By substituting 0, 1, t, or −t into G(α, β, γ, δ; z), we get some interesting cases.
(1) α = 1, β = 1, γ = 1, δ = 1
Setting α = β = γ = δ = 1 in Equation (11), we have G(1, 1, 1, 1; z) = 1 −√
1 − 4z
2z =X
n≥0
Cnzn,
where Cn are the Catalan numbers. The first few terms are
1 + z + 2z2+ 5z3 + 14z4+ 42z5+ 132z6+ 429z7+ · · · .
(2) α = t, β = 1, γ = 1, δ = 1
The generating function of the enumerating polynomials for statistic edes is
G(t, 1, 1, 1; z) =X
n≥0
X
ω∈D(n)
tedes(ω)zn= 1 + (1 − t)z −p1 − 2(1 + t)z + (−3 + 2t + t2)z2
2z . (16)
The first few terms are
1 + z + (1 + t)z2+ (2 + 2t + t2)z3+ (4 + 6t + 3t2+ t3)z4+ (9 + 16t + 12t2+ 4t3+ t4)z5 +(21+45t+40t2+20t3+5t4+t5)z6+(51+126t+135t2+80t3+30t4+6t5+t6)z7+· · · . Let Tn,k = [tkzn]G(t, 1, 1, 1; z). Then Tn,k is the number of Dyck paths of semilength n with k even valleys. It is also the number of Dyck paths of semilength n with k peaks at even height (see [9, A091869]).
n \ k 0 1 2 3 4 5 6
0 1
1 1
2 1 1
3 2 2 1
4 4 6 3 1
5 9 16 12 4 1
6 21 45 40 20 5 1
7 51 126 135 80 30 6 1
In the case of t = 0 in Equation (16), we get the generating function for the number of Dyck paths of semilength n with no even valleys:
G(0, 1, 1, 1; z) = 1 + z −√
1 − 2z − 3z2
2z = 1 +X
n≥1
Mn−1zn,
where Mn are the Motzkin numbers.
Since nl(π) = edes(φ(π)) for all π ∈ N C(n), we have X
n≥0
X
π∈N C(n)
tnl(π)zn = 1 + (1 − t)z −p1 − 2(1 + t)z + (−3 + 2t + t2)z2
2z .
(3) α = 1, β = t, γ = 1, δ = 1
The generating function of the enumerating polynomials for statistic odes is G(1, t, 1, 1; z) =X
n≥0
X
ω∈D(n)
todes(ω)zn
= 1 − (1 − t)z −p1 − 2(1 + t)z + (−3 + 2t + t2)z2
2z[t + (1 − t)z] . (17)
The first few terms are
1+z+2z2+(4+t)z3+(9+4t+t2)z4+(21+15t+5t2+t3)z5+(51+50t+24t2+6t3+t4)z6 +(127 + 161t + 98t2+ 35t3+ 7t4+ t5)z7+ · · · .
Let Tn,k = [tkzn]G(1, t, 1, 1; z). Then Tn,k is the number of Dyck paths of semilength n with k odd valleys. It is also the number of Dyck paths of semilength n with k uuu’s (triple rise) steps (see [9, A092107]).
n \ k 0 1 2 3 4 5
0 1
1 1
2 2
3 4 1
4 9 4 1
5 21 15 5 1
6 51 50 24 6 1
7 127 161 98 35 7 1
In the case of t = 0 in Equation (17), we get the generating function for the number of Dyck paths of semilength n with no odd valleys:
G(1, 0, 1, 1; z) = 1 − z −√
1 − 2z − 3z2
2z2 =X
n≥0
Mnzn.
Since nir(π) = odes(φ(π)) for all π ∈ N C(n), we have X
n≥0
X
π∈N C(n)
tnir(π)zn= 1 − (1 − t)z −p1 − 2(1 + t)z + (−3 + 2t + t2)z2
2z[t + (1 − t)z] .
