A23 Integers 16 (2016) Two Explicit Formulas Of

#A23 INTEGERS 16 (2016)

TWO EXPLICIT FORMULAS OF THE SCHRODER¨ NUMBERS

Feng Qi1 Institute of Mathematics, Henan Polytechnic University, Jiaozuo City, China; College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City, Inner Mongolia Autonomous Region, China; Department of Mathematics, Tianjin Polytechnic University, Tianjin City, China [email protected], [email protected]

Xiao-Ting Shi Department of Mathematics, Tianjin Polytechnic University, Tianjin City, China [email protected], [email protected]

Bai-Ni Guo School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, China [email protected], [email protected]

Received: 1/13/16, Accepted: 4/13/16, Published: 4/22/16

Abstract In the paper, by several methods and approaches, the authors establish two explicit formulas for the large and little Schr¨oder numbers.

1. Introduction

In combinatorics and number theory, there are two kinds of Schröder numbers, the large Schröder numbers Sn and the little Schröder numbers sn. They are named after the German mathematician Ernst Schröder.

A large Schröder number Sn describes the number of paths from the southwest corner (0, 0) of an n n grid to the northeast corner (n, n), using only single steps ⇥ north, northeast, or east, that do not rise above the southwest-northeast diagonal. The first eleven large Schröder numbers S for 0 n 10 are n   1, 2, 6, 22, 90, 394, 1806, 8558, 41586, 206098, 1037718.

In [3, Theorem 8.5.7], it was proved that the large Schr¨oder numbers Sn have the

1Corresponding author INTEGERS: 16 (2016) 2 generating function

1 x px2 6x + 1 1 G(x) = = S xn, (1.1) 2x n n=0 X which can also be rearranged as

px2 + 6x + 1 1 x 1 (x) = G( x) = = ( 1)nS xn. (1.2) G 2x n n=0 X The little Schröder numbers sn form an integer sequence that can be used to count the number of plane trees with a given set of leaves, the number of ways of inserting parentheses into a sequence, and the number of ways of dissecting a convex polygon into smaller polygons by inserting diagonals. The first eleven little Schröder numbers s for 1 n 11 are n   1, 1, 3, 11, 45, 197, 903, 4279, 20793, 103049, 518859. They are also called the small Schröder numbers, the Schröder-Hipparchus numbers, or the Schröder numbers, after Ernst Schröder and the ancient Greek mathematician Hipparchus who appears from evidence in Plutarch to have known of these numbers. They are also called the super-Catalan numbers, after Eugéne Charles Catalan, but are di↵erent from a generalization of the Catalan numbers [8, 18]. In [3, Theorem 8.5.6], it was proved that the little Schröder numbers sn have the generating function

1 + x px2 6x + 1 1 g(x) = = s xn. (1.3) 4 n n=1 X For more information on the large Schr¨oder numbers Sn and the little Schr¨oder numbers sn, please refer to [3, 13, 14, 15, 19, 20] and plenty of references therein. Comparing (1.1) with (1.3), we can reveal

1 1 x2 6x + 1 = 1 + x 4 s xn = 1 x 2 S xn+1, n n n=1 n=0 p X X that is,

1 n 1 1 n 1 n 1 2 s x = 1 2 s x = S x . n n+1 n n=1 n=0 n=0 X X X Accordingly, we acquire Sn = 2sn+1, n N. 2 See also [3, Corollary 8.5.8]. This relation tells us that it is sucient to analytically study the large Schr¨oder numbers Sn. The main aim of this paper is, by several methods and approaches, to establish two explicit formulas for the large and little Schr¨oder numbers Sn and sn. INTEGERS: 16 (2016) 3

Theorem 1. For n N, the large and little Schr¨oder numbers Sn and sn+1 can be 2 computed by

( 1)n 1 n+1 62k 1 k S = 2s = (1.4) n n+1 12 6n k! 2 n k + 1 k= (n+1)/2 ⌧ k✓ ◆ d X e and

n s n r 1 1 n! s q 2(r+j) 1 1 S = 2s = ( 1) 6 k n n+1 12 6n 2 2 q=0 j=0 k j kX=0 r+Xs=k `+Xm=n X X ⌧ ⌧ ` 1 s j m + 2q 1 1 , ⇥ r 1 q n r j + 1 2q 1 r!s!j! ✓ ◆✓ ◆✓ ◆✓ ◆ (1.5) where x stands for the ceiling function which gives the smallest integer not less d e than x and x is the falling factorial deﬁned by h in n 1 x(x 1) (x n + 1), n 1, x n = (x k) = · · · h i (1, n = 0. kY=0

2. Lemmas

In order to prove our main results, we need the following notions and lemmas.

