" is the high number, and ")" is the high number but I am unsure this works.



    Would this be a valid argument? Any ideas?






    combinatorics catalan-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Given a set of numbers {$1,2,...,n$}, I want to find the number of permutations without 3-term subsequence of '1-3-2' pattern



      Now this would be $A_n = n! - U_n$, where $A_n$ is the number of acceptable permutations and $U_n$ is the number of unacceptable permutations.



      I have a vague idea that I should be using the Catalan numbers to consider cases of $U_n$. I could think of this as the parentheses problem where "(" is the low number, "A" is the high number, and ")" is the high number but I am unsure this works.



      Would this be a valid argument? Any ideas?






      combinatorics catalan-numbers







      share|cite|improve this question











      $endgroup$




      Given a set of numbers {$1,2,...,n$}, I want to find the number of permutations without 3-term subsequence of '1-3-2' pattern



      Now this would be $A_n = n! - U_n$, where $A_n$ is the number of acceptable permutations and $U_n$ is the number of unacceptable permutations.



      I have a vague idea that I should be using the Catalan numbers to consider cases of $U_n$. I could think of this as the parentheses problem where "(" is the low number, "A" is the high number, and ")" is the high number but I am unsure this works.



      Would this be a valid argument? Any ideas?






      combinatorics catalan-numbers




      combinatorics catalan-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 25 at 9:51







      Corp. and Ltd.

















      asked Jan 25 at 4:59









      Corp. and Ltd.Corp. and Ltd.

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      16914






















          1 Answer
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          $begingroup$

          This is a classical problem in the theorem of "permutation avoidance".
          I'm not sure what you mean by "acceptable" permutations; ones which have
          no $132$ patterns or ones which have a $132$ pattern. I'll let $C_n$
          be the permutations with no $132$ patterns; clearly $C_n=n!$ for $nle2$.



          For a general permutation on ${1,ldots,n}$, thought of as a word,
          to be $132$-avoiding it must have the form $w,n,w'$ for some $k$ where
          $w'$ is a $132$-avoiding word on ${1,ldots,k}$ and
          $w'$ is a $132$-avoiding word on ${k+1,ldots,n-1}$. For a given $k$
          the number of such words is $C_kC_{n-k-1}$ and so we get the recurrence
          $$C_n=sum_{k=0}^{n-1}C_kC_{n-k-1}$$
          which may be familiar.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @Corp.andLtd. In the jargon of permutation patterns, $243$ is a $132$ pattern. There's an order-isomorphism between ${2,3,4}$ and ${1,2,3}$ taking the first to the second.
            $endgroup$
            – Lord Shark the Unknown
            Jan 25 at 8:15













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          $begingroup$

          This is a classical problem in the theorem of "permutation avoidance".
          I'm not sure what you mean by "acceptable" permutations; ones which have
          no $132$ patterns or ones which have a $132$ pattern. I'll let $C_n$
          be the permutations with no $132$ patterns; clearly $C_n=n!$ for $nle2$.



          For a general permutation on ${1,ldots,n}$, thought of as a word,
          to be $132$-avoiding it must have the form $w,n,w'$ for some $k$ where
          $w'$ is a $132$-avoiding word on ${1,ldots,k}$ and
          $w'$ is a $132$-avoiding word on ${k+1,ldots,n-1}$. For a given $k$
          the number of such words is $C_kC_{n-k-1}$ and so we get the recurrence
          $$C_n=sum_{k=0}^{n-1}C_kC_{n-k-1}$$
          which may be familiar.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @Corp.andLtd. In the jargon of permutation patterns, $243$ is a $132$ pattern. There's an order-isomorphism between ${2,3,4}$ and ${1,2,3}$ taking the first to the second.
            $endgroup$
            – Lord Shark the Unknown
            Jan 25 at 8:15


















          1












          $begingroup$

          This is a classical problem in the theorem of "permutation avoidance".
          I'm not sure what you mean by "acceptable" permutations; ones which have
          no $132$ patterns or ones which have a $132$ pattern. I'll let $C_n$
          be the permutations with no $132$ patterns; clearly $C_n=n!$ for $nle2$.



