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1 1 of 61

2 Inspiration of The Pascal Triangle (Yidong Sun) A Joint Work with Ma Fei and Ma Luping 2 of 61 College of Science of Dalian Maritime University

3 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 3 of 61

4 1. The Pascal Triangle: In mathematics, Pascal s triangle is a triangular array of the binomial coefficients. In much of the Western world, it is named after French mathematician Blaise Pascal ( ), although other mathematicians studied it centuries before him in India, Iran, China, Germany, and Italy. n/k of 61 Table 1. The Pascal triangle P for n and k up to 5.

5 Pascal s triangle was known in China in the early 11th century through the work of the Chinese mathematician Jia Xian ( ). In the 13th century, Yang Hui ( ) presented the triangle and hence it is still called Yang Hui s triangle in China. 5 of 61 Figure 1. Yang Hui s triangle P for n and k up to 9.

6 2. The Well-known Motivation: (1) Gaussian binomial coefficient ( ) n k q = [n]! [k]![n k]! = (1 qn )(1 q n 1 ) (1 q n k+1 ) (1 q k )(1 q k 1 ) (1 q) (x + y) n = (x + y) n = Vector Spaces on Finite Field, The Theory of Partitions, The Theory of Q-series. ( ) n x k y n k, k ( ) n x k y n k, k q xy = yx, qxy = yx. 6 of 61

7 (2) Binomial Inversion Formula A n = Gould-Hsu Inversion Formula, ( ) n B k B n = k Krattenthaler Inversion Formula, Ma Xinrong Inversion Formula. ( ) n ( 1) n k A k. k (3) Congruence Properties of Binomial Coefficients Lucas Congruence, Wolstenholme Congruence, Binomial Congruence. 7 of 61

8 (4) The Theory of Riordan Array P = ( ) 1 1 t, t 1 t A Riordan array R = (R n,k ) n k 0 is an infinite lower triangular matrix with nonzero entries on the main diagonal, such that R n,k = [t n ]g(t)(f(t)) k for n k, namely, R n,k equals the coefficient of t n in the expansion of the series g(t)(f(t)) k, where g(t) = 1 + g 1 t + g 2 t 2 + and f(t) = f 1 t + f 2 t 2 + with f 1 0 are two formal power series. R 0,0 R 1,0 R 1,1 R 2,0 R 2,1 R 2,2 R 3,0 R 3,1 R 3,2 R 3,3... g(t) g(t)f(t) g(t)f(t) 2 g(t)f(t) 3 8 of 61

9 (5) Fibonomial Coefficient Analog ( ) n n! F = k F k! F (n k)! F where n! F = F 1 F 2 F n with F 0 = 0, F 1 = 1 and F n = F n 1 + F n 2. (6) Fractal 9 of 61 Figure 2. Sierpinski triangle.

10 (7) Unimodality of Binomial Coefficients ( ) n 0 ( ) ( ) ( n n 1 [n/2] n [n/2] 1 ) Log-Concavity (a 2 k a k 1a k+1 ), Log-Convexity (a 2 k a k 1a k+1 ), q-log-concavity, q-log-convexity, Infinite Log-Concavity, Infinite Log-Convexity. ( ) n n 10 of 61

11 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 n + 1 n + 2 n n + 1 n + 2 =, k k 1 k + 1 k 1 k + 1 k for any k and n, or equivalently which implies that ( )( )( ) ( )( )( ) n n + 1 n + 2 n n + 1 n + 2 =. k + 1 k k + 2 k k + 2 k of 61

12 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 where N n,k = 1 n( n k N n,k+1 N n+1,k N n+2,k+2 = N n,k N n+1,k+2 N n+2,k+1, )( n k 1) is the Narayana number. n/k of 61 Table 2.1. The Narayana triangle N for n and k up to 5.

