4 Connections on rs and area
4.2 Area of a 2-Motzkin path
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4.2 Area of a 2-Motzkin path
A Motzkin path of length n is a path starting on (1, 0) and ending on (n + 1, 0) but never going below the x-axis, with possible steps u = (1, 1), d = (1, −1), and s = (1, 0). The set of all Motzkin paths of length n is denoted by Motz(n).
A 2-Motzkin path of length n is a path starting on (1, 0) and ending on (n + 1, 0) but never going below the x-axis, with possible steps u = (1, 1), d = (1, −1), and (1, 0), where the level step (1, 0) can be either of two kinds: straight (s) and zigzag (z). The set of all 2-Motzkin paths of length n is denoted by 2-Motz(n).
Let λistand for the i-th step in a path m which is comprised of u, d, s, and z. We denote U (m) = {j | λj = u, λj ∈ m}, D(m) = {j | λj = d, λj ∈ m}, S(m) = {j | λj = s, λj ∈ m}, and Z(m) = {j | λj = z, λj ∈ m}. Note that if m ∈ Motz(n), then Z(m) must be empty.
For an m ∈ 2-Motz(n), let area(m) denote the total area of the regions bounded by the path of m and the x-axis. Note that it is not the same as the area of Dyck paths.
Define a map φ4 from N C(n) to 2-Motz(n − 1) as in [8]. Given π ∈ N C(n), we convert each interval [j, j + 1], 1 6 j < n with λj ∈ {u, d, s, z} as follows:
λj =
u, if j and j + 1 are in different blocks and j is not the largest element in its block;
d, if j and j + 1 are in different blocks and j + 1 is not the smallest element in its block;
s, if j and j + 1 are in the same block;
z, if j and j + 1 are in different blocks, j is the largest element in its block, and j + 1 is the smallest element in its block.
Then let φ4(π) = λ1λ2. . . λn−1. About φ4, Simion and Ullman gave an observation.
Observation 4.1. [8, Observation 2.4] If m = λ1λ2. . . λn−1 is the path associated with π ∈ N C(n) as above, then |U (m)| = |D(m)| and |U (λ1λ2. . . λi)| ≥ |D(λ1λ2. . . λi)| for all i = 1, 2, . . . , n − 1 (in other words, there is a complete parenthesization of the u’s and the d’s).
For π ∈ N C(n), φ4(π) will never go below the x-axis and end on the x-axis by Observation 4.1. Hence φ4(π) ∈ 2-Motz(n − 1).
For the inverse mapping, given m = λ1λ2. . . λn−1 ∈ 2-Motz(n − 1), first we plot 1, 2, ..., n on a horizontal line with a left parenthesis “(” on the left of 1 and a right parenthesis “)” on the right of n. For each λi = u (resp. d), we put a left parenthesis
“(” (resp. a right parenthesis “)”) between i and i + 1. For each λi = z, we put a right parenthesis and a left parenthesis “)(” between i and i + 1. For each λi = s, we do nothing. Since m ∈ 2-Motz(n − 1), the u’s and d’s are completely parenthesized. In addition, inserting “)(” inside a complete parenthesization of pairs of parentheses does not change the completion. So, we now have a complete parenthesization of pairs of parentheses on [n]. Then, we can deduce it as a partition of [n], φ−14 (m), by creating a block of φ−14 (m) for each of the consecutive integers inside the lowest level parenthesis pairs (i.e., parenthesis which pair each other and enclose no others), and removing these lowest level parenthesis pairs and all the integers they enclose. Repeat this procedure until all parentheses have been removed. Clearly the partition φ−14 (m) is non-crossing.
Hence we have the following proposition.
Proposition 4.5. φ4 is bijective.
Example 4.2. Let π = 1 6 7 / 2 3 / 4 5 / 8 14 / 9 / 10 / 11 13 / 12 ∈ N C(14), then φ4(π) = uszsdszuzzudd.
φ4(π) = 1
2 3 4 5
6 7 8
9 10 11
12 13
14 Example 4.3. Let m = uudszudzd ∈ 2-Motz(9), then we have (1(2(3)4 5)(6(7)8)(9)10).
Firstly, create blocks {3}, {7}, and {9} and remove (3), (7), and (9), then we have (1(2 4 5)(6 8)10). Secondly, create blocks {2, 4, 5} and {6, 8} and remove (2 4 5) and (6 8), then we have (1 10). Finally, create a block {1, 10} and remove (1 10), then we get φ−14 (m) = 1 10 / 2 4 5 / 3 / 6 8 / 7 / 9.
Lemma 4.6. Let π ∈ N C(n) and m = φ4(π) = λ1λ2. . . λn−1 ∈ 2-Motz(n − 1). For integers 1 ≤ a < b ≤ n − 1, λa and λb are a pair of matched u and d in m if and only if a and b + 1 are successive elements in the same block of π.
Proof. Let π ∈ N C(n) and m = φ4(π) = λ1λ2. . . λn−1 ∈ 2-Motz(n − 1). If λa and λb are a pair of matched u and d in m, we can get the result easily by operations of φ−14 .
For the converse, let a and b + 1 be successive elements in the same block of π, we can derive that λa = u and λb = d immediately. Since a and b + 1 are successive elements in the same block, the blocks of π in {a + 1, . . . , b} are non-crossing. So there is a complete parenthesization of the u’s and the d’s in λa+1. . . λb−1. Hence λa and λb are matched in m.
