Citation: Feng Qi, Bai-Ni Guo. Several explicit and recursive formulas for generalized Motzkin numbers[J]. AIMS Mathematics, 2020, 5(2): 1333-1345. doi: 10.3934/math.2020091
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The Motzkin numbers Mn enumerate various combinatorial objects. In 1977, Donaghey and Shapiro [3] gave fourteen different manifestations of the Motzkin numbers Mn. In particular, the Motzkin numbers Mn give the numbers of paths from (0,0) to (n,0) which never dip below the x-axis y=0 and are made up only of the steps (1,0), (1,1), and (1,−1).
The first seven Motzkin numbers Mn for 0≤n≤6 are 1,1,2,4,9,21,51. All the Motzkin numbers Mn can be generated by
M(x)=1−x−√1−2x−3x22x2=11−x+√1−2x−3x2=∞∑k=0Mkxk. |
In 2007, Mansour et al [12] introduced the (u,l,d)-Motzkin numbers m(u,l,d)n and obtained [12,Theorem 2.1] that m(u,l,d)n=m(1,l,ud)n,
Mu,l,d(x)=1−lx−√(1−lx)2−4udx22udx2=∞∑n=0m(u,l,d)nxn, | (1.1) |
and
m(u,l,d)n=lnn/2∑j=01j+1(2jj)(n2j)(udl2)j. | (1.2) |
From (1.1) and (1.2), it is easy to see that m(u,l,d)n=m(d,l,u)n.
In 2014, Sun [42] generalized the Motzkin numbers Mn to
Mn(a,b)=⌊n/2⌋∑k=0(n2k)Ckan−2kbk | (1.3) |
for a,b∈N in terms of the Catalan numbers
Cn=1n+1(2nn) | (1.4) |
and established the generating function
Ma,b(x)=1−ax−√(1−ax)2−4bx22bx2=11−ax+√(1−ax)2−4bx2=∞∑k=0Mk(a,b)xk, | (1.5) |
where ⌊λ⌋ denotes the floor function defined by the largest integer less than or equal to λ∈R. Wang and Zhang pointed out in [43] that
Mn(1,1)=Mn,Mn(2,1)=Cn+1,andMn(3,1)=Hn, | (1.6) |
where Hn denote the restricted hexagonal numbers described by Harary and Read [4].
For more information on many results, applications, and generalizations of the Motzkin numbers Mn, please refer to the papers [3,9,10,42,43] and closely related references therein. For more information on many results, applications, and generalizations of the Catalan numbers Cn, please refer to the monograph [5], the newly published papers [11,17,19,26,27,31,36,37,38,40,41], the survey articles [25,29], and closely related references therein.
Comparing (1.1) with (1.5) reveals that Mk(a,b) and m(u,l,d)k are equivalent to each other and satisfy
Mk(a,b)=m(1,a,b)n=m(b,a,1)kandm(u,l,d)k=Mk(l,ud). | (1.7) |
Therefore, it suffices to consider generalized Motzkin numbers Mk(a,b), rather than the (u,l,d)-Motzkin numbers m(u,l,d)n, in this paper.
By the second relation in (1.7), one can reformulated the formula (1.2) as
Mn(a,b)=an⌊n/2⌋∑j=01j+1(2jj)(n2j)(ba2)j. | (1.8) |
Substituting (1.4) into (1.3) recovers (1.8) once again.
In 2015, Wang and Zhang [43,Theorem 1] combinatorially obtained, among other things, the recursive formula
Mn+2(a,b)=aMn+1(a,b)+bn∑ℓ=0Mℓ(a,b)Mn−ℓ(a,b),n≥0. | (1.9) |
It is not difficult to see that the function (1−ax)2−4bx2=(a2−4b)x2−2ax+1 is nonnegative if and only if
1. either b=0 and x∈R,
2. or a2−4b=0, a≠0, and x≤12a,
3. or a2−4b>0, b<0, and x∈R,
4. or a2−4b>0, b>0, and x≥1a−2√b or x≤1a+2√b.