In combinatorial mathematics, the Bell polynomials of the second kind Bn,k are deﬁned by

n k+1 ` n! xi i Bn,k(x1, x2, . . . , xn k+1) = n k+1 ì! i! ì 0 N i=1 i=1 n2X{ }[ Y ⇣ ⌘ i=1 iì=n n ì=k Q P i=1 P for n k 0. See [4, p. 134, Theorem A]. In combinatorial analysis, the Faà di Bruno formula plays an important role and can be described by

n n d (k) (n k+1) [f h(t)] = f (h(t))B h0(t), h00(t), . . . , h (t) (2.1) d tn n,k k=0 X in terms of the Bell polynomials of the second kind Bn,k. See [4, p. 139, Theorem C]. Lemma 1 ([4, p. 135]). For n k 0, we have 2 n k+1 k n Bn,k abx1, ab x2, . . . , ab xn k+1 = a b Bn,k(x1, x2, . . . , xn k+1), (2.2) where a and b are any complex numbers. INTEGERS: 16 (2016) 4

Lemma 2 ([5, Theoem 4.1], [12, Theorem 3.1], and [17, Lemma 2.5]). For n k 0, the Bell polynomials of the second kind Bn,k satisfy

(n k)! n k 2k n B (x, 1, 0, . . . , 0) = x . (2.3) n,k 2n k k n k ✓ ◆✓ ◆ Lemma 3 ([2, p. 40, Exercise 5)], [6, Section 2.2, p. 849], [9, p. 94], and [17, Lemma 2.1]). Let u(x) and v(x) = 0 be two di↵erentiable functions. Let U(n+1) 1(x) 6 (k 1) ⇥ be an (n + 1) 1 matrix whose elements u (x) = u (x) for 1 k n + 1, let ⇥ k,1   V(n+1) n(x) be an (n + 1) n matrix whose elements ⇥ ⇥

i 1 (i j) v (x), i j 0 v (x) = j 1 i,j 8✓ ◆ 0, 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) . ⇥ ⇥ ⇥ u(x) Then the nth derivative of the ratio v(x) can be computed by

n d u(x) W(n+1) (n+1)(x) n ⇥ n = ( 1) n+1 . d x v(x) v (x)  Lemma 4 ([16, Theorem 2.1] and [21]). Let M be a square matrix of order n n ⇥ and partitioned as A B M = . C D ✓ ◆ 1 Let K denote the inverse of an invertible matrix K. 1. If B is a k k matrix and C nonsingular, then the determinant of M can be ⇥ computed by (n+1)k 1 M = ( 1) C B AC D . (2.4) | | | | 2. If A, B, C, D are respectively p p, p q, q p, and q q matrices and if ⇥ ⇥ ⇥ ⇥ D is invertible, then the determinant of M can be computed by

1 M = D A BD C . (2.5) | | | | Lemma 5 ([1, Example 2.6] and [4, p. 136, Eq. [3n]]). The Bell polynomials of the second kind Bn,k satisfy

Bn,k(x1 + y1, x2 + y2, . . . , xn k+1 + yn k+1) n = B`,r(x1, x2, . . . , x` r+1)Bm,s(y1, y2, . . . , ym s+1). (2.6) ` r+Xs=k `+Xm=n ✓ ◆ INTEGERS: 16 (2016) 5

Lemma 6. For n k 0, we have n n 1 B (1!, 2!, . . . , (n k + 1)!) = (n k)! (2.7) n,k k k 1 ✓ ◆✓ ◆ and k 1 k ` k (n + 2` 1)! B (2!, 3!, . . . , (n k + 2)!) = ( 1) . (2.8) n,k k! ` (2` 1)! X`=0 ✓ ◆ Proof. The formula (2.7) can be found in [4, p. 135] or [7, Theorem 1]. In [4, p. 133], it was stated that k 1 1 tm 1 tn xm = Bn,k(x1, x2, . . . , xn k+1) (2.9) k! m! n! m=1 ! X nX=k for k 0. Hence, it follows from (2.9) that k 1 tn 1 1 tm B (2!, 3!, . . . , (n k + 2)!) = (m + 1)! n,k n! k! m! m=1 ! nX=k X k k 1 1 1 k ` k 1 = 1 = ( 1) k! (t 1)2 k! ` (t 1)2`  X`=0 ✓ ◆ and, by di↵erentiating with respect to t,

1 tn m B (2!, 3!, . . . , (n k + 2)!) n,k (n m)! n=m X k 1 k ` k m (2` + m 1)! 1 = ( 1) ( 1) , m k 0. k! ` (2` 1)! (t 1)2`+m X`=0 ✓ ◆ Further letting t 0 in the above equation yields the formula (2.8). !