(4) α = 1, β = 1, γ = t, δ = 1
The generating function of the enumerating polynomials for statistic easc is
G(1, 1, t, 1; z) =X
The generating function of the enumerating polynomials for statistic oasc is
G(1, 1, 1, t; z) =X
The first few terms are
1 + tz + (1 + t2)z2+ (1 + 3t + t3)z3+ (3 + 4t + 6t2+ t4)z4+ (6 + 15t + 10t2+ 10t3+ t5)z5 +(15+36t+45t2+20t3+15t4+t6)z6+(36+105t+126t2+105t3+35t4+21t5+t7)z7+· · · . Let Tn,k = [tkzn]G(1, 1, 1, t; z). Then Tn,k is the number of Dyck paths of semilength n with k odd peaks (see [9, A091867]).
n \ k 0 1 2 3 4 5 6 7
In the case of t = 0 in Equation (18), we get the generating function for the number of Dyck paths of semilength n with no odd peaks:
G(1, 1, 1, 0; z) = 1 + z −√
1 − 2z − 3z2
2z(1 + z) =X
n≥0
Rnzn, (19)
where Rn are the Riordan numbers or ring numbers (see [9, A005043]).
The first few terms of Riordan numbers {Rn} for n ≥ 0 are 1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, · · · .
If we add each two consecutive Riordan numbers, we will get the Motzkin numbers.
That is, Mn= Rn+ Rn+1 for n ≥ 0.
Since nsg(π) = oasc(φ(π)) for all π ∈ N C(n), we have X
n≥0
X
π∈N C(n)
tnsg(π)zn= 1 + (1 − t)z −p1 − 2(1 + t)z + (−3 + 2t + t2)z2
2z[1 + (1 − t)z] .
Note that Equation (19) is the generating function for the number of non-crossing partitions of [n] with no singletons as well.
(6) α = t, β = t, γ = 1, δ = 1
The generating function of the enumerating polynomials for statistic des is
G(t, t, 1, 1; z) =
1 − (1 − t)2z2− q
1 − 2tz + (1 − t)2z22
− 4z2
2tz1 + (1 − t)z .
The first few terms are 1 + z + (1 + t)z2+ (1 + 3t + t2)z3+ (1 + 6t + 6t2+ t3)z4 + (1 + 10t + 20t2+ 10t3 + t4)z5+ (1 + 15t + 50t2+ 50t3+ 15t4+ t5)z6+ (1 + 21t + 105t2+ 175t3+ 105t4+ 21t5+ t6)z7+ · · · .
Let Tn,k = [tkzn]G(t, t, 1, 1; z). Then Tn,k is the number of Dyck paths of similength n with k valleys. It is the Narayana number Nn,k (see [9, A001263]).
n \ k 0 1 2 3 4 5 6
0 1
1 1
2 1 1
3 1 3 1
4 1 6 6 1
5 1 10 20 10 1
6 1 15 50 50 15 1
7 1 21 105 175 105 21 1
(7) α = 1, β = 1, γ = t, δ = t
The generating function of the enumerating polynomials for statistic asc is
G(1, 1, t, t; z) = 1 + (1 − t)z2
− q
1 − 2tz + (1 − t)2z22
− 4z2
2z1 + (1 − t)z .
The first few terms are 1 + tz + (t + t2)z2+ (t + 3t2+ t3)z3+ (t + 6t2+ 6t3+ t4)z4 + (t + 10t2+ 20t3 + 10t4+ t5)z5+ (t + 15t2 + 50t3+ 50t4+ 15t5+ t6)z6+ (t + 21t2 + 105t3+ 175t4+ 105t5+ 21t6+ t7)z7+ · · · .
Let Tn,k = [tkzn]G(1, 1, t, t; z). Then Tn,k is the number of Dyck paths of semilength n with k peaks (see [9, A090181]). It is the Narayana number Nn,k−1 for 1 ≤ k ≤ n.
n \ k 0 1 2 3 4 5 6 7
0 1
1 0 1
2 0 1 1
3 0 1 3 1
4 0 1 6 6 1
5 0 1 10 20 10 1
6 0 1 15 50 50 15 1
7 0 1 21 105 175 105 21 1
Note that G(1, 1, t, t; z) = t
G(t, t, 1, 1; z) − 1
+ 1 because asc(ω) = des(ω) + 1 for ω ∈ D(n), n ≥ 1.