13 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 N n,k+m 1 N n+1,k+m 2 N n+2,k+m 3 N n+m 2,k+1 N n+m 1,k N n+m,k+m = N n,k N n+1,k+m N n+2,k+m 1 N n+m 2,k+3 N n+m 1,k+2 N n+m,k of 61

14 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 = ( A n,k be an infinite lower triangular matrix, for )n k 0 any given integer m 3, if there hold r i=0 p r 1 A n+i,k+r i i=0 A n+p i,k+r+i+1 = r i=0 p r 1 A n+p i,k+p r+i i=0 A n+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 order m. 14 of 61

15 In order to give a more intuitive view on the definition, we present a pictorial description of the generalized rule for the case m = of 61

16 SDR m : the set of SDR-matrices of order m; SDR : the set of SDR-matrices A of order, that is A SDR m for any m 3. By our notation, it is obvious that the Pascal triangle P = (( )) n k n k 0 Narayana triangle N = ( N n+1,k+1 are SDR-matrices of order 3. )n k 0 In fact, both of them will be proved to be SDR-matrices of order. and the 16 of 61

17 2.2 The Basic Properties of SDR-matrices Lemma 0.1 For any A SDR m, B SDR m+i with i 0, there hold A B SDR m, and A ( 1) SDR m 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 = ( ) A n,k n k 0 SDR m, then ( A n+i,k+j )n k 0 SDR m for fixed i, j of 61

18 Lemma 0.3 Given any sequence (a n ) n 0, let A n,k = a k, B n,k = b n k and C n,k = c n for n k 0, then ( ) A n,k n k 0, ( ) B n,k n k 0, ( C n,k )n k 0 SDR. a 0 a 0 a 1 a 0 a 1 a 2, a 0 a 1 a 2 a 3 b 0 b 1 b 0 b 2 b 1 b 0, b 3 b 2 b 1 b 0 c 0 c 1 c 1 c 2 c 2 c 2 c 3 c 3 c 3 c 3 The Pascal triangle P = (( )) n k 0 k n SDR. The Narayana triangle N = ( ( 1 n+1 )( n+1 )) n+1 k k+1 SDR 0 k n. ( The Lah triangle L = (n+1)! n (k+1)!( k) ) SDR. 0 k n 18 of 61

19 2.3 The Main Results of SDR-matrices Theorem 0.1 For any sequences (a n ) n 0, (b n ) n 0 and (c n ) n 0 such that b 0 = 1, a n 0 and c n 0 for n 0, let A = ( a k b n k c n )n k 0, then A 1 SDR. Conjecture 0.1 For any A SDR m, if the inverse A 1 of A exists, then A 1 SDR m. 19 of 61

20 Theorem 0.2 For any sequences (a n ) n 0, (b n ) n 0 with b 0 = 1 and a n 0 for n 0, let A = ( ) a n b n k a 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 P j, N j, L j SDR for any integer j. In general, for A, B SDR m, their matrix product A B is possibly NOT in SDR m. 20 of 61

21 Theorem 0.3 For any A = ( A n,k )n k 0 with A n,k 0 for n k 0, then A SDR m+1 if and only if A SDR m. Remark 0.1 The condition A n,k 0 for n k 0 in Theorem 0.3 is necessary. The following example verifies this claim ( ( n+k) ) 2 n k = SDR 3, but not in SDR 4. n k of 61

22 Recall that the Narayana number N n+1,k+1 can be represented as N n+1,k+1 = 1 ( )( ) ( n ( n ) n + 1 n + 1 = det k) k+1 n + 1 k + 1 k ( n+1 ) ( n+1 ), k k+1 so we can come up with the following definition. Definition 0.2 Let A = ( A n,k be an infinite lower triangular matrix, for )n k 0 any integer m 1, define A [m] = ( A [m] ) n,k n k 0, where A n,k A n,k+m 1 A [m] n,k = det... A n+m 1,k A n+m 1,k+m 1 22 of 61

23 Theorem 0.4 For any sequences (a n ) n 0, (b n ) n 0 and (c n ) n 0 such that b 0 = 1, a n 0 and c n 0 for n 0, let A = ( a k b n k c n )n k 0, then A [m] SDR for any integer m 1. Let a 1 n and b 1 n = b 1 n = c n = n!, a 1 n = b 1 n = n! for n 0 in Theorem 0.4, one has = c n = n!(n+1)! or a 1 n = c n = n!(n+1)! 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 = ( A n,k )n k 0 SDR with A n,k 0 for n k 0, then there exist three sequences (a n ) n 0, (b n ) n 0 and (c n ) n 0 with a n b n c n 0 for n 0 such that A n,k = a k b n k c n. 23 of 61 Ref: Yidong Sun, The star of David rule, Linear Algebra and its Applications, 429 (8-9), (2008),

