Linear Recurrence Relations for Sums of Products of Two Terms

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1 Linear Recurrence Relations for Sums of Products of Two Terms Yan-Ping Mu College of Science, Tianjin University of Technology Tianjin , P.R. China Submitted: Dec 27, 2010; Accepted: Aug 16, 2011; Published: Aug 26, 2011 Mathematics Subject Classification: 68W30, 33F10, 05A19 Abstract For a sum of the form F(n,)G(n,), we set up two systems of equations involving shifts of F(n,) and G(n,). Then we solve the systems by utilizing the recursion of F(n, ) and the method of undetermined coefficients. From the solutions, we derive linear recurrence relations for the sum. With this method, we prove many identities involving Bernoulli numbers and Stirling numbers. 1. Introduction Finding recurrence relations is a basic method for proving identities. Fasenmyer [4 6] proposed a systematic method to find linear recurrence relations for hypergeometric sums. Wilf and Zeilberger [10 13] provided an efficient algorithm, called Zeilberger s algorithm, to construct linear recurrence relations for hypergeometric sums. Chyza [2] extended Zeilberger s algorithm to holonomic systems. Kauers [8] presented algorithms for sums involving Stirling-lie sequences. Chyza, Kauers and Salvy [3] further considered nonholonomic systems. In this paper, we focus on deriving a linear recurrence relation for the sum f(n) = = F(n, )G(n, ), (1.1) where n = (n 1,...,n r ) and F(n, ), G(n, ) are two functions of n and. Our approach can be described as follows. We first tae a finite subset S of Z r+1 as the set of shifts of Supported by the National Natural Science Foundation of China (Project ). the electronic journal of combinatorics 18 (2011), #P170 1

2 the variables n and. Then we mae an ansatz λ α,β (n, )F(n α, β) = 0, (1.2) (α,β) S where λ α,β (n, ) are functions to be determined. Denote and A S = {α Z r : there exists β Z such that (α, β) S} S α = {β Z: (α, β) S}. For each α A S, we tae a finite subset S α of Z r and mae an ansatz β S α λ α,β (n, + β)f(n α, )G(n, + β) = γ S α c α,γ (n)f(n γ, )G(n γ, ), (1.3) where c α,γ (n) are functions which are independent of and need to be determined. The system of equations consisting of (1.2) and (1.3) for all α A S is called a coupling system. Suppose that we obtain a solution (λ α,β (n, ), c α,γ (n)) to the coupling system such that c α,γ (n) are not all zeros. Then (by Lemma 2.1) we are led to a non-trivial linear recurrence relation for f(n): c α,γ (n)f(n γ) = 0. (1.4) α A S γ S α We mainly investigate the case in which F(n, ) is independent of n 1, n 2,..., n s and S α {α + l 1e l s e s : l i Z, i = 1, 2,..., s}, where each e i Z r is the unit vector whose i-th component is 1. In this case, Equation (1.3) reduces to λ α,β (n, + β)g(n, + β) = c α,γ (n)g(n γ, ). (1.5) β S α We call such a coupling system a split system. We will see that both Sister-Celine s method and Zeilberger s algorithm fall into the framewor of split systems. We use the above method to prove identities of sums involving special combinatorial sequences. We first split the summand into a product of two terms F(n, ) and G(n, ) so that F(n, ) depends on as few variables as possible and satisfies a simple recurrence relation. Then by solving the equations (1.2) and (1.5), we get a recurrence relation for the sum. Finally we prove the identity by checing the initial values. Notice that the method given in [3] treats the summand as a whole. The paper is organized as follows. In Section 2, we prove the recursion (1.4) and the existence of non-trivial solutions to a ind of split systems. In Section 3, we consider the case in which λ α,β (n, ) are given a priori. In Section 4, we study the split systems in which λ α,β (n, ) can be expressed in terms of c α,γ (n). γ S α the electronic journal of combinatorics 18 (2011), #P170 2