Consequently, the generating function Ma,b(x) defined by (1.5) in [42] is defined for either b≤0 or a≥2√b>0.
In this paper, we will find two explicit formulas, which are different from (1.8), and recover the recursive formula (1.9) for generalized Motzkin numbers Mn(a,b). Consequently, we will derive two explicit formula and a recursive formula for the Motzkin numbers Mn, the Catalan numbers Cn, and the restricted hexagonal numbers Hn respectively.
We can state our main results as the following three theorems.
Theorem 1. For n≥0, we can compute generalized Motzkin numbers Mn(a,b) by
Mn(a,b)=12b(4b−a22a)n+2n+2∑ℓ=0(2a24b−a2)ℓ(2ℓ−3)!!ℓ!(ℓn−ℓ+2), | (1.10) |
where (pq)=0 for q>p≥0 and the double factorial of negative odd integers −(2n+1) is
[−(2n+1)]!!=(−1)n(2n−1)!!=(−1)n2nn!(2n)!,n=0,1,…. |
Consequently, we can compute the Motzkin numbers Mn and the restricted hexagonal numbers Hn respectively by
Mn=98(32)nn+2∑ℓ=0(23)ℓ(2ℓ−3)!!ℓ!(ℓn−ℓ+2) | (1.11) |
and
Hn=(−1)n2572(56)nn+2∑ℓ=0(−1)ℓ(185)ℓ(2ℓ−3)!!ℓ!(ℓn−ℓ+2). | (1.12) |
Theorem 2. For n≥0, we can compute generalized Motzkin numbers Mn(a,b) by
Mn(a,b)=−(a−2√b)n+22bn+2∑ℓ=0(2ℓ−3)!!(2ℓ)!![2(n−ℓ+2)−3]!![2(n−ℓ+2)]!!(a+2√ba−2√b)ℓ. | (1.13) |
Consequently, we can compute the Motzkin numbers Mn and the restricted hexagonal numbers Hn respectively by
Mn=(−1)n+12n+2∑ℓ=0(−1)ℓ3ℓ(2ℓ−3)!!(2ℓ)!![2(n−ℓ+2)−3]!![2(n−ℓ+2)]!! |
and
Hn=−12n+2∑ℓ=05ℓ(2ℓ−3)!!(2ℓ)!![2(n−ℓ+2)−3]!![2(n−ℓ+2)]!!. |
Theorem 3. For n≥0, generalized Motzkin numbers Mn(a,b) satisfy
M0(a,b)=1,M1(a,b)=a, | (1.14) |
and the recursive formula (1.9). Consequently, for n≥0, the Motzkin numbers Mn, the Catalan numbers Cn, and the restricted hexagonal numbers Hn meet the recursive formulas
Mn+2=Mn+1+n∑ℓ=0MℓMn−ℓ, | (1.15) |
Cn+2=2Cn+1+n∑ℓ=0CℓCn−ℓ, | (1.16) |
and
Hn+2=3Hn+1+n∑ℓ=0HℓHn−ℓ | (1.17) |
respectively.
In order to prove the explicit formula (1.10), we need the following lemmas.