(8) α = t, β = 1, γ = t, δ = 1
Setting α = γ = t, β = δ = 1 in Equation (11), we have
G(t, 1, t, 1; z) = 1 + 2(1 − t)z − 2zp(1 − 2tz)2− 4z2
2z .
The first few terms are 1 + z + 2tz2+ (1 + 4t2)z3+ (6t + 8t3)z4+ (2 + 24t2+ 16t4)z5 + (20t + 80t3+ 32t5)z6+ (5 + 120t2+ 240t4+ 64t6)z7 + · · · .
Let Tn,k = [tkzn]G(t, 1, t, 1; z). Then Tn,k is the number of Dyck paths of similength n with a total of k even peaks and even valleys (see [9, A121448]).
n \ k 0 1 2 3 4 5 6
0 1
1 1
2 0 2
3 1 0 4
4 0 6 0 8
5 2 0 24 0 16
6 0 20 0 80 0 32
7 5 0 120 0 240 0 64
(9) α = 1, β = t, γ = 1, δ = t
Setting α = γ = 1, β = δ = t in Equation (11), we have G(1, t, 1, t; z) = 1 −p(1 − 2tz)2 − 4z2
2zt + (1 − t2)z .
The first few terms are 1 + tz + (1 + t2)z2 + (4t + t3)z3+ (2 + 11t2+ t4)z4+ (15t + 26t3+ t5)z5+ (5 + 69t2+ 57t4 + t6)z6 + (56t + 252t3+ 120t5+ t7)z7+ · · · . Let Tn,k = [tkzn]G(1, t, 1, t; z). Then Tn,k is the number of Dyck paths of similength n with a total of k odd peaks and odd valleys. It is also the number of 2-Motzkin paths of length n with k horizontal steps having no wavy horizontal steps on the x-axis (see [9, A126222]).
n \ k 0 1 2 3 4 5 6 7
0 1
1 0 1
2 1 0 1
3 0 4 0 1
4 2 0 11 0 1
5 0 15 0 26 0 1
6 5 0 69 0 57 0 1
7 0 56 0 252 0 120 0 1
(10) α = t, β = −t, γ = 1, δ = 1
Setting α = t, β = −t, γ = δ = 1 in Equation (11), we have
G(t, −t, 1, 1; z) = 1 − 2tz + (t2− 1)z2− q
1 + (1 − t2)z2)2
− 4z2 2tz − 1 + (1 + t)z .
The first few terms are 1 + z + (1 + t)z2+ (1 + t + t2)z3+ (1 + 2t + 2t2+ t3)z4+ (1 + 2t + 4t2+2t3+t4)z5+(1+3t+6t2+6t3+3t4+t5)z6+(1+3t+9t2+9t3+9t4+3t5+t6)z7+· · · .
Let Tn,k = [tkzn]G(t, −t, 1, 1; z). Then Tn,k for n ≥ 1 is the number of symmetric Dyck paths of semilength n with k valleys (see [9, A088855]).
n\k 0 1 2 3 4 5 6
0 1
1 1
2 1 1
3 1 1 1
4 1 2 2 1
5 1 2 4 2 1
6 1 3 6 6 3 1
7 1 3 9 9 9 3 1
(11) α = 1, β = −t, γ = 1, δ = t
Setting α = γ = 1, β = −t, δ = t in Equation (11), we have G(1, −t, 1, t; z) = 1 − 2tz −√
1 − 4z2 2z − t + (1 + t2)z .
The first few terms are 1+tz+(1+t2)z2+(2t+t3)z3+(2+3t2+t4)z4+(5t+4t3+t5)z5+ (5 + 9t2+ 5t4+ t6)z6+ (14t + 14t3+ 6t5+ t7)z7+ · · · . Let Tn,k= [tkzn]G(1, −t, 1, t; z).