3. Proofs of Theorem 1

We are now in a position to prove our main results.

First proof of the formula (1.4). It is clear that the generating function G(x) is naturally deﬁned on ( , 0) 0, 3 2p2 3 + 2p2 , . It is easy to calculate 1 [ [ 1 that limx 0 (x) = limx 0 G(x) = 2. Therefore, at the point x = 0, both of the ! G ! ⇤ ⇥ functions G(x) and (x) are removably discontinuous. Hence, the function (x) G G can be regarded to be deﬁned on , 3 2p2 2p2 3, ). Since 1 [ 1 1 1/2 ⇤ ⇥ x2 + 6x + 1 = h ik [x(x + 6)]k, x(x + 6) < 1, (3.1) k! | | p kX=0 INTEGERS: 16 (2016) 6 we obtain

1 1 1 1/2 k k 1 k (x) = + h i x (x + 6) , x(x + 6) < 1. (3.2) G 2 2 k! | | kX=1 This implies that the function (x) = G( x) is inﬁnitely di↵erentiable at x = 0. G Utilizing the expansion (3.2) and di↵erentiating give

(m) 1 1/2 k k 1 k (m) 2 (0) = h i lim x (x + 6) G k! x 0 k=1 ! X ⇥ m ⇤ 1 1/2 k m k 1 (`) k (m `) = h i lim x (x + 6) k! x 0 ` k=1 ! `=0 ✓ ◆ X m X ⇥ ⇤ 1 1/2 k m k 1 (`) k (m `) = h i lim x lim (x + 6) x 0 x 0 k! ` ! ! kX=1 X`=0 ✓ ◆ m+1 ⇥ ⇤ 1/2 k m (m k+1) = h i (k 1)! lim (x + 6)k x 0 k! k 1 ! kX=1 ✓ ◆ m+1 ⇥ ⇤ 1/2 k m (m k+1) = h i lim (x + 6)k k k 1 x 0 k= (m+1)/2 ✓ ◆ ! d X e ⇥ ⇤ m+1 1/2 k m 2k m 1 = h i k m k+1 lim(x + 6) k k 1 h i x 0 k= (m+1)/2 ✓ ◆ ! d X e m+1 1/2 k m 2k m 1 = h i k m k+16 k k 1 h i k= (m+1)/2 ✓ ◆ d X e for m 1. From (1.2), it follows that (m)(0) = ( 1)mm!S , m 1. (3.3) G m Consequently,

(m) m m+1 m (0) 1 ( 1) 1/2 k m 2k m 1 Sm = ( 1) G = h i k m k+16 m! 2 m! k k 1 h i k= (m+1)/2 ✓ ◆ d X e for m 1, which can be rewritten as the formula (1.4). The ﬁrst proof of the formula (1.4) is complete.

Second proof of the formula (1.4). By Fa`a di Bruno’s formula (2.1), the mth deriva- INTEGERS: 16 (2016) 7 tive of the function px2 + 6x + 1 can be computed by

m 2 (m) (k) x + 6x + 1 = pu Bm,k(2x + 6, 2, 0, . . . , 0) k=0 p Xm 1 1/2 k = u B (2x + 6, 2, 0, . . . , 0) 2 m,k k=0⌧ k Xm 1 1/2 k = x2 + 6x + 1 B (2x + 6, 2, 0, . . . , 0) 2 m,k k=0⌧ k X m 1 Bm,k(6, 2, 0, . . . , 0) ! 2 k kX=0⌧ as x 0, where u = u(x) = x2 + 6x + 1 and m 0. By (2.2) and (2.3), we obtain ! m (m) 1 1/2 k x2 + 6x + 1 = x2 + 6x + 1 (m k)! 2 k=0⌧ k p X m k 2k m [2(x + 3)] (3.4) ⇥ k m k ✓ ◆✓ ◆ m 1 m k 2k m (m k)! 6 ! 2 k k m k kX=0⌧ ✓ ◆✓ ◆ as x 0 for m 0. The equation (1.2) can be rewritten as ! 1 x2 + 6x + 1 1 x = 2 ( 1)nS xn+1 n n=0 p X which implies that

