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• 201825-8

Inspiration of The Pascal Triangle

(Yidong Sun)

A Joint Work with Ma Fei and Ma Luping College of Science of Dalian Maritime University

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• Contents
• Pascal Triangle
• The Well-known Motivation
• The First Starting Point and Problems
• The Second Starting Point and Problems
• The Third Starting Point and Problems
• New Discovery
• References
• Thanks

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• 1. The Pascal Triangle:

In mathematics, Pascal’s triangle is a triangular array of the binomial coeffi-

• cients. In much of the Western world, it is named after French mathematician

Blaise Pascal (1623-1662), although other mathematicians studied it centuries before him in India, Iran, China, Germany, and Italy. n/k 0 1 2 3 4 5 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1 Table 1. The Pascal triangle P for n and k up to 5.

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• Pascal’s triangle was known in China in the early 11th century through the work
• f the Chinese mathematician Jia Xian (1010-1070). In the 13th century, Yang

Hui (1238-1298) presented the triangle and hence it is still called Yang Hui’s triangle in China. Figure 1. Yang Hui’s triangle P for n and k up to 9.

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• 2. The Well-known Motivation:

(1) Gaussian binomial coefficient

n k

• q

= [n]! [k]![n − k]! = (1 − qn)(1 − qn−1) · · · (1 − qn−k+1) (1 − qk)(1 − qk−1) · · · (1 − q) (x + y)n =

n

• k=0

n k

• xkyn−k,

xy = yx, (x + y)n =

n

• k=0

n k

• q

xkyn−k, qxy = yx.

• Vector Spaces on Finite Field,
• The Theory of Partitions,
• The Theory of Q-series.

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• (2) Binomial Inversion Formula

An =

n

• k=0

n k

• Bk ⇔ Bn =

n

• k=0

(−1)n−k n k

• Ak.
• Gould-Hsu Inversion Formula,
• Krattenthaler Inversion Formula,
• Ma Xinrong Inversion Formula.

(3) Congruence Properties of Binomial Coefficients

• Lucas Congruence,
• Wolstenholme Congruence,
• Binomial Congruence.

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• (4) The Theory of Riordan Array

P =

• 1

1 − t, t 1 − t

• A Riordan array R = (Rn,k)n≥k≥0 is an infinite lower triangular matrix with

nonzero entries on the main diagonal, such that Rn,k = [tn]g(t)(f(t))k for n ≥ k, namely, Rn,k equals the coefficient of tn in the expansion of the se- ries g(t)(f(t))k, where g(t) = 1 + g1t + g2t2 + · · · and f(t) = f1t + f2t2 + · · · with f1 = 0 are two formal power series.               R0,0 R1,0 R1,1 R2,0 R2,1 R2,2 R3,0 R3,1 R3,2 R3,3 · · · · · · · · · · · · ... g(t) g(t)f(t) g(t)f(t)2 g(t)f(t)3 · · ·              

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• (5) Fibonomial Coefficient Analog

n k

• F

= n!F k!F(n − k)!F where n!F = F1F2 · · · Fn with F0 = 0, F1 = 1 and Fn = Fn−1 + Fn−2.

(6) Fractal

Figure 2. Sierpinski triangle.

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• (7) Unimodality of Binomial Coefficients

n

• ≤

n 1

• ≤ · · · ≤

n [n/2]

• ≥
• n

[n/2] − 1

• ≥ · · · ≥

n n

• Log-Concavity (a2

k ≥ ak−1ak+1), Log-Convexity (a2 k ≤ ak−1ak+1),

• q-Log-Concavity, q-Log-Convexity,
• Infinite Log-Concavity, Infinite Log-Convexity.

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• 2. The First Starting Points and Problems:

2.1 The Rule of David Star

The Star of David rule, originally stated by Gould in 1972, is given by n k n + 1 k − 1 n + 2 k + 1

• =
• n

k − 1 n + 1 k + 1 n + 2 k

• ,

for any k and n, or equivalently which implies that

• n

k + 1 n + 1 k n + 2 k + 2

• =

n k n + 1 k + 2 n + 2 k + 1

• .

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• In 2003, the author observed in his Master dissertation that if multiplying the

above two identities and dividing by n(n + 1)(n + 2), one can arrive at Nn,k+1Nn+1,kNn+2,k+2 = Nn,kNn+1,k+2Nn+2,k+1, where Nn,k = 1

n

n

k

n

k−1

• is the Narayana number.

n/k 0 1 2 3 4 5 1 1 1 1 2 1 3 1 3 1 6 6 1 4 1 10 20 10 1 5 1 15 50 50 15 1 Table 2.1. The Narayana triangle N for n and k up to 5.

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• In the summer of 2006, the author asked Mansour for a combinatorial proof of

the above Narayana identity to be found. Later, by Chen’s bijective algorithm for trees, Li and Mansour provided a combinatorial proof of a general identity Nn,k+m−1Nn+1,k+m−2Nn+2,k+m−3 · · · Nn+m−2,k+1Nn+m−1,kNn+m,k+m = Nn,kNn+1,k+mNn+2,k+m−1 · · · Nn+m−2,k+3Nn+m−1,k+2Nn+m,k+1.