          For a general permutation on ${1,ldots,n}$, thought of as a word,
          to be $132$-avoiding it must have the form $w,n,w'$ for some $k$ where
          $w'$ is a $132$-avoiding word on ${1,ldots,k}$ and
          $w'$ is a $132$-avoiding word on ${k+1,ldots,n-1}$. For a given $k$
          the number of such words is $C_kC_{n-k-1}$ and so we get the recurrence
          $$C_n=sum_{k=0}^{n-1}C_kC_{n-k-1}$$
          which may be familiar.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @Corp.andLtd. In the jargon of permutation patterns, $243$ is a $132$ pattern. There's an order-isomorphism between ${2,3,4}$ and ${1,2,3}$ taking the first to the second.
            $endgroup$
            – Lord Shark the Unknown
            Jan 25 at 8:15
















          1












          1








          1





          $begingroup$

          This is a classical problem in the theorem of "permutation avoidance".
          I'm not sure what you mean by "acceptable" permutations; ones which have
          no $132$ patterns or ones which have a $132$ pattern. I'll let $C_n$
          be the permutations with no $132$ patterns; clearly $C_n=n!$ for $nle2$.



          For a general permutation on ${1,ldots,n}$, thought of as a word,
          to be $132$-avoiding it must have the form $w,n,w'$ for some $k$ where
          $w'$ is a $132$-avoiding word on ${1,ldots,k}$ and
          $w'$ is a $132$-avoiding word on ${k+1,ldots,n-1}$. For a given $k$
          the number of such words is $C_kC_{n-k-1}$ and so we get the recurrence
          $$C_n=sum_{k=0}^{n-1}C_kC_{n-k-1}$$
          which may be familiar.






          share|cite|improve this answer











          $endgroup$



          This is a classical problem in the theorem of "permutation avoidance".
          I'm not sure what you mean by "acceptable" permutations; ones which have
          no $132$ patterns or ones which have a $132$ pattern. I'll let $C_n$
          be the permutations with no $132$ patterns; clearly $C_n=n!$ for $nle2$.



          For a general permutation on ${1,ldots,n}$, thought of as a word,
          to be $132$-avoiding it must have the form $w,n,w'$ for some $k$ where
          $w'$ is a $132$-avoiding word on ${1,ldots,k}$ and
          $w'$ is a $132$-avoiding word on ${k+1,ldots,n-1}$. For a given $k$
          the number of such words is $C_kC_{n-k-1}$ and so we get the recurrence
          $$C_n=sum_{k=0}^{n-1}C_kC_{n-k-1}$$
          which may be familiar.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 25 at 8:18

























          answered Jan 25 at 5:13









          Lord Shark the UnknownLord Shark the Unknown

          106k1161133




          106k1161133








          • 1




            $begingroup$
            @Corp.andLtd. In the jargon of permutation patterns, $243$ is a $132$ pattern. There's an order-isomorphism between ${2,3,4}$ and ${1,2,3}$ taking the first to the second.
            $endgroup$
            – Lord Shark the Unknown
            Jan 25 at 8:15
















          • 1




            $begingroup$
            @Corp.andLtd. In the jargon of permutation patterns, $243$ is a $132$ pattern. There's an order-isomorphism between ${2,3,4}$ and ${1,2,3}$ taking the first to the second.
            $endgroup$
            – Lord Shark the Unknown
            Jan 25 at 8:15










          1




          1




          $begingroup$
          @Corp.andLtd. In the jargon of permutation patterns, $243$ is a $132$ pattern. There's an order-isomorphism between ${2,3,4}$ and ${1,2,3}$ taking the first to the second.
          $endgroup$
          – Lord Shark the Unknown
          Jan 25 at 8:15






          $begingroup$
          @Corp.andLtd. In the jargon of permutation patterns, $243$ is a $132$ pattern. There's an order-isomorphism between ${2,3,4}$ and ${1,2,3}$ taking the first to the second.
          $endgroup$
          – Lord Shark the Unknown
          Jan 25 at 8:15




















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