n 2 (n+1) 2( 1) (n + 1)!Sn = lim x + 6x + 1 1 x , n 0. x 0 ! p Consequently, we obtain

n 1 ( 1) 2 (n+1) Sn = lim x + 6x + 1 x 0 2 (n + 1)! ! n n+1 p (3.5) 1 ( 1) 1 n + 1 k 2k n 1 = (n k + 1)! 6 2 (n + 1)! 2 k k n k + 1 kX=0⌧ ✓ ◆✓ ◆ for n 1. The formula (1.4) is thus proved again. Third proof of the formula (1.4). Applying Lemma 3 to the functions

u(x) = x2 + 6x + 1 1 x p INTEGERS: 16 (2016) 8 and v(x) = x yields

u(x) x 0 0 0 0 0 0 · · · u0(x) 1 x 0 0 0 0 0 · · · u00(x) 0 2 x 0 0 0 0 · · · u(3)(x) 0 0 3 0 0 0 0 ( 1)n (n) ...... ·.·.·...... (x) = n+1 . (3.6) G 2x (n 3) u (x) 0 0 0 n 3 x 0 0 (n 2) · · · u (x) 0 0 0 0 n 2 x 0 (n 1) · · · u (x) 0 0 0 0 0 n 1 x · · · u(n)(x) 0 0 0 0 0 0 n · · · By the formulas (2.4) and (2.5) and by induction, we have

x 0 0 0 0 0 0 1 x 0 · · · 0 0 0 0 0 · · · 1 0 2 x 0 0 0 0 n+(n+2)n · · · ( 1) B0 0 3 0 0 0 0C (n)(x) = u(n)(x) B · · · C G 2xn+1 B...... C B C B0 0 0 n 3 x 0 0C B · · · C B0 0 0 0 n 2 x 0C B · · · C B0 0 0 0 0 n 1 xC B · · · C @ A u(x) u0(x) 0 1 u00(x) 1 B u(3)(x) C B C 0 0 0 0 0 0 n u(n)(x) B...... C · · · B (n 3) C Bu (x)C B C Bu(n 2)(x)C B C Bu(n 1)(x)C B C @ A x 0 0 0 0 0 nu(x) · · · 1 x 0 0 0 0 nu0(x) · · · 0 2 x 0 0 0 nu00(x) · · · 1 0 0 3 0 0 0 nu(3)(x) = · · · 2xn+1 ...... (n 3) 0 0 0 n 3 x 0 nu (x) · · · (n 2) 0 0 0 0 n 2 x nu (x) · · · (n) (n 1) 0 0 0 0 0 n 1 xu (x) nu (x) · · · 1 = 2xn+1 INTEGERS: 16 (2016) 9

m x 0 0 0 ( 1) n mu(x) · · · m h i 1 x 0 0 ( 1) n mu0(x) · · · m h i 0 2 x 0 ( 1) n mu00(x) · · · m h i (3) 0 0 3 0 ( 1) n mu (x) · · · h i ...... m (n m 1) 0 0 0 x ( 1) n mu (x) · · · m h i 0 0 0 n m ( 1)k n xm ku(n k)(x) k=0 k · · · h i P n 1 1 x ( 1) n n 1u(x) = n+1 n 1 k h in k 1 (n k) 2x 1 ( 1) n x u (x) k=0 h ik n 1 1 P k n k (n k) n 1 = n+1 ( 1) n kx u (x) ( 1) n n 1u(x) 2x " h i h i # kX=0 n (n+1) 1 k n k (n k) lim ( 1) n kx u (x) ! 2(n + 1)! x 0 h i ! "k=0 # nX 1 n k k (k) (n+1) = lim ( 1) n n k x u (x) x 0 2(n + 1)! ! h i kX=0 n n+1 ⇥ ⇤ 1 n k n + 1 k (`) (n+k `+1) = ( 1) n n k lim x u (x) 2(n + 1)! h i ` x 0 k=0 `=0 ✓ ◆ ! h i Xn X 1 n k n + 1 (n+1) = ( 1) n n k k!u (0) 2(n + 1)! h i k k=0 ✓ ◆ X n 1 (n+1) n k n + 1 = u (0) ( 1) n n k k! 2(n + 1)! h i k kX=0 ✓ ◆ 1 = u(n+1)(0) 2(n + 1) as x 0, where 3 m n 2. This means, by considering (3.3), that !   1 ( 1)n S = u(n+1)(0), n 2 (n + 1)! which is same as the ﬁrst line in (3.5). Combining this with (3.4) recovers the formula (1.4). INTEGERS: 16 (2016) 10