24 3. The Second Starting Points and Problems: 3.1 The Determinant Transformation of Pascal Triangle Catalan numbers C n = 1 n+1( 2n n ), [1, 1, 2, 5, 14, 42, 132, 429,...] are the row sums of the Narayana triangle N = (N n+1,k+1 ) n k 0, C n+1 = N n+1,k+1, ( where N n+1,k+1 = 1 n+1 )( n+1 n+1 k k+1). Their close relations are reflected according to the two ways, ( )( ) ( n ( n ) 1 n + 1 n + 1 = det k) k+1 n + 1 k k + 1 ( n+1 ) ( n+1 ), k 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 ) ), 24 of 61 where det( ) denotes the determinant of a square matrix.

25 There exists a close connection between the Pascal triangle P = ( ( n and the Narayana triangle N = ( ( 1 n+1 )( n+1 )) n+1 k k+1 n k 0, k) )n k 0 n/k = n/k row sums Table 3.1. The Pascal triangle P and Narayana triangle N for n and k up to of 61

26 ( Shapiro s Catalan triangle B = (B n,k ) n k 0 with B n,k = k+1 2n+2 ) n+1 n k. n/k Table 3.2. The values of B n,k for n and k up to 5. Let X = (X n,k ) n k 0 be the infinite lower triangles defined on B by X n,k = det B n,k B n,k+1 B n+1,k B n+1,k of 61

27 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 row sums alternating row sums = = = = = Table 3.3. The values of X n,k for n and k up to 4, together with the row sums and alternating row sums. 27 of 61

28 Question: Let A = (A n,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 A p = ( A (p) n,k (m, r, l)) n k 0, where A n,k A n+r,k+l A n+pr,k+pl A (p) A n,k (m, r, l) = det n+m,k A n+m+r,k+l A n+m+pr,k+pl... A n+pm,k A n+pm+r,k+l A n+pm+pr,k+pl Then how to determine the explicit expression for the n-th row sum or alternating row sums of A p, p+1 S (p) n,m,r,l (A p) = A (p) n,k (m, r, l), T (p) n,m,r,l (A p) = ( 1) n k A (p) n,k (m, r, l)? 28 of 61 In the case p = 1, for some special infinite lower triangular matrices related to weighted partial Motzkin paths, it can produce several surprising results.

29 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). 29 of 61

30 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. 30 of 61

31 A weighted partial Motzkin path is a partial Motzkin path with the weight assignment 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 ) = xy 2. 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. 31 of 61

32 Let M n,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 M n,k (x, y), M n,0 (x, y) = xm n 1,0 (x, y) + M n 1,1 (x, y), (n 1), M n,k (x, y) = M n 1,k 1 (x, y) + ym n 1,k (x, y) + M n 1,k+1 (x, y), (n k 1), with M 0,0 (x, y) = 1 and M n,k (x, y) = 0 if n < k or k < of 61

33 The first few entries of the triangle M = (M n,k (x, y)) n k 0. n/k x 1 2 x x + y 1 3 x 3 + 2x + y x 2 + xy + y x + 2y 1 4 x 4 + 3x 2 + 2xy + y x 3 + x 2 y + xy 2 + 3x + y 3 + 5y x 2 + 2xy + 3y x + 3y 1 Table 4. The values of M n,k (x, y) for n and k up to of 61

34 Denote M k (x, y; t) = n k M n,k(x, y)t n to be the generating function of weighted partial Motzkin paths ending at level k. Lemma 0.4 Let M = (M n,k (x, y)) n k 0, then M = (M 0 (x, y; t), tm 0 (y, y; t)) is a Riordan array, i.e., M k (x, y; t) = M 0 (x, y; t)(tm 0 (y, y; t)) k, where M 0 (y, y; t) = 1 yt (1 yt) 2 4t 2 2t 2, M 0 (x, y; t) = 1 1 xt t 2 M 0 (y, y; t) 1 2xt + yt (1 yt)2 4t = 2 2(y x)(1 xt)t + 2t 2. We can NOT expect to obtain a simple and explicit expressions from M. 34 of 61 M n,k (x, y) = [t n ]M 0 (x, y; t)(tm 0 (y, y; t)) k.