Lemma 1 ([1,p. 40,Exercise 5)], [16,Section 2.2,p. 849], [22,p. 94], [34,Lemma 3], and [44,Lemma 2.1]). Let u(x) and v(x)≠0 be two differentiable functions. Let U(n+1)×1(x) be an (n+1)×1 matrix whose elements uk,1(x)=u(k−1)(x) for 1≤k≤n+1, let V(n+1)×n(x) be an (n+1)×n matrix whose elements
vi,j(x)={(i−1j−1)v(i−j)(x),i−j≥00,i−j<0 |
for 1≤i≤n+1 and 1≤j≤n, and let |W(n+1)×(n+1)(x)| denote the determinant of the (n+1)×(n+1) matrix
W(n+1)×(n+1)(x)=(U(n+1)×1(x)V(n+1)×n(x)). |
Then the nth derivative of the ratio u(x)v(x) can be computed by
dndxn[u(x)v(x)]=(−1)n|W(n+1)×(n+1)(x)|vn+1(x). |
Lemma 2 ([2,p. 134,Theorem A and p. 139,Theorem C]). The Faà di Bruno formula can be described in terms of the Bell polynomials of the second kind
Bn,k(x1,x2,…,xn−k+1)=∑1≤i≤n−k+1ℓi∈{0}∪N∑n−k+1i=1iℓi=n∑n−k+1i=1ℓi=kn!∏n−k+1i=1ℓi!n−k+1∏i=1(xii!)ℓi |
for n≥k≥0 by
dndtn[f∘h(t)]=n∑k=0f(k)(h(t))Bn,k(h′(t),h″(t),…,h(n−k+1)(t)) | (2.1) |
for n≥0.
Lemma 3 ([2,p. 135]). The Bell polynomials of the second kind Bn,k satisfy
Bn,k(abx1,ab2x2,…,abn−k+1xn−k+1)=akbnBn,k(x1,x2,…,xn−k+1) | (2.2) |
for n≥k≥0.
Lemma 4. For n≥k≥0, we have
Bn,k(x,1,0,…,0)=(n−k)!2n−k(nk)(kn−k)x2k−n. | (2.3) |
More generally, for n≥k≥0 and λ,α∈C, we have
Bn,k(1,1−λ,(1−λ)(1−2λ),…,n−k∏ℓ=0(1−ℓλ))=(−1)kk!k∑ℓ=0(−1)ℓ(kℓ)n−1∏q=0(ℓ−qλ) | (2.4) |
or, equivalently,
Bn,k(⟨α⟩1,⟨α⟩2,…,⟨α⟩n−k+1)=(−1)kk!k∑ℓ=0(−1)ℓ(kℓ)⟨αℓ⟩n. | (2.5) |
Proof. The formula (2.3) can be found in [24,Theorem 5.1], [35,p. 7,(19)], [39,Section 3], and [44,Lemma 2.5]. The explicit formula (2.4) was first established in [30,Remark 1] and then was applied in [18,Section 2], [20,First proof of Theorem 2], [21,Lemma 2.2], [24,Remark 6.1], [28,Lemma 4], and [32,Lemma 2.6]. The formula (2.5) and the equivalence were presented in [33,Theorems 2.1 and 4.1].
Remark 1. In recent years, there have been some literature such as [6,7,8,13,14,15,23,24,30,35,45,46,47,48] devoting to deep investigation and extensive applications of the Bell polynomials of the second kind Bn,k(x1,x2,…,xn−k+1). Specially, in the papers [13,14], the generalized Dyck paths (namely, various type of Motzkin paths) and the Bell polynomials were connected closely.
We are now in a position to prove our main results.
Proof of Theorem 1. By virtue of (2.1), (2.2), and (2.3), we obtain for k≥0 that
[√(1−ax)2−4bx2](k+2)=k+2∑ℓ=0⟨12⟩ℓ[(1−ax)2−4bx2]1/2−ℓ×Bk+2,ℓ(−2[a+(4b−a2)x],2(a2−4b),0,…,0)→k+2∑ℓ=0⟨12⟩ℓBk+2,ℓ(−2a,2(a2−4b),0,…,0)=k+2∑ℓ=0⟨12⟩ℓ[2(a2−4b)]ℓBk+2,ℓ(a4b−a2,1,0,…,0)=k+2∑ℓ=0⟨12⟩ℓ[2(a2−4b)]ℓ(k−ℓ+2)!2k−ℓ+2(k+2ℓ)(ℓk−ℓ+2)(a4b−a2)2ℓ−k−2 | (3.1) |
as x→0, where
⟨x⟩n={x(x−1)⋯(x−n+1),n≥11,n=0 |
denotes the falling factorial of x∈R.