Fourth proof of the formula (1.4). Employing (2.5) to compute (3.6) gives

1 x 0 0 0 0 0 · · · 0 2 x 0 0 0 0 · · · n 0 0 3 0 0 0 0 (n) ( 1) · · · (x) = ...... G 2xn+1 0 0 0 0 n 2 x 0 · · · 0 0 0 0 0 n 1 x · · · 0 0 0 0 0 0 n · · · T 1 x 1 x 0 0 0 0 u0(x) · · · 0 0 2 x 0 0 0 u00(x) · · · (3) 0 0 0 1 00 0 3 0 0 01 0 u (x) 11 · · · B B C B...... C B...... CC Bu(x) B· · ·C B C B (n 3) CC ⇥ B B 0 C B0 0 0 x 0 0C Bu (x)CC B B C B · · · C B (n 2) CC B B 0 C B0 0 0 n 2 x 0C Bu (x)CC B B C B · · · C B (n 1) CC B B 0 C B0 0 0 0 n 1 xC Bu (x)CC B B C B · · · C B CC B B 0 C B0 0 0 0 0 nC B u(n)(x) CC B B C B · · · C B CC @ @ A @ A @ AA u0(x) u00(x) 0 0 u(3)(x) 11 ( 1)nn! B B...... CC = Bu(x) x 0 0 0 0 0 0 P B (n 3) CC 2xn+1 B · · · Bu (x)CC B B (n 2) CC B Bu (x)CC B B (n 1) CC B Bu (x)CC B B CC B B (n) CC B B u (x) CC @ @ AA u0(x) u00(x) n 0 0 (3) 11 ( 1) n! x x2 x3 ( 1)n 1xn u (x) = n+1 u(x) 1 2 3 n 2x B h i1 h i2 h i3 · · · h in B...... CC B B (n 1) CC B ⇣ ⌘ Bu (x)CC B B CC B B u(n)(x) CC B B CC n @ n k @ AA ( 1) n! k+1 x (k) = n+1 u(x) ( 1) u (x) 2x k k " k=1 # X h i n+1 n k+1 ( 1) ( 1) (n+1) lim xku(k)(x) ! 2(n + 1) k! x 0 k=0 ! X ⇥ ⇤ n+1 n k+1 n+1 ( 1) ( 1) n + 1 k (`) (n `+k+1) = lim x u (x) 2(n + 1) k! x 0 ` k=0 ! `=0 ! X X n ( 1)n+1 ( 1)k+1 n + 1 = k!u(n+1)(0) 2(n + 1) k! k k=0 ! X n ( 1)n+1 n + 1 = u(n+1)(0) ( 1)k+1 2(n + 1) k k=0 ! X 1 = u(n+1)(0) 2(n + 1) INTEGERS: 16 (2016) 11 as x 0, where KT stands for the transpose of a matrix K and ! 2 n 3 n 3 n 2 n 2 n 1 n 1 1 x x ( 1) x ( 1) x ( 1) x 1 1 2 2 3 3 n 2 n 2 n 1 n 1 n n h i h i h i · · · h n i4 n 4 h n i3 n 3 hn i2 n 2 1 x ( 1) x ( 1) x ( 1) x 0 0 1 2 1 3 2 n 2 n 3 n 1 n 2 n n 1 h i h i · · · h n i5 n 5 h n i4 n 4 h ni 3 n 3 1 ( 1) x ( 1) x ( 1) x 0 0 B 3 1 n 2 n 4 n 1 n 3 n n 2 C B · · · C B...... h. .i...... h. .. .i...... h. .. .i...... h. i...... C P = B 2 3 C . B 0 0 0 x x x C B n 2 2 n 1 3 n 4 C B · · · h i h i h 2 i C B 0 0 0 1 x x C B · · · n 2 1 n 1 2 n 3 C B 0 0 0 h 0 i h 1 i h ix C B n 1 1 n 2 C B · · · h i h1 i C B 0 0 0 0 0 n C B · · · h i1 C @ A Similar to the above argument, we verify the formula (1.4) once again.