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• This motivates the author to reconsider the Star of David rule and to propose

a new concept called SDR-matrix which obeys the generalized rule of David star. Definition 0.1 Let A =

• An,k
• n≥k≥0 be an infinite lower triangular matrix, for

any given integer m ≥ 3, if there hold

r

• i=0

An+i,k+r−i

p−r−1

• i=0

An+p−i,k+r+i+1 =

r

• i=0

An+p−i,k+p−r+i

p−r−1

• i=0

An+i,k+p−r−i−1, for all 2 ≤ p ≤ m − 1 and 0 ≤ r ≤ p − 1, then A is called an SDR-matrix of

• rder m.

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• In order to give a more intuitive view on the definition, we present a pictorial

description of the generalized rule for the case m = 5.

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• SDRm: the set of SDR-matrices of order m;

SDR∞: the set of SDR-matrices A of order ∞, that is A ∈ SDRm for any m ≥ 3. By our notation, it is obvious that the Pascal triangle P = n

k

• n≥k≥0 and the

Narayana triangle N =

• Nn+1,k+1
• n≥k≥0 are SDR-matrices of order 3.

In fact, both of them will be proved to be SDR-matrices of order ∞.

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• 2.2 The Basic Properties of SDR-matrices

Lemma 0.1 For any A ∈ SDRm, B ∈ SDRm+i with i ≥ 0, there hold

A ◦ B ∈ SDRm, and A ◦(−1) ∈ SDRm if it exists, where A ◦ B is the

Hadamard product of A and B, A ◦(−1) is the Hadamard inverse of A . Lemma 0.2 For any A =

• An,k
• n≥k≥0 ∈ SDRm, then
• An+i,k+j
• n≥k≥0 ∈

SDRm for fixed i, j ≥ 0.

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• Lemma 0.3 Given any sequence (an)n≥0, let An,k = ak, Bn,k = bn−k and

Cn,k = cn for n ≥ k ≥ 0, then

• An,k
• n≥k≥0,
• Bn,k
• n≥k≥0,
• Cn,k
• n≥k≥0 ∈

SDR∞.            a0 a0 a1 a0 a1 a2 a0 a1 a2 a3 · · ·            ,            b0 b1 b0 b2 b1 b0 b3 b2 b1 b0 · · ·            ,            c0 c1 c1 c2 c2 c2 c3 c3 c3 c3 · · ·           

• The Pascal triangle P =

n

k

• 0≤k≤n ∈ SDR∞.
• The Narayana triangle N =

1

n+1

n+1

k

n+1

k+1

• 0≤k≤n ∈ SDR∞.
• The Lah triangle L =
• (n+1)!

(k+1)!

n

k

• 0≤k≤n ∈ SDR∞.

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• 2.3 The Main Results of SDR-matrices

Theorem 0.1 For any sequences (an)n≥0, (bn)n≥0 and (cn)n≥0 such that b0 = 1, an = 0 and cn = 0 for n ≥ 0, let A =

• akbn−kcn
• n≥k≥0, then A −1 ∈ SDR∞.

Conjecture 0.1 For any A ∈ SDRm, if the inverse A −1 of A exists, then

A −1 ∈ SDRm.

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• Theorem 0.2 For any sequences (an)n≥0, (bn)n≥0 with b0 = 1 and an = 0 for

n ≥ 0, let A =

• anbn−ka−1

k

• n≥k≥0, then the matrix power A j ∈ SDR∞ for

any integer j. Corollary 0.1 For P, N and L , then Pj, N j, L j ∈ SDR∞ for any integer j.

• In general, for A , B ∈ SDRm, their matrix product A B is possibly NOT

in SDRm.

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• Theorem 0.3 For any A =
• An,k
• n≥k≥0 with An,k = 0 for n ≥ k ≥ 0, then

A ∈ SDRm+1 if and only if A ∈ SDRm.

Remark 0.1 The condition An,k = 0 for n ≥ k ≥ 0 in Theorem 0.3 is neces-

• sary. The following example verifies this claim.

n+k

2 n−k 2

• n≥k≥0 =

                 1 0 1 1 0 1 0 2 1 1 0 3 0 1 0 3 4 0 1 · · ·                  ∈ SDR3, but not in SDR4.

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• Recall that the Narayana number Nn+1,k+1 can be represented as

Nn+1,k+1 =

1 n + 1 n + 1 k + 1 n + 1 k

• = det

  n

k

• n

k+1

• n+1

k

n+1

k+1

• 

 , so we can come up with the following definition. Definition 0.2 Let A =

• An,k
• n≥k≥0 be an infinite lower triangular matrix, for

any integer m ≥ 1, define A[m] =

• A[m]

n,k

• n≥k≥0, where

A[m]

n,k = det

      An,k · · · An,k+m−1 . . . · · · . . . An+m−1,k · · · An+m−1,k+m−1       .