3 2. Split systems We follow the notation of Section 1. The following lemma is valid for a general coupling system. Lemma 2.1 Suppose that λ α,β (n, ) and c α,γ (n) satisfy (1.2) and (1.3). Then (1.4) holds. Proof. By direct calculation, we see that 0 = = (α,β) S = α A S λ α,β (n, )F(n α, β)g(n, ) λ α,β (n, + β)f(n α, )G(n, + β) = β S α = c α,γ (n)f(n γ, )G(n γ, ) α A S = α A S = γ S α γ S α c α,γ (n)f(n γ). From now on until the end of the paper, we always assume that the coupling systems are split. Consider the split system with r = 1, F(n, ) = 1, S = {(j, j): 0 j J} and S j = {0, 1, 2,..., I}, 0 j J, where I, J are two non-negative integers. Then (1.5) reduces to λ j,j (n, + j)g(n, + j) = I c j,i (n)g(n i, ), i=0 so that λ j,j (n, ) = I c j,i (n)g(n i, j)/g(n, ). i=0 Now substitute the above equation into (1.2). We finally obtain an equation J j=0 I c j,i (n)g(n i, j) = 0, i=0 which is the same as the one appearing in Sister-Celine s method. Now consider the split system with r = 1, F(n, ) = 1, S = {(0, 0), (0, 1)} and S 0 = {0, 1,..., I}. the electronic journal of combinatorics 18 (2011), #P170 3

4 By equation (1.2), we derive that λ 0,0 (n, ) = λ 0,1 (n, ). Thus, (1.5) reduces to λ 0,1 (n, + 1)G(n, + 1) λ 0,1 (n, )G(n, ) = I c i (n)g(n i, ), i=0 which is the sew recurrence relation appearing in Zeilberger s algorithm. Therefore, both Sister-Celine s method and Zeilberger s algorithm fall into the framewor of split systems. The following theorem ensures the existence of non-trivial solutions to a ind of split systems. Theorem 2.2 Suppose that F(n, ) is independent of n 1,...,n s and satisfies a nontrivial linear recurrence relation F(n, ) = a α,β (n, )F(n α, β), (2.1) where R is a finite subset of (α,β) R {(0,...,0, n s+1,...,n r, ) Z r+1 } and a α,β (n, ) are rational functions of n and. Assume that G(n, ) is a proper hypergeometric term (see [10] for the definition). Then there exist S and S α such that the corresponding split system has a non-trivial solution (λ α,β (n, ), c α,γ (n)) with λ α,β (n, ) being polynomials in. Proof. We will set up a system of linear equations on λ α,β (n, ) and c α,γ (n) according to (1.2) and (1.5). Then the theorem follows from the fact that the number of unnowns is larger than that of equations. Assume that D λ α,β (n, ) = λ α,β,l (n) l. (2.2) We tae and l=0 S = {(0,..., 0, n s+1,...,n r, ): 0 I 0, 0 n j I j, j = s + 1,..., r} S α = {α + l 1 e l s e s : 0 l i I i, i = 1, 2,..., s}. Then the corresponding split system leads to a system of linear equations on λ α,β,l (n) and c α,γ (n). We will derive an upper bound for the number of equations in terms of D and I 0, I 1,...,I r. We first consider (1.2). Without loss of generality, we may assume that each element in R is strictly greater than the zero vector in the lexicographic order. Let S R = {(α, β) S : there exists (α, β ) R such that (α, β) + (α, β ) S} (2.3) the electronic journal of combinatorics 18 (2011), #P170 4

5 be the boundary set with respect to the recurrence relation (2.1). By iterating use of the recursion (2.1), we can express the terms F(n α, β) with (α, β) S in terms of those with (α, β) S R. Therefore, λ α,β (n, )F(n α, β) = µ α,β (n, )F(n α, β), (2.4) (α,β) S (α,β) S R where µ α,β (n, ) = λ α,β (n, ) + A α,β,α,β (n, )λ α,β (n, ), (2.5) (α,β ) S\S R and A α,β,α,β (n, ) are linear combinations of terms of the form a αi,β i (n δ i, δ i ). i By setting all µ α,β (n, ) to zeros, we reduce (1.2) to a system of linear equations on λ α,β (n, ). The number of equations of the system equals the cardinality of S R, which is bounded by a multi-variable polynomial P 1 in I 0, I s+1,...,i r of total degree r s. By multiplying the common denominators, we transfer the coefficients of the system into polynomials in. The degrees of these polynomials are bounded by P 2 = C(I 0 +1)(I s+1 + 1) (I r +1), where C is a constant. Now substituting (2.2) into the system and equating the coefficient of each power of to zero, we finally obtain a system of linear equations on λ α,β,l (n). The number of equations of the new system is bounded by P 1 (P 2 + D + 1). We next consider (1.5). Dividing G(n, ) on both sides and multiplying the common denominators, we are led to a system of linear equations on λ α,β (n, ) and c α,γ (n). The number of equations of the system is equal to the cardinality of A S, which is (I s+1 + 1) (I r +1). Since G(n, ) is a proper hypergeometric term, the coefficients of the system are polynomials in whose degrees are bounded by a linear function P 3 of I 0, I 1,..., I r (see [11]). Once again, we substitute (2.2) into the system and equate the coefficient of each power of to zero. This leads to a system of linear equations on λ α,β,l (n) and c α,γ (n). The number of equations of the new system is bounded by (P 3 +D+1)(I s+1 +1) (I r +1). Finally, we combine the equations deduced from (1.2) and (1.5) together. The total number of equations is bounded by E = P 1 (P 2 + D + 1) + (P 3 + D + 1)(I s+1 + 1) (I r + 1). While the total number of unnowns is U = (I 0 + 1)(I s+1 + 1) (I r + 1)(D + 1) + (I 1 + 1) (I r + 1). The leading coefficient of E in variable D is a polynomial in I 0, I s+1,...,i r of total degree r s. While the leading coefficient of U in variable D is (I 0 + 1)(I s+1 + 1) (I r + 1). Hence U > E holds for sufficiently large I 0, I 1,..., I r and D, which implies that the split system has a non-trivial solution. Remar. For s > 1, the product (I 1 + 1) (I r + 1) which is the number of unnowns c α,γ (n) will be larger than E for sufficiently large I 1,..., I s. Thus we derive that c α,γ (n) are not all zeros. But for s = 1, we could not ensure that c α,γ (n) are not all zeros. the electronic journal of combinatorics 18 (2011), #P170 5