35 We give a short list in Table 5, where C(t) = 1 1 4t 2t and M(t) = 1 t 1 2t 3t 2 2t 2 are generating functions respectively for Catalan numbers C n and Motzkin numbers M n. (x, y) M 0 (x, y; t) Sequences M (0, 0) C(t 2 ) A = C n 2 A (0, 1) 1+t 1 2t 3t 2 2t(1+t) A = Riordan numbers R n A (0, 2) 1 1 4t 3 1 4t (1, 0) 1 1 t t 2 C(t 2 ) A = F ine numbers F n A = ( ) n n/2 A A (1, 1) M(t) A = Motzkin numbers M n A of 61 (1, 2) C(t) A = Catalan numbers C n A (2, 2) C 2 (t) A = Catalan numbers C n+1 A C(t) (3, 2) 1 4t A = ( ) 2n+1 A n Table 5. The specializations of (x, y), where C n 2 is set to be zero when n is odd.

36 3.3. Main Results and Consequences Theorem 0.5 Let M = (M n,k (x, y)) n k 0, for any integers n, r 0 and m N r l 0, set N r = min{n + r + 1, m + r l}. Then there hold det M n,k(x, y) M m,k+l+1 (x, y) r = M n+i,0 (x, y)m m+r i,l (y, y).(1) M n+r+1,k (x, y) M m+r+1,k+l+1 (x, y) i=0 36 of 61 The cases r = 2, l = 4 and m = n + 2, n or n 2.

37 (Proof Sketch) Define A (r) n,m,k,l (x, y) = {(P, Q) P M n,k(x, y), Q M m+r+1,k+l+1 (x, y)}, B (r) n,m,k,l (x, y) = {(P, Q) P M n+r+1,k(x, y), Q M m,k+l+1 (x, y)}, and C (r,i) (r) n,m,k,l (x, y) = {(P, Q) A n,m,k,l (x, y) Q = Q 1UQ 2 } with Q 1 M i,k (x, y) and Q 2 M m+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 N r Nr B(r) n,m,k,l (x, y). A (r) n,m,k,l (x, y)/ r i=0 i C (r,i) n,m,k,l (x, y) and 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). 37 of 61

38 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). 38 of 61

39 39 of 61

40 The special case r = 0 in (1) produces the following result. Theorem 0.6 Let M = (M n,k (x, y)) n k 0 be given in Section 2. For any integers n 0 and m l 0, set N 0 = min{n + 1, m l}. Then there holds N 0 det M n,k(x, y) M m,k+l+1 (x, y) M n+1,k (x, y) M m+1,k+l+1 (x, y) = M n,0 (x, y)m m,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 = (M n,k (x, y)) n k 0 be given in Section 2. Then there holds det M n,k(y, y) M n,k+1 (y, y) = 2M n,0 (y, y)m n+1,0 (y, y). (3) M n+2,k (y, y) M n+2,k+1 (y, y) 40 of 61

41 Example (i) When (x, y) = (1, 2), M = (M n,k (1, 2)) n k 0 = (C(t), tc 2 (t)). By the series expansion, C(t) α = n 0 α 2n + α ( 2n + α n ) t n, (4) we have M n,k (1, 2) = [t n ]C(t)(tC 2 (t)) k = [t n k ]C(t) 2k+1 = 2k + 1 ( ) 2n + 1. (5) 2n + 1 n k Then, after some routine simplifications, (2) produces the following result. l + 1 m + 1 ( ) 2m + 2 C n = m l N 0 ( )( ) (2k + 1)(2k + 2l + 1)α n,k (m, l) 2n + 3 2m + 3, (6) (2n + 1) 3 (2m + 1) 3 n k + 1 m k l where α n,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 = of 61