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• Theorem 0.4 For any sequences (an)n≥0, (bn)n≥0 and (cn)n≥0 such that b0 = 1,

an = 0 and cn = 0 for n ≥ 0, let A =

• akbn−kcn
• n≥k≥0, then A[m] ∈ SDR∞

for any integer m ≥ 1. Let a−1

n = b−1 n = cn = n!, a−1 n = b−1 n = cn = n!(n+1)! or a−1 n = cn = n!(n+1)!

and b−1

n = n! for n ≥ 0 in Theorem 0.4, one has

Corollary 0.2 For P, N and L , then P[m], N[m], L[m] ∈ SDR∞ for any integer m ≥ 1. Conjecture 0.2 If A ∈ SDR∞, then A[m] ∈ SDR∞ for any integer j ≥ 1. Conjecture 0.3 If A =

• An,k
• n≥k≥0 ∈ SDR∞ with An,k = 0 for n ≥ k ≥ 0,

then there exist three sequences (an)n≥0, (bn)n≥0 and (cn)n≥0 with anbncn = 0 for n ≥ 0 such that An,k = akbn−kcn. Ref: Yidong Sun, The star of David rule, Linear Algebra and its Applications, 429 (8-9), (2008), 1954–1961.

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• 3. The Second Starting Points and Problems:

3.1 The Determinant Transformation of Pascal Triangle

• Catalan numbers Cn =

1 n+1

2n

n

• , [1, 1, 2, 5, 14, 42, 132, 429, ...] are the row

sums of the Narayana triangle N = (Nn+1,k+1)n≥k≥0, Cn+1 =

n

• k=0

Nn+1,k+1, where Nn+1,k+1 =

1 n+1

n+1

k

n+1

k+1

• .

Their close relations are reflected according to the two ways,

• 1

n + 1 n + 1 k n + 1 k + 1

• = det

  n

k

• n

k+1

• n+1

k

n+1

k+1

• 

,

• 1

n + 1 n + 1 k n + 1 k + 1

• = det

  n

k

• n+1

k+1

• n+1

k

n+2

k+1

• 

, where det(·) denotes the determinant of a square matrix.

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• There exists a close connection between the Pascal triangle P = (

n

k

• )n≥k≥0

and the Narayana triangle N = 1

n+1

n+1

k

n+1

k+1

• n≥k≥0,

n/k 0 1 2 3 4 5 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1

= ⇒

• 3 3

4 6

• n/k 0

1 2 3 4 5 row sums 1 1 1 1 1 2 2 1 3 1 5 3 1 6 6 1 14 4 1 10 20 10 1 42 5 1 15 50 50 15 1 132 Table 3.1. The Pascal triangle P and Narayana triangle N for n and k up to 5.

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• Shapiro’s Catalan triangle B = (Bn,k)n≥k≥0 with Bn,k = k+1

n+1

2n+2

n−k

• .

n/k 1 2 3 4 5 1 1 2 1 2 5 4 1 3 14 14 6 1 4 42 48 27 8 1 5 132 165 110 44 10 1 Table 3.2. The values of Bn,k for n and k up to 5.

• Let X = (Xn,k)n≥k≥0 be the infinite lower triangles defined on B by

Xn,k = det   Bn,k Bn,k+1 Bn+1,k Bn+1,k+1   .

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• The triangle deduced from Shapiro’s Catalan triangle. It indicates that the row

sums have close relation with the first column of the triangle B. n/k 1 2 3 4 row sums alternating row sums 1 1 = 12 1 1 3 1 4 = 22 2 2 14 10 1 25 = 52 5 3 84 90 21 1 196 = 142 14 4 594 825 308 36 1 1764 = 422 42 Table 3.3. The values of Xn,k for n and k up to 4, together with the row sums and alternating row sums.

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• Question:

Let A = (An,k)n≥k≥0 be an infinite lower triangular matrix with nonzero entries on the main diagonal. Given integers m, r, ℓ, p with m, ℓ, p ≥ 0, define a transformation on A by Ap =

• A(p)

n,k(m, r, ℓ)

• n≥k≥0, where

A(p)

n,k(m, r, ℓ) = det

        An,k An+r,k+ℓ · · · An+pr,k+pℓ An+m,k An+m+r,k+ℓ · · · An+m+pr,k+pℓ . . . . . . · · · . . . An+pm,k An+pm+r,k+ℓ · · · An+pm+pr,k+pℓ        

p+1

Then how to determine the explicit expression for the n-th row sum or alternat- ing row sums of Ap, S(p)

n,m,r,ℓ(Ap) = n

• k=0

A(p)

n,k(m, r, ℓ),

T (p)

n,m,r,ℓ(Ap) = n

• k=0

(−1)n−kA(p)

n,k(m, r, ℓ)?

• In the case p = 1, for some special infinite lower triangular matrices related to

weighted partial Motzkin paths, it can produce several surprising results.

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• 3.2. Weighted Partial Motzkin Paths

A Motzkin path is a lattice path from (0, 0) to (n, 0) in the XOY -plane that does not go below the X-axis and consists of (1) : up steps u = (1, 1), (2) : down steps d = (1, −1), and (3) : horizontal steps h = (1, 0).

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• A partial Motzkin path, also called a Motzkin path from (0, 0) to (n, k), is just

a Motzkin path but without the requirement of ending on the X-axis.