6 3. Split systems with given λ In this section, we consider split systems in which λ α,β (n, ) are given a priori. Assume that F(n, ) is independent of n 1,...,n s and satisfies a linear recurrence relation of the form (2.1). We tae S = R {0} and set λ α,β (n, ) = { aα,β (n, ), (α, β) R, 1, (α, β) = 0. It is straightforward to see that those λ α,β (n, ) satisfy (1.2). The remaining tas is to solve (1.5). Noting that G(n, ) is clearly equal to itself, we need only solve the following equations: a α,β (n, + β)g(n, + β) = c α,γ (n)g(n γ, ), α A R, (3.1) β R α where γ S α A R = {α: there exists β Z such that (α, β) R}, and R α = {β Z: (α, β) R}. Each solution {c α,γ (n)} to (3.1) leads to a recurrence relation f(n) = c α,γ (n)f(n γ) for f(n) = α A R γ S α = F(n, )G(n, ). We illustrate the method by an identity involving Bernoulli numbers B. Example 3.1 We have m =0 ( ) m B n+ = ( 1) m+n n =0 ( ) n B m+. (3.2) Chen and Sun [1] found a recurrence relation satisfied by both sides based on the integral representation of B. Here we provide a proof in the framewor of split systems. Proof. Let ( ) m F(n, m, ) = and G(n, m, ) = B n+. We see that F(n, m, ) is independent of n and F(n, m, ) = F(n, m 1, ) + F(n, m 1, 1). the electronic journal of combinatorics 18 (2011), #P170 6

7 In this case, (3.1) becomes G(n, m, ) + G(n, m, + 1) = l c l (n, m)g(n l, m 1, ). (3.3) Observing that G(n, m, ) = G(n, m 1, ) and G(n, m, + 1) = G(n + 1, m 1, ), we obtain a solution c 0 = c 1 = 1 to (3.3). Therefore, the sum satisfies the recurrence relation f(n, m) = m =0 ( ) m B n+ f(n, m) = f(n, m 1) + f(n + 1, m 1). (3.4) Similarly, by taing F(n, m, ) = ( 1) n ( n ) and G(n, m, ) = ( 1) m B m+, we derive that the sum satisfies which is equivalent to (3.4). Finally, from the identity we see that g(n, m) = ( 1) m+n n =0 ( ) n B m+ g(n, m) = g(n 1, m) + g(n 1, m + 1), x n B n n! = x e x 1 = x e x e x 1, n=0 B n = ( 1) n n =0 ( ) n B, that is, (3.2) holds for m = 0. Hence by the recurrence relation (3.4), (3.2) holds for all m, n 0. By a similar argument as above, we prove most of the identities in [1]. We always tae F to be a binomial coefficient which satisfies a triangular recurrence relation. As another example, we derive the recurrence relation satisfied by both sides of a convolution identity for Bernoulli numbers. the electronic journal of combinatorics 18 (2011), #P170 7