Fifth proof of the formula (1.4). Combining the equation (1.2) with (3.1) leads to

1 1/2 1 h ik [x(x + 6)]k 1 x = 2 ( 1)kS xk+1. k! k kX=0 kX=0 This equality can be reformulated as

k 1 1/2 k k k ` k ` 1 k 1 k h i x 6 x x = 2 ( 1) Sk 1x , k! ` kX=1 X`=0 ✓ ◆ kX=1 k 1 1/2 k k ` k k+` 1 k 1 k h i 6 x x = 2 ( 1) Sk 1x , k! ` kX=1 X`=0 ✓ ◆ kX=1 2k 1 1/2 k 2k m k m 1 k 1 k h i 6 x x = 2 ( 1) Sk 1x , k! m k k=1 m=k ✓ ◆ k=1 X mX X 1 1/2 k 2k m k m 1 m 1 m h i 6 x x = 2 ( 1) Sm 1x , k! m k m=1 k= m/2 ✓ ◆ m=1 X Xd e X m 1/2 k 2k m k m 1 h i 6 = 2( 1) Sm 1, k! m k k= m/2 ✓ ◆ Xd e m 1 m ( 1) 1/2 k 2k m k Sm 1 = h i 6 , 2 k! m k k= m/2 ✓ ◆ Xd e m m+1 ( 1) 1/2 k 2k m 1 k S = h i 6 , m 1. m 2 k! m k + 1 k= (m+1)/2 ✓ ◆ d X e The explicit formula (1.4) is thus proved once again. INTEGERS: 16 (2016) 12

Proof of the formula (1.5). The generating function (1.1) can be rearranged as

1 1 6 1 G(x) = 1 1 + , x > 0. 2 x x x2 ✓ r ◆ Then, by virtue of the formulas (2.1) and (2.6), we see that the nth derivative of G(x) for n N equals 2 n n (n) ( 1) n! (k) (n k+1) 2G (x) = pu B u0(x), u00(x), . . . , u (x) xn+1 n,k kX=0 n 1/2 k ( 1)nn! 1 6 1 = n+1 1 + 2 x 2 k x x kX=0⌧ ✓ ◆ 6 2 12 6 n k 6(n k + 1)! B , + , . . . , ( 1) ⇥ n,k x2 x3 x3 x4 xn k+2 ✓ n k+1 (n k + 2)! + ( 1) xn k+3 ◆ n 1/2 k ( 1)nn! 1 6 1 = n+1 1 + 2 x 2 k x x kX=0⌧ ✓ ◆ n 6 12 ` r 6(` r + 1)! B`,r 2 , 3 , . . . , ( 1) ` r+2 ⇥ ` x x x r+Xs=k `+Xm=n ✓ ◆ ✓ ◆ 2 6 m s+1 (m s + 2)! B , , . . . , ( 1) , ⇥ m,s x3 x4 xm s+3 ✓ ◆ 6 1 where u(x) = 1 x + x2 . Further making use of the formulas (2.2), (2.7), and (2.8) results in INTEGERS: 16 (2016) 13