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• A weighted partial Motzkin path is a partial Motzkin path with the weight as-

signment that the all up steps and down steps are weighted by 1, the horizontal steps are endowed with a weight x if they are lying on X-axis, and endowed with a weight y if they are not lying on X-axis.

• The weight w(P) of a path P is the product of the weight of all its steps. The

above example indicates that w(P) = xy2.

• The weight of a set of paths is the sum of the total weights of all the paths.
• Another type of weighted partial Motzkin paths is used by Chen, Li, Shapiro

and Yan to derive many nice matrix identities related to a class of Riordan arrays.

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• Let Mn,k(x, y) denote the set of weighted partial Motzkin paths P from (0, 0)

to (n, k), according to the last step u, h or d of P, one can easily deduce the following recurrences for Mn,k(x, y), Mn,0(x, y) = xMn−1,0(x, y) + Mn−1,1(x, y), (n ≥ 1), Mn,k(x, y) = Mn−1,k−1(x, y) + yMn−1,k(x, y) + Mn−1,k+1(x, y), (n ≥ k ≥ 1), with M0,0(x, y) = 1 and Mn,k(x, y) = 0 if n < k or k < 0.

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• The first few entries of the triangle M = (Mn,k(x, y))n≥k≥0.

n/k 1 2 3 4 1 1 x 1 2 x2 + 1 x + y 1 3 x3 + 2x + y x2 + xy + y2 + 2 x + 2y 1 4 x4 + 3x2 + 2xy + y2 + 2 x3 + x2y + xy2 + 3x + y3 + 5y x2 + 2xy + 3y3 + 3 x + 3y 1 Table 4. The values of Mn,k(x, y) for n and k up to 4.

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• Denote Mk(x, y; t) =

n≥k Mn,k(x, y)tn to be the generating function of

weighted partial Motzkin paths ending at level k. Lemma 0.4 Let M = (Mn,k(x, y))n≥k≥0, then M = (M0(x, y; t), tM0(y, y; t)) is a Riordan array, i.e., Mk(x, y; t) = M0(x, y; t)(tM0(y, y; t))k, where M0(y, y; t) = 1 − yt −

• (1 − yt)2 − 4t2

2t2 , M0(x, y; t) = 1 1 − xt − t2M0(y, y; t) = 1 − 2xt + yt −

• (1 − yt)2 − 4t2

2(y − x)(1 − xt)t + 2t2 .

• We can NOT expect to obtain a simple and explicit expressions from M.

Mn,k(x, y) = [tn]M0(x, y; t)(tM0(y, y; t))k.

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• We give a short list in Table 5, where C(t) = 1−√1−4t

2t

and M(t) = 1−t−

√ 1−2t−3t2 2t2

are generating functions respectively for Catalan numbers Cn and Motzkin num- bers Mn. (x, y) M0(x, y; t) Sequences M (0, 0) C(t2) A126120 = Cn

2

A053121 (0, 1)

1+t− √ 1−2t−3t2 2t(1+t)

A005043 = Riordan numbers Rn A089942 (0, 2)

1−√1−4t 3−√1−4t

A000957 = Fine numbers Fn A126093 (1, 0)

1 1−t−t2C(t2)

A001405 =

• n

⌊n/2⌋

• A061554

(1, 1) M(t) A001006 = Motzkin numbers Mn A064189 (1, 2) C(t) A000108 = Catalan numbers Cn A039599 (2, 2) C2(t) A000108 = Catalan numbers Cn+1 A039598 (3, 2)

C(t) √1−4t

A001700 = 2n+1

n

• A111418

Table 5. The specializations of (x, y), where Cn

2 is set to be zero when n is odd.

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• 3.3. Main Results and Consequences

Theorem 0.5 Let M = (Mn,k(x, y))n≥k≥0, for any integers n, r ≥ 0 and m ≥ ℓ ≥ 0, set Nr = min{n + r + 1, m + r − ℓ}. Then there hold

Nr

• k=0

det   Mn,k(x, y) Mm,k+ℓ+1(x, y) Mn+r+1,k(x, y) Mm+r+1,k+ℓ+1(x, y)  =

r

• i=0

Mn+i,0(x, y)Mm+r−i,ℓ(y, y).(1)

• The cases r = 2, ℓ = 4 and m = n + 2, n or n − 2.

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• (Proof Sketch) Define

A (r)

n,m,k,ℓ(x, y) = {(P, Q)|P ∈ Mn,k(x, y), Q ∈ Mm+r+1,k+ℓ+1(x, y)},

B(r)

n,m,k,ℓ(x, y) = {(P, Q)|P ∈ Mn+r+1,k(x, y), Q ∈ Mm,k+ℓ+1(x, y)},

and C (r,i)

n,m,k,ℓ(x, y) = {(P, Q) ∈ A (r) n,m,k,ℓ(x, y)|Q = Q1UQ2} with Q1 ∈

Mi,k(x, y) and Q2 ∈ Mm+r−i,ℓ(y, y), where the U-step (along the path) is just

the (k + 1)-th R-visible up step of Q, which is also the last (ℓ + 1)-th R-visible up step of Q. Find a bijection between Nr

k=0 A (r) n,m,k,ℓ(x, y)/ r i=0

i

k=0 C (r,i) n,m,k,ℓ(x, y) and

Nr

k=0 B(r) n,m,k,ℓ(x, y).

