Mixing
Catalan numbers and numbers of subsequence
$begingroup$
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
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
$endgroup$
add a comment |
$begingroup$
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
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
$endgroup$
add a comment |
$begingroup$
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
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
$endgroup$
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
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
edited Jan 25 at 9:51
Corp. and Ltd.
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
asked Jan 25 at 4:59
Corp. and Ltd.Corp. and Ltd.
16914
16914
add a comment |
add a comment |
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.
answered Jan 25 at 5:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
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
add a comment |
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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.
answered Jan 25 at 5:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jan 25 at 5:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jan 25 at 5:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
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.
answered Jan 25 at 5:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
edited Jan 25 at 8:18
answered Jan 25 at 5:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 25 at 5:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
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
add a comment |
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
add a comment |
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