Letting u(x)=1−ax−√(1−ax)2−4bx2 and v(x)=x2 in Lemma 1 gives
dnMa,b(x)dxn=12b(−1)nx2(n+1)|u(x)(00)x20⋯000u′(x)2(10)x(11)x2⋯000u″(x)2(20)2(21)x⋯000u(3)(x)02(31)⋯000u(4)(x)00⋯000⋯⋯⋯⋱⋯⋯⋯u(n−2)(x)00⋯2(n−2n−3)x(n−2n−2)x20u(n−1)(x)00⋯2(n−1n−3)2(n−1n−2)x(n−1n−1)x2u(n)(x)00⋯02(nn−2)2(nn−1)x|=12b(−1)nx2(n+1)[(−1)nu(n)(x)|(00)x200⋯0002(10)x(11)x20⋯0002(20)2(21)x(22)x2⋯00002(31)2(32)x⋯000002(42)⋯000⋯⋯⋯⋱⋯⋯⋯000⋯2(n−2n−3)x(n−2n−2)x20000⋯2(n−1n−3)2(n−1n−2)x(n−1n−1)x2|+2(nn−1)x|u(x)(00)x2000⋯00u′(x)2(10)x(11)x200⋯00u″(x)2(20)2(21)x(22)x20⋯00u(3)(x)02(31)2(32)x(33)x2⋯00u(4)(x)002(42)2(43)x⋯00⋯⋯⋯⋯⋯⋱⋯⋯u(n−2)(x)0000⋯2(n−2n−3)x(n−2n−2)x2u(n−1)(x)0000⋯2(n−1n−3)2(n−1n−2)x|−2(nn−2)(n−1n−1)x2|u(x)(00)x200⋯00u′(x)2(10)x(11)x20⋯00u″(x)2(20)2(21)x(22)x2⋯00u(3)(x)02(31)2(32)x⋯00u(4)(x)002(42)⋯00⋯⋯⋯⋯⋱⋯⋯u(n−3)(x)000⋯2(n−3n−4)x(n−3n−3)x2u(n−2)(x)000⋯2(n−2n−4)2(n−2n−3)x|]=12bu(n)(x)x2−2nx12b(−1)n−1x2n|u(x)(00)x20⋯00u′(x)2(10)x(11)x2⋯00u″(x)2(20)2(21)x⋯00u(3)(x)02(31)⋯00u(4)(x)00⋯00⋯⋯⋯⋱⋯⋯u(n−2)(x)00⋯2(n−2n−3)x(n−2n−2)x2u(n−1)(x)00⋯2(n−1n−3)2(n−1n−2)x|−n(n−1)1x212b(−1)n−2x2(n−1)|u(x)(00)x200⋯00u′(x)2(10)x(11)x20⋯00u″(x)2(20)2(21)x(22)x2⋯00u(3)(x)02(31)2(32)x⋯00u(4)(x)002(42)⋯00⋯⋯⋯⋯⋱⋯⋯u(n−3)(x)000⋯2(n−3n−4)x(n−3n−3)x2u(n−2)(x)000⋯2(n−2n−4)2(n−2n−3)x|=12bu(n)(x)x2−2nxdn−1Ma,b(x)dxn−1−n(n−1)x2dn−2Ma,b(x)dxn−2=1x2[u(n)(x)2b−2nxdn−1Ma,b(x)dxn−1−n(n−1)dn−2Ma,b(x)dxn−2]. |
Therefore, by L'Hôspital's rule, we have
limx→0dnMa,b(x)dxn=limx→0{1x2[u(n)(x)2b−2nxdn−1Ma,b(x)dxn−1−n(n−1)dn−2Ma,b(x)dxn−2]}=limx→0{12x[u(n+1)(x)2b−2nxdnMa,b(x)dxn−n(n+1)dn−1Ma,b(x)dxn−1]}=12limx→0[u(n+2)(x)2b−2nxdn+1Ma,b(x)dxn+1−n(n+3)dnMa,b(x)dxn]=12[limx→0u(n+2)(x)2b−n(n+3)limx→0dnMa,b(x)dxn] |
which is equivalent to
limx→0dnMa,b(x)dxn=1(n+1)(n+2)limx→0u(n+2)(x)2b=12b(n+1)(n+2)limx→0u(n+2)(x). |
Considering
limx→0dnMa,b(x)dxn=n!Mn(a,b), |
making use of (3.1), and simplifying lead to the explicit formula (1.10).