In order to give a more intuitive view on the bijection φ, we present a pictorial description of φ for the case (P, Q) ∈ A (4)

10,12,2,2(x, y) − 4 i=2 C (4,i) 10,12,2,2(x, y) and

φ(P, Q) = (P ∗, Q∗) ∈ B(4)

10,12,1,2(x, y).

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• In order to give a more intuitive view on the bijection φ, we present a pictorial

description of φ for the case (P, Q) ∈ A (4)

10,12,2,2(x, y) − 4 i=2 C (4,i) 10,12,2,2(x, y) and

φ(P, Q) = (P ∗, Q∗) ∈ B(4)

10,12,1,2(x, y).

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• The special case r = 0 in (1) produces the following result.

Theorem 0.6 Let M = (Mn,k(x, y))n≥k≥0 be given in Section 2. For any inte- gers n ≥ 0 and m ≥ ℓ ≥ 0, set N0 = min{n + 1, m − ℓ}. Then there holds

N0

• k=0

det   Mn,k(x, y) Mm,k+ℓ+1(x, y) Mn+1,k(x, y) Mm+1,k+ℓ+1(x, y)   = Mn,0(x, y)Mm,ℓ(y, y). (2)

• The special case r = 1, ℓ = 0 and m = n in (1) produces the following result.

Theorem 0.7 Let M = (Mn,k(x, y))n≥k≥0 be given in Section 2. Then there holds

n

• k=0

det   Mn,k(y, y) Mn,k+1(y, y) Mn+2,k(y, y) Mn+2,k+1(y, y)   = 2Mn,0(y, y)Mn+1,0(y, y). (3)

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• Example (i) When (x, y) = (1, 2), M = (Mn,k(1, 2))n≥k≥0 = (C(t), tC2(t)).

By the series expansion, C(t)α =

• n≥0

α 2n + α 2n + α n

• tn,

(4) we have Mn,k(1, 2) = [tn]C(t)(tC2(t))k = [tn−k]C(t)2k+1 = 2k + 1 2n + 1 2n + 1 n − k

• . (5)

Then, after some routine simplifications, (2) produces the following result.

ℓ + 1 m + 1 2m + 2 m − ℓ

• Cn =

N0

• k=0

(2k + 1)(2k + 2ℓ + 1)αn,k(m, ℓ) (2n + 1)3(2m + 1)3 2n + 3 n − k + 1 2m + 3 m − k − ℓ

• , (6)

where αn,k(m, ℓ) = 6(m−n)(n+1)(m+1)+(ℓ+1)(2k+ℓ+2)(2n+1)(2n+ 2) − 2(m − n)k(k + 1)(2n + 2m + 3) and (x)k = x(x + 1) · · · (x + k − 1) for k ≥ 1 and (x)0 = 1.

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• Taking ℓ = 0 and m = n − 1, n or n + 1 into account, we have

αn,k(n − 1, 0) = (n + k + 3)(8nk + 2n + 2k + 2), αn,k(n, 0) = (2k + 2)(2n + 1)(2n + 2), αn,k(n + 1, 0) = (n − k + 1)(8nk + 14n + 10k + 16). Then in these three cases, after shifting n to n + 1 in the case m = n − 1, (6) generates Corollary 0.3 For any integer n ≥ 0, there hold C2

n+1 = n

• k=0

(2k + 1)(2k + 3)(8nk + 2n + 10k + 4) (2n + 1)(2n + 2)(2n + 3)(2n + 4)(2n + 5) 2n + 2 n − k 2n + 5 n − k + 2

• ,

CnCn+1 =

n

• k=0

(2k + 1)(2k + 2)(2k + 3) (2n + 1)(2n + 2)(2n + 3)2 2n + 3 n − k 2n + 3 n − k + 1

• ,

(7) CnCn+2 =

n

• k=0

(2k + 1)(2k + 3)(8nk + 14n + 10k + 16) (2n + 1)(2n + 2)(2n + 3)(2n + 4)(2n + 5) 2n + 2 n − k 2n + 5 n − k + 1

• .

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• Example (ii) When (x, y) = (2, 2), M = (Mn,k(2, 2))n≥k≥0 is Shapiro’s Catalan

triangle, where Mn,k(2, 2) = 2k + 2 2n + 2 2n + 2 n − k

• .

(8) Then, we have Corollary 0.4 For any integer n ≥ 0, there hold Cn+1Cn+2 =

n

• k=0

(2k + 2)(2k + 4)(8nk + 6n + 14k + 12) (2n + 2)(2n + 3)(2n + 4)(2n + 5)(2n + 6) 2n + 3 n − k 2n + 6 n − k + 2

• ,

C2

n+1 = n

• k=0

(2k + 2)(2k + 3)(2k + 4) (2n + 2)(2n + 3)(2n + 4)2 2n + 4 n − k 2n + 4 n − k + 1

• ,

(9) Cn+1Cn+2 =

n

• k=0

(2k + 2)(2k + 4)(8nk + 18n + 14k + 30) (2n + 2)(2n + 3)(2n + 4)(2n + 5)(2n + 6) 2n + 3 n − k 2n + 6 n − k + 1

• .