n n 1/2 k ( 1) n! 1 6 1 2G(n)(x) = 1 + xn+1 2 x x2 k=0 k X⌧ ✓ ◆ n r `+r 1 6 ( 1) B`,r(1!, 2!, . . . , (` r + 1)!) ⇥ ` x`+r r+s=k `+m=n ! X X m 1 ( 1) Bm,s(2!, 3!, . . . , (m s + 2)!) ⇥ xm+2s n n 1/2 k ( 1) n! 1 6 1 n = 1 + 6r( 1)m+`+r xn+1 2 x x2 ` k=0 k r+s=k `+m=n ! X⌧ ✓ ◆ X X s 1 ` ` 1 1 s q s (m + 2q 1)! (` r)! ( 1) ⇥ xm+2s+`+r r r 1 s! q (2q 1)! ! ! q=0 ! X n n n 1/2 k ( 1) n! ( 1) 1 6 1 1 n = 1 + 6r xn+1 xn 2 x x2 xk ` k=0 k r+s=k `+m=n ! X⌧ ✓ ◆ X X r s ( 1) ` ` 1 1 s q s (m + 2q 1)! (` r)! ( 1) ⇥ xs r r 1 s! q (2q 1)! ! ! q=0 ! X n n ( 1) 1 2 1/2 k k n r r = n! x 6x + 1 x 6 ( 1) xn+1 2 ` " k=0 k r+s=k `+m=n ! X⌧ X X s 1 ` ` 1 1 s q s (m + 2q 1)! (` r)! ( 1) ⇥ xs r r 1 s! q (2q 1)! ! ! q=0 ! # X n ( 1)n 1 n ` ` 1 lim n! 6r( 1)r ! (n + 1)! x 0+ 2 ` r r 1 " k=0 k r+s=k `+m=n ! ! ! ! X⌧ X X s (n+1) (` r)! s q s (m + 2q 1)! 2 1/2 k k s ( 1) x 6x + 1 x ⇥ s! q (2q 1)! q=0 ! # X n ( 1)n+1 1 n ` ` 1 (` r)! = 6r( 1)r (n + 1)! 2 ` r r 1 s! k=0 k r+s=k `+m=n ! ! ! X⌧ X X s s q s (m + 2q 1)! 2 1/2 k k s (n+1) ( 1) lim x 6x + 1 x ⇥ q (2q 1)! x 0+ q=0 ! ! X ⇥ ⇤ n+1 n s ( 1) 1 n r r ` ` 1 (` r)! s q = 6 ( 1) ( 1) (n + 1)! 2 ` r r 1 s! k=0 k r+s=k `+m=n ! ! ! q=0 X⌧ X X X n+1 s (m + 2q 1)! n + 1 2 1/2 k (n p+1) k s (p) lim x 6x + 1 x ⇥ q (2q 1)! x 0+ p ! ! p=0 ! X ⇥ ⇤ n+1 n s ( 1) 1 n r r ` ` 1 (` r)! s q = 6 ( 1) ( 1) (n + 1)! 2 ` r r 1 s! k=0 k r+s=k `+m=n ! ! ! q=0 X⌧ X X X s (m + 2q 1)! n + 1 2 1/2 k (n k+s+1) (k s)! lim x 6x + 1 ⇥ q (2q 1)! k s x 0+ ! ! ! ⇥ ⇤ INTEGERS: 16 (2016) 14 as t 0+. By the formulas (2.1), (2.2), and (2.3), we have ! 1/2 k (n k+s+1) lim x2 6x + 1 x 0+ ! ⇥n k+s 1 ⇤ 1/2 k (j) = lim u Bn k+s+1,j(2x 6, 2, 0 . . . , 0) x 0+ ! j=0 X n k+s 1 1 1/2 k j = lim k u Bn k+s+1,j(2x 6, 2, 0 . . . , 0) x 0+ 2 j=0 j ! X ⌧ n k+s 1 1 = k Bn k+s+1,j( 6, 2, 0 . . . , 0) 2 j=0 j X ⌧ n k+s 1 2j n+k s 1 1 (n k j + s + 1)! = ( 1) k 2 6n k 2j+s+1 j=0 j X ⌧ n k + s + 1 j , ⇥ j n k j + s + 1 ✓ ◆✓ ◆ where u = u(x) = x2 6x + 1. As a result, from the generating function (1.1), it follows that 1 S = G(n)(0) n n! 1 ( 1)n+1 n 1 n ` ` 1 (` r)! = 6r( 1)r 2 (n + 1)! 2 k ` r r 1 s! kX=0⌧ r+Xs=k `+Xm=n ✓ ◆ ✓ ◆✓ ◆ s n k+s 1 s q s (m + 2q 1)! n + 1 2j n+k s 1 ( 1) (k s)! ( 1) ⇥ q (2q 1)! k s q=0 j=0 X ✓ ◆ ✓ ◆ X 1 (n k j + s + 1)! n k + s + 1 j k ⇥ 2 6n k 2j+s+1 j n k j + s + 1 ⌧ j ✓ ◆✓ ◆ which can be simpliﬁed as the formula (1.5). The proof of Theorem 1 is complete.

Remark 1. This paper is a slightly revised version of the preprint [11] and a companion of the preprint [10].

Acknowledgements. The authors thank Professor Wiwat Wanicharpichat at Naresuan University in Thailand for recommending his paper [16] through Research- Gate on 3 January 2016. The authors appreciate Professor Bruce Landman for his careful corrections in English and mathematics to the original version of this paper. INTEGERS: 16 (2016) 15

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