42 Taking l = 0 and m = n 1, n or n + 1 into account, we have α n,k (n 1, 0) = (n + k + 3)(8nk + 2n + 2k + 2), α n,k (n, 0) = (2k + 2)(2n + 1)(2n + 2), α n,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 ( )( ) Cn+1 2 (2k + 1)(2k + 3)(8nk + 2n + 10k + 4) 2n + 2 2n + 5 =, (2n + 1)(2n + 2)(2n + 3)(2n + 4)(2n + 5) n k n k + 2 ( )( ) (2k + 1)(2k + 2)(2k + 3) 2n + 3 2n + 3 C n C n+1 = (2n + 1)(2n + 2)(2n + 3) 2, (7) n k n k + 1 ( )( ) (2k + 1)(2k + 3)(8nk + 14n + 10k + 16) 2n + 2 2n + 5 C n C n+2 =. (2n + 1)(2n + 2)(2n + 3)(2n + 4)(2n + 5) n k n k of 61

43 Example (ii) When (x, y) = (2, 2), M = (M n,k (2, 2)) n k 0 is Shapiro s Catalan triangle, where M n,k (2, 2) = 2k + 2 ( ) 2n + 2. (8) 2n + 2 n k Then, we have Corollary 0.4 For any integer n 0, there hold ( )( ) (2k + 2)(2k + 4)(8nk + 6n + 14k + 12) 2n + 3 2n + 6 C n+1 C n+2 =, (2n + 2)(2n + 3)(2n + 4)(2n + 5)(2n + 6) n k n k + 2 ( )( ) Cn+1 2 (2k + 2)(2k + 3)(2k + 4) 2n + 4 2n + 4 = (2n + 2)(2n + 3)(2n + 4) 2, (9) n k n k + 1 ( )( ) (2k + 2)(2k + 4)(8nk + 18n + 14k + 30) 2n + 3 2n + 6 C n+1 C n+2 =. (2n + 2)(2n + 3)(2n + 4)(2n + 5)(2n + 6) n k n k of 61

44 Example (iii) When (x, y) = (3, 2), M = (M n,k (3, 2)) n k 0, where ( ) 2n + 1 M n,k (3, 2) =. (10) n k Then, we have Corollary 0.5 For any integer n 0, there hold ( ) 2n + 3 ( )( ) (8nk + 6n + 14k + 12) 2n + 3 2n + 4 C n+1 =, n + 1 (2n + 2)(2n + 3)(2n + 4) n k n k + 2 ( ) 2n + 1 ( )( ) (2k + 2) 2n + 3 2n + 3 C n+1 =, (11) n (2n + 2)(2n + 3) n k n k + 1 ( ) 2n + 1 ( )( ) (8nk + 10n + 14k + 6) 2n + 4 2n + 3 C n+2 =. n (2n + 2)(2n + 3)(2n + 4) n k n k of 61

45 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 2n + 2 n + 1 n l ( 2n n n + 1 n l ( 2n + 2 n l ) C n = ) C n+1 = ) C n = n l n l n l ( (2k + 1)(2k + l + 2)(2k + 2l + 3) 2n + 3 (2n + 1)(2n + 2)(2n + 3) 2 n k l ( 2n + 4 (2k + 2)(2k + l + 3)(2k + 2l + 4) (2n + 2)(2n + 3)(2n + 4) 2 (2k + l + 2) (2n + 1)(2n + 2)(2n + 3) )( ) 2n + 3, (12) n k + 1 )( ) 2n + 4, (13) n k l n k + 1 ). (14) ( )( 2n + 3 2n + 3 n k l n k of 61 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.

46 Recently, J.M. Gutierrez et al. (2008), Miana and Romero (2007), Chen and Chu (2009), and Guo and Zeng (2010) studied the binomial sums related to the classical Catalan triangle. (k + 1) m ( 2n + 2 n k Zhang and Pang (2010) also considered some alternating cases. Miana and Romero (2010) investigated another binomial sums ( ) 2 2n + 1 (2k + 1) m. n k 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 ( ) 2n (n + 2k + 2 i)b n,k B n,n+k i = (n + 2) C n+1, (0 i n). i ) 2 46 of 61