Letting (a,b)=(1,1) and (a,b)=(3,1) respectively in (1.10) and considering the three relations in (1.6) derive (1.11) and (1.12) immediately. The proof of Theorem 1 is complete.
Proof of Theorem 2. From (1.5), it is derived that
√(1−ax)2−4bx2=1−ax−2b∞∑k=0Mk(a,b)xk+2. |
This implies that
Mk(a,b)=−12b1(k+2)!limx→0[√(1−ax)2−4bx2](k+2),k≥0. | (3.2) |
It is easy to see that
1. when a2−4b>0 and x≤min{1a+2√b,1a−2√b}=1a+2√b, we have
[√(1−ax)2−4bx2](k+2)=[√(a2−4b)(x−1a+2√b)(x−1a−2√b)](k+2)=√a2−4b(√1a+2√b−x√1a−2√b−x)(k+2)=√a2−4bk+2∑ℓ=0(k+2ℓ)(√1a+2√b−x)(ℓ)(√1a−2√b−x)(k−ℓ+2)=(−1)k√a2−4bk+2∑ℓ=0(k+2ℓ)⟨12⟩ℓ(1a+2√b−x)1/2−ℓ⟨12⟩k−ℓ+2(1a−2√b−x)ℓ−k−3/2→(−1)k√a2−4bk+2∑ℓ=0(k+2ℓ)⟨12⟩ℓ(1a+2√b)1/2−ℓ⟨12⟩k−ℓ+2(1a−2√b)ℓ−k−3/2=(k+2)!(a−2√b)k+2k+2∑ℓ=0(2ℓ−3)!!(2ℓ)!![2(k−ℓ+2)−3]!![2(k−ℓ+2)]!!(a+2√ba−2√b)ℓ |
as x→0;
2. when a2−4b<0 and
1a+2√b=max{1a+2√b,1a−2√b}>x>min{1a+2√b,1a−2√b}=1a−2√b, |
we have
[√(1−ax)2−4bx2](k+2)=[√(4b−a2)(1a+2√b−x)(x−1a−2√b)](k+2)=√4b−a2(√1a+2√b−x√x−1a−2√b)(k+2)=√4b−a2k+2∑ℓ=0(k+2ℓ)(√1a+2√b−x)(ℓ)(√x−1a−2√b)(k−ℓ+2)=√4b−a2k+2∑ℓ=0(k+2ℓ)(−1)ℓ⟨12⟩ℓ(1a+2√b−x)1/2−ℓ⟨12⟩k−ℓ+2(x−1a−2√b)1/2−(k−ℓ+2)→√4b−a2k+2∑ℓ=0(k+2ℓ)(−1)ℓ⟨12⟩ℓ(1a+2√b)1/2−ℓ⟨12⟩k−ℓ+2(12√b−a)1/2−(k−ℓ+2)=(2√b−a)k+1√4b−a2k+2∑ℓ=0(k+2ℓ)(2ℓ−3)!!2ℓ(a+2√b2√b−a)ℓ−1/2(−1)k−ℓ[2(k−ℓ+2)−3]!!2k−ℓ+2=(2√b−a)k+2k+2∑ℓ=0(k+2ℓ)(2ℓ−3)!!2ℓ(a+2√b2√b−a)ℓ(−1)k−ℓ[2(k−ℓ+2)−3]!!2k−ℓ+2=(a−2√b)k+2k+2∑ℓ=0(k+2ℓ)(2ℓ−3)!!2ℓ[2(k−ℓ+2)−3]!!2k−ℓ+2(a+2√ba−2√b)ℓ=(k+2)!(a−2√b)k+2k+2∑ℓ=0(2ℓ−3)!!(2ℓ)!![2(k−ℓ+2)−3]!![2(k−ℓ+2)]!!(a+2√ba−2√b)ℓ |
as x→0.