Then Tn,kis the number of Motzkin paths of length n with k horizontal steps having no horizontal steps at positive heights (see [9, A053121]).
n \ k 0 1 2 3 4 5 6 7
0 1
1 0 1
2 1 0 1
3 0 2 0 1
4 2 0 3 0 1
5 0 5 0 4 0 1
6 5 0 9 0 5 0 1
7 0 14 0 14 0 6 0 1
4 Bijection between 2-Motzkin paths and Dyck paths
Natural q-analogues of the Catalan numbers and the Narayana numbers are given by
F¨urlinger and Hofbauer [8] proved that X
In this section, we would like to find some similar results for the statistics emaj, omaj, ecomaj, and ocomaj. To this end, we encode emaj (omaj, ecomaj, ocomaj, respectively) and edes (odes, easc, oasc, respectively) by q and t, and write down the enumerating polynomials according to these statistics by
H1(n; q, t) := X
Using the first return decomposition, we get the following relations.
Lemma 4.1. Let ω ∈ D(n), n ≥ 1, and ω = uµdν be the first return decomposition
Proof. Let µ+ = uµd ∈ D(k). Since the locations of even valleys of µ+ are the locations of odd valleys of µ plus one, we have emaj(µ+) = omaj(µ) + odes(µ). And there is an even valley at the location 2k if ν 6= ∅. Also, the locations of even valleys of ν plus 2k are the locations of even valleys of ω, thus we complete the proof of Equation (20). The others can be proved similarly.
Using Lemma 4.1 and 3.1, we have the following lemma.
Lemma 4.2. For n ≥ 1, we have
Proof. Here we will only prove Equation (24) and (25), and the others can be proved similarly. Using Equation (20) and (7), we have
H1(n; q, t)
qomaj(µ)+odes(µ)+2k+emaj(ν)+2k edes(ν)todes(µ)+1+edes(ν)
+ X
µ∈D(n−1)
qomaj(µ)+odes(µ)todes(µ)
=
Using Equation (21) and (8), we have
H2(n; q, t) =
qemaj(µ)+edes(µ)+omaj(ν)+2k odes(ν)tedes(µ)+odes(ν)
=
However, we cannot get the generating function of H2(n; q, t) from the above equa-tion. H1(n; q, t), H3(n; q, t), and H4(n; q, t) are also in a similar situation. So we need other tools.
Here we establish a bijection between 2-Motzkin paths of length n − 1 and Dyck paths of semilength n to achieve our purpose (see [5, 6]).
Define ψ : T M(n − 1) −→ D(n) where σ = σ1σ2· · · σn−1 ∈ T M(n − 1) is mapping
We take n = 3 as an example to illustrate the bijection ψ in Figure 10.
Let σ = σ1σ2· · · σn−1 ∈ T M(n − 1). We define Str(σ) and Wav(σ) as follows:
Str(σ) = {i: σi = s}, and Wav(σ) = {i: σi = w}.
And we define the number of and the sum of the elements of them by wav(σ) := | Wav(σ)|, mwav(σ) := X
According to the bijection ψ, we have the following relations between 2-Motzkin paths and Dyck paths.
σ ∈ T M(n − 1) ω = ψ(σ) ∈ D(n)
i ∈ Str(σ) ←→ 2i ∈ eAsc(ω) (28)
i ∈ Wav(σ) ←→ 2i ∈ eDes(ω) (29)
Now we are going to discuss the statistics emaj and ecomaj using the bijection ψ.
2-Motzkin paths
Figure 10: Bijection between T M(2) and D(3).
4.1 The statistics emaj and ecomaj
In this subsection, we compute the enumerating polynomials H1(n; q, t) and H3(n; q, t).
We will use the q-Binomial Theorem (see, for example [1]).
Theorem 4.3. (The q-Binomial Theorem) For n ≥ 0,
Using Theorem 4.3 and relations (28) and (29) between 2-Motzkin paths and Dyck paths, we get the following results.
Theorem 4.4. For n ≥ 1, we have
remaining n − 1 − k steps form a Motzkin path of length n − k − 1. So the number
And by the q-Binomial Theorem, P
S⊆[n−1]
Since, in T M(n − 1), steps s and steps w are interchangeable. Hence, we have the same result for the statistics mstr(σ) and ecomaj(ω).
Theorem 4.5. For n ≥ 1, we have