An algebraic approach to a class of combinatorial sums Donatella Merlini, Renzo Sprugnoli, M. Cecilia Verri Dipartimento di Sistemi e Informatica Via Lombroso 6/17|Firenze (Italy) e-mail:
[email protected] Abstract
We obtain a theorem that allows us to nd the generating function for some combinatorial sums related to non proper Riordan arrays. This function can be used to obtain a closed form for the sum (or an asymptotic evaluation). We give several examples to illustrate some practical applications of the theorem.
1 Introduction
A Riordan array D = fdn;k gn;k2N is an in nite lower triangular array characterized by a pair (d(t); h(t)) of formal power series. This concept was formally introduced under the name of Riordan array by Shapiro et al. in 1991 (see e.g. [13]), but a similar concept can be found in previous works, such as [1, 2, 7, 8, 9, 10, 11, 12]. In [14], Riordan arrays are used as a valid tool for solving combinatorial sums, i.e., for nding a closed form or an asymptotic evaluation by means of generating functions. More speci cally, Riordan arrays allow us to translate a sum P1 k dn;k fk into a transformation of the generating function f (t) for the sequence ffk gk2N and the pair of formal power series d(t) and h(t) de ning the Riordan array D. By applying the Lagrange Inversion Formula (see [4]), it is possible to extend the class of combinatorial sums that can be solved by means of Riordan arrays. In Section 2,Pwe begin byh introducing some notations and refer to a result illustrated in [14], where a sum P1k dn;k f^k with dn;k as above and f^kh coecients of the function f^ h (y) = f (t)jy th t = 1k f^kh yk , is transformed into an expression involving f (t) and the function d(t) of the Riordan array. This transformation allows us to prove the classical identities of Abel and Gould and many other interesting combinatorial sums in a very simple way but it only holds if h(t) is a formal power series such that h(0) 6= 0. This transformation is so important that, in Section 3, we extend it to functions h(t) having h(0) = 0; in particular, we prove Theorem 3.3 which generalizes the results of [14] and constitutes this paper's most important result. Finally, in Section 4, we provide some applications of this formula that are mostly related to some particular classes of numbers such as Catalan, Motzkin, Schroder and Stirling numbers. =0
[ ]
=0
[ ]
[ ]
365
=
( )
=0
[ ]
The present technics can be considered as an alternative to existing methods such as the \snake oil" by Wilf [16] and the Zeilberger's algorithm [17].
2 Riordan arrays and combinatorial sums
Let ffk gk2N be a sequence of P (real) numbers. The generating function f (t) of the sequence k n n is de ned as f (t) = Gffkg = 1 k fk t . The notation [t ]f (t) denotes the coecient of t in the Taylor development of f (t) around t = 0. A Riordan array (see Sprugnoli [14]) is an in nite lower triangular array fdn;k gn;k2N de ned by a pair (d(t); h(t)) of formal power series in the sense that dn;k = [tn]d(t)(th(t))k. We usually assume that d(0) 6= 0; when we also have h(0) 6= 0, the Riordan array is called proper. Fr having r 0 denotes the set of formal power series whose rst non-zero element is in position r. In this paper, we write [f (y) j y = g(t)] instead of the more traditional f (y)jy g t (g(0) = 0); as usual, n;m denotes the Kronecker delta. When dealing with combinatorial sums, Riordan arrays are a very powerful tool, as can be seen in the following theorem, (see, e.g., [14]): Theorem 2.1 Let D = (d(t); h(t)) be a Riordan array and let f (t) be the generating function of the sequence ffn gn2N : Then: 1 X dn;k fk = [tn]d(t)f (th(t)): =0