47 Specially, in the case l = 1, replacing n + 1 by n, after some routine simplifications, (12)-(14) lead respectively to the following identities, (2k + 1) 3 ( ) 2 2n + 1 (2n + 1) 2 = n k (k + 1) 3 (n + 1) 2 (2k + 1) (2n + 1) ( ) 2 2n, [Miana and Romero, Remark 11] (15) n ( ) 2 ( )( ) 2n + 2 2n 2n + 1 =, [Gutierrez et al. Coro. 6] (16) n k n n ( ) 2 ( ) 2 2n + 1 2n =. (17) n k n Note that these identities can be regarded as companion ones obtained by (2k + 1) 2 ( ) 2n + 1 = 4 n, [Deng and Yan], (2n + 1) n k (k + 1) 2 ( ) 2n + 2 = 4 n, [Cameron and Nkwanta]. (n + 1) n k 47 of 61

48 The case r = 2 in (3), that is det M n,k(y, y) M n,k+1 (y, y) = 2M n,0 (y, y)m n+1,0 (y, y), M n+2,k (y, y) M n+2,k+1 (y, y) together with (8), after some routine computations, produces Corollary 0.7 For any integer n 0, there holds ( )( ) (2k + 2)(2k + 3)(2k + 4) 2n + 5 2n + 5 C n C n+1 =. (2n + 2)(2n + 3)(2n + 4)(2n + 5) n k n k of 61

49 3.4. Alternating Row Sums In this section, some alternating sums related to M = (M n,k (x, y)) n k 0 are considered. Despite it has no general and unified results as in the previous section, but in several isolated cases, mainly by the creative telescoping algorithm, we also obtain some interesting results. Theorem 0.8 Let M = (M n,k (x, y)) n k 0 be given in Section 2. Then there holds ( 1) n k det M n,k(0, 0) M n,k+1 (0, 0) = C n+1, (18) M n+1,k (0, 0) M n+1,k+1 (0, 0) or equivalently, C 2m+1 = C 2m+2 = m j=0 m j=0 (2j + 2) 2 ( ) 2 2m + 2 (2m + 2) 2, (19) m j (2j + 2) 2 (2m + 2)(2m + 4) ( 2m + 3 m j )( 2m + 4 m j + 1 ). (20) 49 of 61

50 Theorem 0.9 Let M = (M n,k (x, y)) n k 0 be given in Section 2. Then for j = 1 or 2, there holds ( 1) k det M n,k(2, 2) M n,k+1 (2, 2) M n+j,k (2, 2) M n+j,k+1 (2, 2) = 4 j 1 C n+1, or equivalently, for j = 1 there has ( )( ) k (2k + 2)(2k + 3)(2k + 4) 2n + 4 2n + 4 C n+1 = ( 1) (2n + 2)(2n + 3)(2n + 4) 2,(21) n k n k + 1 and for j = 2 there has ( 1) k ( )( ) (2k + 2)(2k + 3)(2k + 4) 2n + 5 2n + 5 2C n+1 =. (22) (2n + 2)(2n + 3)(2n + 4)(2n + 5) n k n k of 61 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)

51 4. The Third Starting Points and Problems: 4.1 The Permanent Transformation of Pascal Triangle n/k per = n/k row sums = = = = = = Table 4.1. The permanent transformation of P for n and k up to of 61

52 4.2 The Main Results Theorem 0.10 For any integers m, n, r with m n 0, there holds m per M n,k(y, y) M n+r,k+1 (y, y) = M m+n+r,1 (y, y) + H n,m (r), (23) M m,k (y, y) M m+r,k+1 (y, y) where H n,m (r) = r 1 i=0 M n+i,0(y, y)m m+r i 1,0 (y, y), if r 1, 0, if r = 0, r i=1 M n i,0(y, y)m m r +i 1,0 (y, y), if r of 61

53 The case y = 0 deduce the following result. ( Theorem 0.11 For any integers m, n, p with m n 0, let C n,k = k+1 2n k ) n+1 n be the ballot number, then there hold m per C n+k,2k where C m+k,2k m per C n+k,2k+1 C m+k,2k+1 F n,m (p) = C n+p+k,2k+1 C m+p+k,2k+1 C n+p+k+1,2k+2 C m+p+k+1,2k+2 = C m+n+p,1 + F n,m (p), (24) p 1 i=0 C n+ic m+p i 1, if p 1, = C m+n+p,1 + F n,m (p), (25) 0, if p = 0, p i=1 C n ic m p +i 1, if p of 61