By virtue of (3.2), we obtain the formula (1.13) readily.
Letting (a,b)=(1,1) and (a,b)=(3,1) respectively in (1.13) and making use of the first and third relations in (1.6) lead to (1.11) and (1.12) immediately. The proof of Theorem 2 is complete.
Proof of Theorem 3. From (1.5), it is derived that
√(1−ax)2−4bx2=1−ax−2b∞∑k=0Mk(a,b)xk+2. |
Squaring on both sides of the above equation gives
(1−ax)2−4bx2=1−2ax+(a2−4b)x2=[1−ax−2b∞∑k=0Mk(a,b)xk+2]2=1+a2x2+4b2[∞∑k=0Mk(a,b)xk+2]2−2ax−4b∞∑k=0Mk(a,b)xk+2+4abx∞∑k=0Mk(a,b)xk+2=1−2ax+a2x2+4b2x4∞∑k=0[k∑ℓ=0Mℓ(a,b)Mk−ℓ]xk−4b∞∑k=2Mk−2(a,b)xk+4ab∞∑k=3Mk−3(a,b)xk=1−2ax+a2x2−4b∞∑k=2Mk−2(a,b)xk+4ab∞∑k=3Mk−3(a,b)xk+4b2∞∑k=4[k−4∑ℓ=0Mℓ(a,b)Mk−ℓ−4(a,b)]xk=1−2ax+a2x2−4b[M0(a,b)x2+M1(a,b)x3]+4abM0(a,b)x3−4b∞∑k=4Mk−2(a,b)xk+4ab∞∑k=4Mk−3(a,b)xk+4b2∞∑k=4[k−4∑ℓ=0Mℓ(a,b)Mk−ℓ−4(a,b)]xk=1−2ax+[a2−4bM0(a,b)]x2+4b[aM0(a,b)−M1(a,b)]x3−4b∞∑k=4[Mk−2(a,b)−aMk−3(a,b)−bk−4∑ℓ=0Mℓ(a,b)Mk−ℓ−4(a,b)]xk |
which means that
a2−4b=a2−4bM0(a,b),4b[aM0(a,b)−M1(a,b)]=0, |
and
Mk−2(a,b)−aMk−3(a,b)−bk−4∑ℓ=0Mℓ(a,b)Mk−ℓ−4(a,b)=0,k≥4. |
Consequently, the identities in (1.14) and the recursive formula (1.9) follow.
Taking (a,b)=(1,1), (a,b)=(2,1), and (a,b)=(3,1) respectively in (1.9) and considering the three relations in (1.6) lead to (1.15), (1.16), and (1.17) immediately. The proof of Theorem 3 is complete.
Remark 2. From the proof of Theorem 1, we can conclude that
x2dnMa,b(x)dxn+2nxdn−1Ma,b(x)dxn−1+n(n−1)dn−2Ma,b(x)dxn−2=u(n)(x)2b,n≥2. |
This implies that the generating function Ma,b(x) expressed in (1.5) is an explicit solution of the linear ordinary differential equations
x2f(n)(x)+2nxf(n−1)(x)+n(n−1)f(n−2)(x)=Fn;a,b(x) |
for all n≥2, where, by (2.2) and (2.3) or (2.4),
Fn;a,b(x)=n!(4b−a2)n2n+1b√(1−ax)2−4bx2[a+(4b−a2)x]nn∑ℓ=12ℓ(2ℓ−3)!!ℓ!(4b−a2)ℓ(ℓn−ℓ)[a+(4b−a2)x]2ℓ[(1−ax)2−4bx2]ℓ. |
Remark 3. This paper is a continuation of the article [49] and a revised version of the preprint [28].
The authors declare that they have no conflict of interest in this paper.
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