Proposition 4.7. Let π ∈ N C(n) and m = φ4(π) ∈ 2-Motz(n − 1), then area(m) = rs(π).
Proof. Given m = λ1λ2. . . λn−1 ∈2-Motz(n − 1), we denote the points on m by (i, y(i))
Remark that m ∈ 2-Motz(n − 1), hence the u’s and d’s in m are completely parenthesized.
For integer i on the line y = y(i), it means that there are y(i) matched pairs of λa = u and λb = d such that a < i and b ≥ i. Translate to non-crossing partition, a and b + 1 are successive elements in the same block of π by Lemma 4.6. So there are y(i) blocks with their first elements less than i and their last elements greater than i. That is y(i) is equal to #{wj | wj < wi, j > i}. Hence
Example 4.4 (continued). We have rs(π) = 0+1+1+1+1+0+0+0+1+1+1+2+1+0 = 10. And area(φ4(π)) = 0.5 + 1 + 1 + 1 + 0.5 + 0 + 0 + 0.5 + 1 + 1 + 1.5 + 1.5 + 0.5 = 10. The rank function for the lattice of non-crossing partitions of [n] is well understood.
For instance, the Narayana numbers Nn,n−k−1 = 1n nk n
k+1 count the number of non-crossing partitions of [n] with rank k. One of the more interesting results about this function is the following identity [5].
X
π∈N C(n)
xrank(π)= X
0≤j≤(n−1)2
γn−1,jxj(1 + x)n−1−2j, (10)
where the γn−1,j are nonnegative integers. (In fact, γn−1,j = X
m∈Motz(n−1)
|U (m)|=j
1 =n − 1 2j
Cj.)
Now we consider the joint distribution of rs and rank on non-crossing partitions.
Proposition 4.7 links N C(n) and 2-Motz(n − 1) by the statistics area and rs. In [1, 8], they proved that for a non-crossing partition π, rank(π) = |U (φ4(π))| + |S(φ4(π))|. With those properties, we can prove a result which is similar to Equation (10).
There is an onto mapping ρ from 2-Motz(n) to Motz(n) such that U (ρ(m)) = U (m), D(ρ(m)) = D(m), and S(ρ(m)) = S(m)S Z(m) for m ∈ 2-Motz(n).
Theorem 4.8. We have X
π∈N C(n)
qrs(π)xrank(π)= X
0≤j≤bn−12 c
γn−1,j(q)xj(1 + x)n−1−2j,
where γn−1,j(q) = X
m∈Motz(n−1)
|U (m)|=j
qarea(m).
Proof. By proposition 4.7, we have X
π∈N C(n)
qrs(π)xrank(π)= X
m∈2-Motz(n−1)
qarea(m)x|U (m)|+|S(m)|
= X
m∈2-Motz(n−1)
qarea(m)x|U (m)|x|S(m)|1|Z(m)|
= X
m∈Motz(n−1)
qarea(m)x|U (m)|(1 + x)|S(m)|
= X
m∈Motz(n−1)
qarea(m)x|U (m)|(1 + x)n−1−2|U (m)|
= X
0≤j≤bn−12 c
γn−1,j(q)xj(1 + x)n−1−2j.
5 Conclusions
The following table describes the connections proved in Section 2 and 4 between non-crossing partitions and Dyck paths.
Bijection maj(p) area(p)
φ1
2rb(π) + rs(π) 2lb(π) + 2ls(π) − rs(π) (1) φ2
lb(π) + 2ls(π) 2rb(π) + lb(π) (2)
φ3 (3) 2rs(π) + n − bk(π)
Of course we are interested in the following question.
Problem 5.1. Can (1), (2), and (3) be expressed by rb, ls, rs, and lb on non-crossing partitions?
In fact, we can express (1) by other statistics on non-crossing partitions.
Proposition 5.2. Let π ∈ N C(n, k) with f (π) = (f1, f2, . . . , fk), and l0(π) = (l01, l02, . . . , l0k). p = φ1(π). First, we construct k isosceles right triangles with the length of the leg of the i-th triangles is (fi+1− fi)√
2 for i = 1, 2, . . . , k, where fk+1 = n + 1. Secondly, we put these triangles on the lattice one by one such that the right angle is at the top and the hypotenuse is on the x-axis. Note that the i-th triangle covers fi+12−fi diamonds.
Thirdly, we put a rectangle with area (l0i− fi+1+ 1)√
2 × (fi+1− fi)√
2 on the top right of the right leg of the i-th triangle for i = 1, 2, . . . , k.
Summing the diamonds of all the covers, then we get area(p) = i-th triangle (resp. rectangle) by Ti (resp. Ri).
p =
In Section 2, we have deduced that X
Also, by Equation (1) we have X
Problem 5.3. Are there any other combinations of rb, ls, rs, and lb on non-crossing partitions with closed-form?
We are also interested in the next problem.
Problem 5.4. Are there any formulas for
X
In Theorem 4.8, we proved the equidistribution result X
π∈N C(n)
qrs(π)xrank(π) = X
m∈2-Motz(n−1)
qarea(m)x|U (m)|+|S(m)|.
Problem 5.5. Can we find a statistic st on 2-Motz(n − 1) such that
X
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