= ( )
k=0
Conversely, if f dn;k j n; k 2 N g is an in nite lower triangular array such that for every sequence ffk gk2N we have:
G
(X 1
k=0
)
dn;k fk = d(t)f (th(t));
where f (t) is the generating function of the sequence ffk gk2N , and d(t); h(t) are two formal power series not depending on f (t), then the triangle de ned by the Riordan array D = (d(t); h(t)) coincides with fdn;k g:
If f (t) and h(t) are formal power series and h(t) 2 F , we can de ne the following function: ^f h (y) = f (t) y = th(t) : (2.1) 0
[ ;0]
Since h(t) 2 F , the Lagrange Inversion Formula assures us that the functional equation y = th(t) has only one solution in a neighbourhood of t = 0, and thus f^ h (y) exists and is uniquely determined. The Lagrange Inversion Formula also allows us to compute the coecients f^kh of the series f^ h (y) = P1k f^kh yk in terms of f (t) and h(t). Consequently, we obtain f^ h = f , and: 0 f^kh = [yk]f^ h (y) = [yk] f (t) y = th(t) = k1 [tk? ] hf((tt))k ; k 6= 0: 0
[ ;0]
[ ;0]
[ ;0] 0
[ ;0]
[ ;0]
[ ;0]
=0
0
[ ;0]
1
366
Example 2.1 Let us consider f (t) = (1 ? t)? and h(t) = (1 ? t)? . By solving the equation th(t) = y for t; we nd: p1 + 4y 1 + 2 y ? t(y) = : 1
Thus we have:
2
2y
" 1 + 2y ? p1 + 4y # 1 + p1 + 4y 1 h ^ = : f (y) = 1 ? t t = 2y 2 [ ;0]
Moreover:
! 2(k ? 1) : k?1
0 k? f^kh = k1 [tk? ] hf((tt))k = k1 [tk? ](1 ? t) k? = (?1)k [ ;0]
1
1
2
1
2
The function f^ h (y) satis es the following very important property (which we base our results on): h ^ Lemma 2.2 If f (y) = f (t) y = th(t) and h(t) 2 F ; then [ ;0]
[ ;0]
0
f^ h (th(t)) = f (t):
(2.2)
[ ;0]
Proof: We have:
f^ h (th(t)) = [[ f (y)j z = yh(y)]j z = th(t)] = [ f (y)j th(t) = yh(y)] : But the equation th(t) = yh(y), having h(t) 2 F , only has one solution y(t) = t in a neighbourhood of t = 0, and so we obtain our proof. It is easy to verify that the relation (2.2) holds for the functions f (t) and h(Pt) in Example 2.1. Let us now consider a proper Riordan array D = (d(t); h(t)) and the sum 1k dn;k f^kh : We obtain the following result (see also [15]) from Theorem 2.1 and Lemma 2.2: Theorem 2.3 Let D = (d(t); h(t)) be a proper Riordan array and f (t) a formal power series, then 1 X dn;k f^kh = [tn]d(t)f (t): (2.3) [ ;0]
0
[ ;0]
=0
[ ;0]
k=0
Example 2.2 Let D be the proper Riordan array de ned by the functions d(t) = (1 ? t)? and h(t) = (1 ? t)? , and let f (t) = (1 ? t)? : We point out that h(t) and f (t) are the same 1
2
1
functions we discussed in Example 2.1. We easily nd: ! k 1 t n + k n dn;k = [t ] 1 ? t (1 ? t) k = 2k : By applying Theorem 2.3, we then get the following closed formula: 1 n + k ! 2(k ? 1) ! (?1)k? X n] 1 = [ t 1+ k (1 ? t) = n + 1: 2k k ?1 k 2
1
2
=1
367