8 Example 3.2 We have [1, Theorem 4.4] n ( ) n B +j B m+n j j where j=0!m! = (m + + 1)! (n + δ(m, )(m + + 1))B m+n+ m+ ( ) ( + ( 1) r B m++1 r r m r ( 1) r + 1 n r=0 m+ ( ) ( + ( 1) r B m++1 r m + 1 m + 1 r m r ( 1)m r m + 1 n r=0 rm ) B n+r r m + 1 1, if (m, ) = (0, 0), δ(m, ) = 0, if m = 0 but (m, ) (0, 0), 1, otherwise. We first consider the sum L(m, n, ) on the left hand side. Let ( ) n F(m, n,, j) = and G(m, n,, j) = B +j B m+n j. j ) B n+r 1, (3.5) We have Observing that F(m, n,, j) = F(m, n 1,, j) + F(m, n 1,, j 1). G(m, n,, j) = G(m + 1, n 1,, j) and G(m, n,, j + 1) = G(m, n 1, + 1, j), we derive that which is equivalent to r=0 L(m, n, ) = L(m + 1, n 1, ) + L(m, n 1, + 1), L(m, n + 1, ) = L(m + 1, n, ) + L(m, n, + 1). Now let us consider the right hand side of (3.5). We split the first sum into the difference of m+ ( ) n ( 1) r B m++1 r m r ( 1) B n+r 1 = n R 1 (m, n, ) r and ( 1) r B m++1 r m r ( 1) m+ m r=0 ( ) B n+r 1 = m R 2 (m, n, ). r 1 the electronic journal of combinatorics 18 (2011), #P170 8

9 By taing F = ( ) ( r and F = r 1), respectively, we derive that R i (m, n, ) = R i (m + 1, n, 1) + R i (m, n + 1, 1), i = 1, 2, which is equivalent to Similarly, let we have Let R i (m, n + 1, ) = R i (m + 1, n, ) + R i (m, n, + 1), i = 1, 2. m+ ( ) R 3 (m, n, ) = ( 1) r B m++1 r m m r ( 1)m B n+r 1, r r=0 m+ ( ) R 4 (m, n, ) = ( 1) r B m++1 r m m r ( 1)m B n+r 1, r 1 r=0 R i (m, n + 1, ) = R i (m + 1, n, ) + R i (m, n, + 1), i = 3, 4.!m! R 5 (m, n, ) = (m + + 1)! (n + δ(m, )(m + + 1))B m+n+, by direct calculation, we derive that R 5 (m, n + 1, ) = R 5 (m + 1, n, ) + R 5 (m, n, + 1). Finally, let R(m, n, ) be the right hand side of (3.5). Replacing the index of summation r in R 2 (m + 1, n, ) and R 4 (m, n, + 1) by r + 1, we see that R(m, n + 1, ) R(m + 1, n, ) R(m, n, + 1) = R 1 (m, n + 1, ) + R 2 (m + 1, n, ) + R 3 (m, n + 1, ) + R 4 (m, n, + 1) = 0. Hence both sides of (3.5) satisfy the same recurrence relation. We can also apply this method to sums involving Stirling numbers (Eulerian numbers, respectively). Let S 1 (n, ) and S 2 (n, ) be the Stirling numbers of the first ind and the second ind, respectively. It is well-nown that S 1 (n, ) = (n 1)S 1 (n 1, ) + S 1 (n 1, 1), S 2 (n, ) = S 2 (n 1, ) + S 2 (n 1, 1). With these recursions, we prove identities (6.15) (6.19), (6.28), (6.29), (6.38), and (6.39) of [7]. Here are two examples. the electronic journal of combinatorics 18 (2011), #P170 9

10 Example 3.3 Find a recurrence relation for the sum [7, Identity (6.28)] Let We have F(l, m, n, ) = In this case, (3.1) becomes f(l, m, n) = ( ) n n m =l and ( ) n S 2 (, l)s 2 (n, m). G(l, m, n, ) = S 2 (, l)s 2 (n, m). F(l, m, n, ) = F(l, m, n 1, ) + F(l, m, n 1, 1). G(l, m, n, ) + G(l, m, n, + 1) = i,j c i,j (l, m, n)g(l i, m j, n 1, ). (3.6) Substituting and S 2 (n, m) = ms 2 (n 1, m) + S 2 (n 1, m 1) S 2 ( + 1, l) = ls 2 (, l) + S 2 (, l 1) into the left hand side of (3.6), we find a solution Thus, c 0,0 = (m + l), c 1,0 = c 0,1 = 1. f(l, m, n) = (m + l)f(l, m, n 1) + f(l, m 1, n 1) + f(l 1, m, n 1). Example 3.4 We have [7, Identity (6.19)] m =0 ( ) m n ( 1) m = m!s 2 (n, m). (3.7) Proof. Let F(m, n, ) = n and G(m, n, ) = ( 1) m ( m ). We have F(m, n, ) = F(m, n 1, ). We tae S 0,1 = {(0, 1), (1, 1)} so that (3.1) becomes G(m, n, ) = c 0 (m, n)g(m, n 1, ) + c 1 (m, n)g(m 1, n 1, ). the electronic journal of combinatorics 18 (2011), #P170 10