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• Example (iii) When (x, y) = (3, 2), M = (Mn,k(3, 2))n≥k≥0, where

Mn,k(3, 2) = 2n + 1 n − k

• .

(10) Then, we have Corollary 0.5 For any integer n ≥ 0, there hold 2n + 3 n + 1

• Cn+1 =

n

• k=0

(8nk + 6n + 14k + 12) (2n + 2)(2n + 3)(2n + 4) 2n + 3 n − k 2n + 4 n − k + 2

• ,

2n + 1 n

• Cn+1 =

n

• k=0

(2k + 2) (2n + 2)(2n + 3) 2n + 3 n − k 2n + 3 n − k + 1

• ,

(11) 2n + 1 n

• Cn+2 =

n

• k=0

(8nk + 10n + 14k + 6) (2n + 2)(2n + 3)(2n + 4) 2n + 4 n − k 2n + 3 n − k + 1

• .

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• The cases (x, y) = (1, 2), (2, 2) or (3, 2) in (2), when m = n, generate the

following results respectively. Corollary 0.6 For any integers n ≥ ℓ ≥ 0, there hold

1 n + 1 2n + 2 n − ℓ

• Cn =

n−ℓ

• k=0

(2k + 1)(2k + ℓ + 2)(2k + 2ℓ + 3) (2n + 1)(2n + 2)(2n + 3)2 2n + 3 n − k − ℓ 2n + 3 n − k + 1

• ,

(12) 1 n + 1 2n + 2 n − ℓ

• Cn+1 =

n−ℓ

• k=0

(2k + 2)(2k + ℓ + 3)(2k + 2ℓ + 4) (2n + 2)(2n + 3)(2n + 4)2 2n + 4 n − k − ℓ 2n + 4 n − k + 1

• ,

(13) 1 n + 1 2n + 2 n − ℓ

• Cn =

n−ℓ

• k=0

(2k + ℓ + 2) (2n + 1)(2n + 2)(2n + 3) 2n + 3 n − k − ℓ 2n + 3 n − k + 1

• .

(14)

• It should be pointed out that despite (12)-(14) are all correct for any integer

ℓ ≤ −1 if one notices that they hold trivially for any integer ℓ > n and both sides of them can be transferred into polynomials on ℓ.

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• Recently, J.M. Gutierrez et al. (2008), Miana and Romero (2007), Chen and

Chu (2009), and Guo and Zeng (2010) studied the binomial sums

n

• k=0

(k + 1)m 2n + 2 n − k 2 related to the classical Catalan triangle.

• Zhang and Pang (2010) also considered some alternating cases.
• Miana and Romero (2010) investigated another binomial sums

n

• k=0

(2k + 1)m 2n + 1 n − k 2 .

• Setting p = k + 1, ℓ = n − i + 1, and then replacing n by n − 2, (13) reduces

to the main identity obtained by Gutirrez et al. [Theorem 5].

i

• k=0

(n + 2k + 2 − i)Bn,kBn,n+k−i = (n + 2) 2n i

• Cn+1, (0 ≤ i ≤ n).

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• Specially, in the case ℓ = −1, replacing n + 1 by n, after some routine simplifi-

cations, (12)-(14) lead respectively to the following identities,

n

• k=0

(2k + 1)3 (2n + 1)2 2n + 1 n − k 2 = 2n n 2 , [Miana and Romero, Remark 11] (15)

n

• k=0

(k + 1)3 (n + 1)2 2n + 2 n − k 2 = 2n n 2n + 1 n

• , [Gutierrez et al. Coro. 6] (16)

n

• k=0

(2k + 1) (2n + 1) 2n + 1 n − k 2 = 2n n 2 . (17) Note that these identities can be regarded as companion ones obtained by

n

• k=0

(2k + 1)2 (2n + 1) 2n + 1 n − k

• = 4n, [Deng and Yan],

n

• k=0

(k + 1)2 (n + 1) 2n + 2 n − k

• = 4n, [Cameron and Nkwanta].

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• The case r = 2 in (3), that is

n

• k=0

det   Mn,k(y, y) Mn,k+1(y, y) Mn+2,k(y, y) Mn+2,k+1(y, y)   = 2Mn,0(y, y)Mn+1,0(y, y), together with (8), after some routine computations, produces Corollary 0.7 For any integer n ≥ 0, there holds CnCn+1 =

n

• k=0

(2k + 2)(2k + 3)(2k + 4) (2n + 2)(2n + 3)(2n + 4)(2n + 5) 2n + 5 n − k 2n + 5 n − k + 2

• .