54 The case p = 0 in (24) and (25), after some routine computation, gives Corollary 0.8 For any integers m n 1, there hold C n+m = n 1 C n+m = (2k + 1)(2k + 2)(4mn 2(m + n)k) (2n)(2n + 1)(2m)(2m + 1) ( 2n + 1 n k (2k + 2)(2k + 3)(4mn + 4m + 4n + 2(m + n)k) (2n)(2n + 1)(2m)(2m + 1) Specially, the m = n case produces C 2n = n 1 n 1 C 2n = ( (2k + 1)(2k + 2) n(2n + 1) ( (2k + 2)(2k + 3) n(2n + 1) )( 2m + 1 m k ), ( 2n + 1 n k 1 )( ) 2n 2n + 1, n k 1 n k )( ) 2n 2n + 1. n k 1 n k 1 )( ) 2m + 1. m k 1 54 of 61

55 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 ( )( (2k + 1)(2k + 2)η n,m (k) 2n + 2 2m + 2 C n+m+1 + C n C m = (2n + 1)(2n + 2)(2m + 1)(2m + 2) n k m k ( )( (2k + 2)(2k + 3)ρ n,m (k) 2n + 2 2m + 2 C n+m+1 C n C m = (2n + 1)(2n + 2)(2m + 1)(2m + 2) n k m k ), ), where η n,m (k) = 4mn + 5(m + n) + 2k(m + n + 1) + 4 and ρ n,m (k) = 4mn + m + n 2k(m + n + 1). Specially, the m = n case produces ( )( ) C 2n+1 + Cn 2 (2k + 1)(2k + 2) 2n + 1 2n + 2 =, (n + 1)(2n + 1) n k n k ( )( ) C 2n+1 Cn 2 (2k + 2)(2k + 3) 2n + 1 2n + 2 =. (n + 1)(2n + 1) n k 1 n k 55 of 61

56 In the case y = 2 and r = p in (23), together with the relations B n,k = M n,k (2, 2) and B n,0 = C n+1, similar to the proof of (24), we obtain a result on Shapiro s Catalan triangle. Theorem 0.12 For any integers m, n, p with m n 0, there holds m per B n,k B n+p,k+1 = B m+n+p,1 + F n+1,m+1 (p), (26) where B m,k F n,m (p) = B m+p,k+1 p 1 i=0 C n+ic m+p i 1, if p 1, 0, if p = 0, p i=1 C n ic m p +i 1, if p of 61

57 The case m = n + 1 and p = 0. n/k n/k row sums = of 61 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.

58 5. New Discovery The Schröder triangle S = (S i,j ) i j 0 is defined by S i,j = S i 1,j 1 + 3S i 1,j + 2S i 1,j+1 with S i,i = 1 for i 0 and S i,j = 0 for i < j or i, j < 0. n/k = n/k vector W RS of =

59 6. Referrences N. Cameron and A. Nkwanta, On some (pseudo) involutions in the Riordan group, J. Integer Seq., 8 (2005), Article X. Chen and W. Chu, Moments on Catalan number, J. Math. Anal. Appl., 349 (2) (2009), W. Y. C. Chen, A general bijective algorithm for trees, Proc. Natl. Acad. Sci. USA 87 (1990), 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), E.Y.P. Deng and W.-J. Yan, Some identities on the Catalan, Motzkin and Schröder numbers, Discrete Applied Mathematics, 156(14), 28 (2008), V.J.W. Guo and J. Zeng, Factors of binomial sums from Catalan triangle, J. Number Theory, 130 (1) (2010), J.M. Gutierrez, M.A. Hernndez, P.J. Miana, N. Romero, New identities in the Catalan triangle, J. Math. Anal. Appl. 341(1) (2008), of 61

60 Referrences (continuous) N. Y. Li and T. Mansour, Identities involving Narayana numbers, Europ. J. Comb., 29:3 (2008), P.J. Miana and N. Romero, Computer proofs of new identities in the Catalan triangle, Biblioteca de la Revista Matemtica Iberoamericana, in: Proceedings of 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), L.W. Shapiro, A Catalan triangle, Discrete Math., 14 (1976), Z. Zhang and B. Pang, Several identities in the Catalan triangle, Indian J. Pure Appl. Math., 41(2) (2010), of 61

61 Thanks for your attentions! 61 of 61