11 Dividing both sides by G(m, n, ), we derive that which has a solution Therefore, the sum satisfies = c 0 (m, n) c 1 (m, n) m m, c 0 (m, n) = c 1 (m, n) = m. f(m, n) = m =0 ( ) m n ( 1) m f(m, n) = mf(m, n 1) + mf(m 1, n 1). It is easy to chec that m!s 2 (n, m) satisfies the same recursion and that The proof follows by induction on n. f(m, 0) = δ m,0 = m!s 2 (0, m). 4. Partially λ-free split systems In this section, we consider a ind of split systems in which λ α,β (n, ) can be expressed in terms of c α,γ (n). Suppose that F(n, ) is independent of n 1,...,n s and satisfies a recurrence relation of the form (2.1). The proof of Theorem 2.2 provides us a method to construct a solution to (1.2). Given an arbitrary shift set S, let S R be the boundary set defined by (2.3). For each (α, β) S \ S R, let λ α,β (n, ) be an arbitrary function. For each (α, β) S R, we set λ α,β (n, ) = A α,β,α,β (n, )λ α,β (n, ), (α,β ) S\S R where A α,β,α,β (n, ) is given as in (2.4) and (2.5). Then these λ α,β(n, ) form a solution to (1.2). Denote the free functions λ α,β (n, ), (α, β) S \ S R by λ 1, λ 2,.... Suppose that the equations corresponding to (1.5) can be ordered such that the first equation involves only λ 1 and there is exactly one term containing λ 1, the second equation involves only λ 1, λ 2 and there is exactly one term containing λ 2, and so on. Then we can express λ i s as linear combinations of c α,γ (n). Finally, substituting these expressions for λ i into the rest equations of (1.5), we obtain a system of linear equations on c α,γ (n). We call such a split system a partially λ-free split system. We illustrate the method by a summation involving Stirling numbers. Example 4.1 Find a recurrence relation for the sum [7, Identity (6.26)] m ( )( ) m n m + n f(n, m) = S 1 (m +, ). m + n + = n the electronic journal of combinatorics 18 (2011), #P170 11

12 Let Then Taing F(n, m, ) = S 1 (m +, ) and G(n, m, ) = ( m n m + )( m + n n + F(n, m, ) = (m + 1)F(n, m 1, ) + F(n, m, 1). (4.1) S = {(0, 0, 0), (0, 1, 0), (0, 0, 1)}, we see that S R = {(0, 1, 0), (0, 0, 1)} and there is exactly one free function λ 0,0,0 (n, m, ). Denote the function by λ() for short. We have λ 0,1,0 (n, m, ) = (m + 1)λ() and λ 0,0,1 (n, m, ) = λ(). Then (1.5) becomes { (m + 1)λ()G(n, m, ) = i c i(n, m)g(n i, m 1, ) λ()g(n, m, ) λ( + 1)G(n, m, + 1) = j d j(n, m)g(n j, m, ). From the first equation, we derive that ). λ()g(n, m, ) = i c i (n, m)g(n i, m 1, )/(m + 1). Substituting this expression into the second equation, we obtain ( ) G(n i, m 1, ) G(n i, m 1, + 1) c i (n, m) m + 1 m + i = j d j (n, m)g(n j, m, ). By setting i, j { 1, 0, 1}, we find a non-trivial solution and (m n)(m n 1) c 1 = 0, c 0 = C, c 1 = (m n)c, m + n 1 d 1 = n m m + n 1 C, d 0 = 2n 1 m + n 1 C, d 1 = C, λ() = (m + )( m + ) (n + 1)(m + n 1)(m + n) C, where C is an arbitrary constant with respect to. Hence, the sum f(n, m) satisfies the recurrence relation (m n)(m n 1)f(n, m 1) + (m n)(m + n 1)f(n 1, m 1) (2n 1)f(n, m) + (n m)f(n + 1, m) + (m + n 1)f(n 1, m) = 0. We conclude by an example involving Stirling numbers and harmonic numbers. the electronic journal of combinatorics 18 (2011), #P170 12