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• 3.4. Alternating Row Sums

In this section, some alternating sums related to M = (Mn,k(x, y))n≥k≥0 are

• considered. Despite it has no general and unified results as in the previous sec-

tion, but in several isolated cases, mainly by the creative telescoping algorithm, we also obtain some interesting results. Theorem 0.8 Let M = (Mn,k(x, y))n≥k≥0 be given in Section 2. Then there holds

n

• k=0

(−1)n−k det   Mn,k(0, 0) Mn,k+1(0, 0) Mn+1,k(0, 0) Mn+1,k+1(0, 0)   = Cn+1, (18)

• r equivalently,

C2m+1 =

m

• j=0

(2j + 2)2 (2m + 2)2 2m + 2 m − j 2 , (19) C2m+2 =

m

• j=0

(2j + 2)2 (2m + 2)(2m + 4) 2m + 3 m − j 2m + 4 m − j + 1

• .

(20)

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• Theorem 0.9 Let M = (Mn,k(x, y))n≥k≥0 be given in Section 2. Then for j = 1
• r 2, there holds

n

• k=0

(−1)k det   Mn,k(2, 2) Mn,k+1(2, 2) Mn+j,k(2, 2) Mn+j,k+1(2, 2)   = 4j−1Cn+1,

• r equivalently, for j = 1 there has

Cn+1 =

n

• k=0

(−1)k (2k + 2)(2k + 3)(2k + 4) (2n + 2)(2n + 3)(2n + 4)2 2n + 4 n − k 2n + 4 n − k + 1

• ,(21)

and for j = 2 there has 2Cn+1 =

n

• k=0

(−1)k(2k + 2)(2k + 3)(2k + 4) (2n + 2)(2n + 3)(2n + 4)(2n + 5) 2n + 5 n − k 2n + 5 n − k + 2

• . (22)

Question: Give a bijective proof of Theorem 0.9. Ref: Yidong Sun and Luping Ma, Minors of a class of Riordan arrays related to weighted partial Motzkin paths, European Journal of Combinatorics, 39 (2014) 157-169.

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• 4. The Third Starting Points and Problems:

4.1 The Permanent Transformation of Pascal Triangle

n/k 0 1 2 3 4 5 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1

= ⇒ per    

3 3 4 6

   

n/k 1 2 3 4 5 row sums 1 1 = 1 × 1 1 3 1 4 = 2 × 2 2 5 9 1 15 = 3 × 5 3 7 30 18 1 56 = 4 × 14 4 9 70 100 30 1 210 = 5 × 42 5 11 135 350 250 45 1 792 = 6 × 132

Table 4.1. The permanent transformation of P for n and k up to 5.

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• 4.2 The Main Results

Theorem 0.10 For any integers m, n, r with m ≥ n ≥ 0, there holds

m

• k=0

per   Mn,k(y, y) Mn+r,k+1(y, y) Mm,k(y, y) Mm+r,k+1(y, y)   = Mm+n+r,1(y, y) + Hn,m(r), (23) where Hn,m(r) =          r−1

i=0 Mn+i,0(y, y)Mm+r−i−1,0(y, y), if r ≥ 1,

0, if r = 0, − |r|

i=1 Mn−i,0(y, y)Mm−|r|+i−1,0(y, y), if r ≤ −1.

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• The case y = 0 deduce the following result.

Theorem 0.11 For any integers m, n, p with m ≥ n ≥ 0, let Cn,k = k+1

n+1

2n−k

n

• be the ballot number, then there hold

m

• k=0

per   Cn+k,2k Cn+p+k,2k+1 Cm+k,2k Cm+p+k,2k+1   = Cm+n+p,1 + Fn,m(p), (24)

m

• k=0

per   Cn+k,2k+1 Cn+p+k+1,2k+2 Cm+k,2k+1 Cm+p+k+1,2k+2   = Cm+n+p,1 + Fn,m(p), (25) where Fn,m(p) =          p−1

i=0 Cn+iCm+p−i−1, if p ≥ 1,

0, if p = 0, − |p|

i=1 Cn−iCm−|p|+i−1, if p ≤ −1.

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• The case p = 0 in (24) and (25), after some routine computation, gives

Corollary 0.8 For any integers m ≥ n ≥ 1, there hold

Cn+m =

n

• k=0

(2k + 1)(2k + 2)(4mn − 2(m + n)k) (2n)(2n + 1)(2m)(2m + 1) 2n + 1 n − k 2m + 1 m − k

• ,

Cn+m =

n−1

• k=0

(2k + 2)(2k + 3)(4mn + 4m + 4n + 2(m + n)k) (2n)(2n + 1)(2m)(2m + 1) 2n + 1 n − k − 1 2m + 1 m − k − 1

• .

Specially, the m = n case produces C2n =

n−1

• k=0

(2k + 1)(2k + 2) n(2n + 1)

• 2n

n − k − 1 2n + 1 n − k

• ,

C2n =

n−1

• k=0

(2k + 2)(2k + 3) n(2n + 1)

• 2n

n − k − 1 2n + 1 n − k − 1

• .