13 Example 4.2 Let H = for the following sum [9] f(m, n) = i=1 m =1 1 be the -th harmonic number. Find a recurrence relation i ( ) m H m (m )!( 1) m +1 S 1 ( 1, n). 1 Let F(m, n, ) = S 1 ( 1, n) and ( ) m G(m, n, ) = H m (m )!( 1) m Then Taing we see that F(m, n, ) = F(m, n 1, 1) ( 2)F(m, n, 1). S = {(0, 0, 0), (0, 0, 1) (0, 1, 0), (0, 1, 1), (0, 2, 1)}, S R = {(0, 0, 1), (0, 1, 1), (0, 2, 1)}. Hence, we have two free functions λ 0,0,0 (m, n, ) and λ 0,1,0 (m, n, ). Denote them by λ() and µ() for short. We have λ 0,0,1 (m, n, ) = ( 2)λ(), λ 0,1,1 (m, n, ) = λ() + ( 2)µ(), and λ 0,2,1 (m, n, ) = µ(). Then (1.5) becomes µ( + 1)G(m, n, + 1) = i c i(m, n)g(m i, n 2, ), µ()g(m, n, ) + ( λ( + 1) + ( 1)µ( + 1))G(m, n, + 1) λ()g(m, n, ) + ( 1)λ( + 1)G(m, n, + 1) = j d j(m, n)g(m j, n 1, ), = l e l(m, n)g(m l, n, ). Set i, j, l { 1, 0, 1}, express H m +t in terms of H m, and compare the coefficients of H m. We find a non-trivial solution c 1 = 0, c 0 = 0, c 1 = C, d 1 = 0, d 0 = 2C, d 1 = 2(m 1)C, e 1 = C, e 0 = (2m 1)C, e 1 = (m 1) 2 C, and ( 1)C µ() = m, λ() = ( 2)H m 2m + 2 ( 1)C, mh m (m + 2) the electronic journal of combinatorics 18 (2011), #P170 13

14 where C is an arbitrary constant with respect to. Therefore, the sum f(m, n) satisfies the recurrence relation (2m 1)f(m, n) + f(m 1, n 2) 2f(m, n 1) 2(m 1)f(m 1, n 1) + f(m + 1, n) + (m 1) 2 f(m 1, n) = S 1 (m 1, n). The non-homogenous part S 1 (m 1, n) comes from the boundary values. Acnowledgments. The author would lie to than Professor Qing-Hu Hou and the referee for useful comments. References [1] W.Y.C. Chen and L.H. Sun, Extended Zeilberger s algorithm for identities on Bernoulli and Euler polynomials, J. Number Theory 129 (2009) [2] F. Chyza, Holonomic systems and automatic proofs of identities, Research Report 2371, Institut National de Recherche en Informatique et en Automatique, [3] F. Chyza, M. Kauers, and B. Salvy, A non-holonomic systems approach to special function identities, In: ISSAC 09: Proceedings of the 22nd International Symposium on Symbolic and Algebraic Computation, pp , ACM, New Yor, USA, [4] M.C. Fasenmyer, Some generalized hypergeometric polynomials, Ph.D. Thesis, University of Michigan, November, [5] M.C. Fasenmyer, Some generalized hypergeometric polynomials, Bull. Amer. Math. Soc. 53 (1947) [6] M.C. Fasenmyer, A note on pure recurrence relations, Amer. Math. Monthly 56 (1949) [7] R.L. Graham, D.E. Knuth, and O. Patashni, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, Reading, Massachusetts, [8] M. Kauers, Summation algorithms for Stirling number identities, J. Symbolic Comput. 42 (2007) [9] M. Kauers and C. Schneider, Automated proofs for some Stirling number identities, Electron. J. Combin. 15(1) (2008) R2. [10] M. Petovše, H.S. Wilf, and D. Zeilberger, A=B, A.K. Peters, Wellesley, MA, [11] H. Wilf and D. Zeilberger, An algorithmic proof theory for hypergeometric (ordinary and q ) multisum/integral identities, Invent. Math. 108 (1992) [12] D. Zeilberger, A fast algorithm for proving terminating hypergeometric identities, Discrete Math. 80 (1990) [13] D. Zeilberger, The method of creative telescoping, J. Symbolic Comput. 11 (1991) the electronic journal of combinatorics 18 (2011), #P170 14