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• The cases p = 1 in (24) and p = −1 in (25), replacing n and m in (25) by n + 1

and m + 1, after some routine computation, yield Corollary 0.9 For any integers m ≥ n ≥ 0, there hold Cn+m+1 + CnCm =

n

• k=0

(2k + 1)(2k + 2)ηn,m(k) (2n + 1)(2n + 2)(2m + 1)(2m + 2) 2n + 2 n − k 2m + 2 m − k

• ,

Cn+m+1 − CnCm =

n

• k=0

(2k + 2)(2k + 3)ρn,m(k) (2n + 1)(2n + 2)(2m + 1)(2m + 2) 2n + 2 n − k 2m + 2 m − k

• ,

where ηn,m(k) = 4mn + 5(m + n) + 2k(m + n + 1) + 4 and ρn,m(k) = 4mn + m + n − 2k(m + n + 1). Specially, the m = n case produces C2n+1 + C2

n = n

• k=0

(2k + 1)(2k + 2) (n + 1)(2n + 1) 2n + 1 n − k 2n + 2 n − k

• ,

C2n+1 − C2

n = n

• k=0

(2k + 2)(2k + 3) (n + 1)(2n + 1) 2n + 1 n − k − 1 2n + 2 n − k

• .

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• In the case y = 2 and r = p in (23), together with the relations Bn,k =

Mn,k(2, 2) and Bn,0 = Cn+1, similar to the proof of (24), we obtain a result

• n Shapiro’s Catalan triangle.

Theorem 0.12 For any integers m, n, p with m ≥ n ≥ 0, there holds

m

• k=0

per   Bn,k Bn+p,k+1 Bm,k Bm+p,k+1   = Bm+n+p,1 + Fn+1,m+1(p), (26) where Fn,m(p) =          p−1

i=0 Cn+iCm+p−i−1, if p ≥ 1,

0, if p = 0, − |p|

i=1 Cn−iCm−|p|+i−1, if p ≤ −1.

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• The case m = n + 1 and p = 0.

n/k 1 2 3 4 5 6 1 1 2 1 2 5 4 1 3 14 14 6 1 4 42 48 27 8 1 5 132 165 110 44 10 1 6 429 572 429 208 65 12 1 = ⇒ n/k 1 2 row sums 1 1 1 13 1 14 2 126 38 1 165 Table 4.2. Shapiro’s Catalan triangle and its permanent transformation. Ref: Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, The Electronic Journal of Combinatorics, 21(1) (2014), #P1.33.

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• 5. New Discovery

The Schr¨

• der triangle S = (Si,j)i≥j≥0 is defined by

Si,j = Si−1,j−1 + 3Si−1,j + 2Si−1,j+1 with Si,i = 1 for i ≥ 0 and Si,j = 0 for i < j or i, j < 0. n/k 1 2 3 4 5 1 1 3 1 2 11 6 1 3 45 31 9 1 4 197 156 60 12 1

= ⇒

• 6

1 31 9

• n/k

1 2 3 vector WRS 1 1 1 1 7 1 2 32 2 71 23 1 4 112 3 913 456 48 1 8 452 112 = 71 ∗ 1 + 23 ∗ 2 + 1 ∗ 4.

◭◭ ◮◮ ◭ ◮ 59 of 61

• 6. Referrences
• N. Cameron and A. Nkwanta, On some (pseudo) involutions in the Riordan

group, J. Integer Seq., 8 (2005), Article 05.3.7.

• X. Chen and W. Chu, Moments on Catalan number, J. Math. Anal. Appl., 349

(2) (2009), 311-316.

• W. Y. C. Chen, A general bijective algorithm for trees, Proc. Natl. Acad. Sci.

USA 87 (1990), 9635–9639.

• W.Y.C. Chen, N.Y. Li, L.W. Shapiro and S.H.F. Yan, Matrix identities on

weighted partial Motzkin paths, Europ. J. Combin., 28 (2007), 1196-1207.

• E.Y.P. Deng and W.-J. Yan, Some identities on the Catalan, Motzkin and

Schr¨

• der numbers, Discrete Applied Mathematics, 156(14), 28 (2008), 2781-

2789.

• V.J.W. Guo and J. Zeng, Factors of binomial sums from Catalan triangle, J.

Number Theory, 130 (1) (2010), 172-186.

• J.M. Gutierrez, M.A. Hernndez, P.J. Miana, N. Romero, New identities in the

Catalan triangle, J. Math. Anal. Appl. 341(1) (2008), 52-61.

◭◭ ◮◮ ◭ ◮ 60 of 61

• Referrences (continuous)
• N. Y. Li and T. Mansour, Identities involving Narayana numbers, Europ. J.

Comb., 29:3 (2008), 672–675.

• P.J. Miana and N. Romero, Computer proofs of new identities in the Catalan

triangle, Biblioteca de la Revista Matemtica Iberoamericana, in: Proceedings

• f the ”Segundas Jornadas de Teora de Nmeros”, (2007), 1-7.
• P.J. Miana and N. Romero, Moments of combinatorial and Catalan numbers,
• J. Number Theory, 130 (2010), 1876-1887.
• L.W. Shapiro, A Catalan triangle, Discrete Math., 14 (1976), 83-90.
• Z. Zhang and B. Pang, Several identities in the Catalan triangle, Indian J. Pure
• Appl. Math., 41(2) (2010), 363-378.

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• Thanks for your attentions!