3 The extended method Theorem 2.3 is a very important tool in computing combinatorial sums, and some other more signi cant examples of its applications are presented in Sprugnoli [14]. In this section, we generalize formula (2.3) to case h(t) 2 Fs? with s 1; while in the next section, we illustrate some of the formula's applications to some well-known classes of numbers. When s > 1; the equation y = th(t) has exactly s solutions t ; t ; : : : ; ts in a neighbourhood of t = 0 (see, e.g., Henrici [6]); these solutions have the following form: 1 X j = 1; : : : ; s; tj = m!sjm ym=s 1
1
2
m=1
where
m = m1 f[xm? ](1 + Q(x))?m=sg(hs? )?m=s 1
and
1
p
Q(x) = h?s? (hs x + hs x + ) !s = e 2si (i = ?1): k Let f (t) = P1 k fk t bePa formal power series; then for all j = 1; : : : ; s; we can denote the jk k=s composition f (tj (y)) as 1 k k !s y , where k onlyPdepends on k and fk . Each function f (t (y)) is not a formal power series but their sum s f (t (y)) is, in fact: 1 1
2
+1
=0
=0
j
1
s
j =1
j
1 s s X 1 s 1X 1X jk k=s 1 X X ! jk y k=s = f ( t j (y )) = k !s y = k s sj sj k sk j =1
=1
= 1s
=0
1 X m=0
=0
sms ym =
1 X m=0
=1
ms ym;
s if k = ms P s jk because j !s = 0 otherwise . We can therefore de ne the formal power series: s X f^ h s? (y) = 1s f (tj (y)); j =1
[ ;
1]
=1
(3.1)
where tj (y)h(tj (y)) = y for j = 1; : : : ; s. A similar approach is proposed in [3]. Example 3.1 Let us consider h(t) = t=(1 ? t)? 2 F (with s = 2) and f (t) = (1 ? t)p, with real p. By solving the equation th(t) = y for t; we obtain the following two values: py ?py t (y) = 1 + py ; t (y) = 1 ? py : 2
1
1
2
Consequently, we have: # " p p g (py) + g(?py) 1 1 ^f h (y) = (1 ? t (y)) + (1 ? t (y)) = 1 ; 2 2 (1 + py)p + (1 ? py)p = 2 [ ;1]
1
2
368
p n? with g(y) = (1 + y)?p = P1 (?y)n. Finally, by applying the well-known bisection n n formula, we obtain: 1 2n + p ? 1 1 X X yn: f^ h (y) = g n yn = 2 n n n =0
+
1
[ ;1]
2
=0
=0
Let us now evaluate f^ h s? (th(t)) = s Psj f (tj (th(t))): we begin by evaluating tj (th(t)). For all integers j = 1; : : :; s, tj (z) is a solution to the functional equation z = th(t) and so yj (t) = tj (th(t)) is a solution to the equation [ ;
1
1]
=1
yh(y) = th(t): (3.2) This, in turn, has exactly s solutions in a neighbourhood of t = 0 when h(t) belongs to Fs? (see Henrici [6], Th. 2.4f and Cor. 2.4g). Then, by solving equation 3.2, we get the s solutions yj (t), j = 1; : : :; s; such that: s X f^ h s? (th(t)) = 1s f (yj (t)): (3.3) j 1
[ ;
1]
=1
In Example 3.1, we obtained exactly s solutions to the equation 3.2. Unfortunately, there are some functions h(t) for which the solution to the equation 3.2, from a formal point of view, produces a number n s of functions. Therefore, the problem is to characterize the s solutions we are looking for. A very trivial check-over can be made by substituting the n functions in the equation 3.2 and by then discarding the n ? s ones that don't verify the equation. In the following theorem, we propose an alternative method for selecting the s solutions yj (t) from the n functions : Theorem 3.1 Let h(t) = ts? g(t) = ts? (hs? + hs t + hs t + : : :) be a formal power series belonging to Fs? , with s 1; the functional equation yh(y) = th(t) has s solutions yj (t), j = 1; : : :; s in a neighbourhood of t = 0. The solutions have the following properties: yj (t) is a formal power series belonging to F ; 1
1
1
+1
2
1
1
yj (t) = !sj hs?=s t + O(t ) where s hs? = 1; i.e., !sj hs?=s is a sth root of unity. Proof: For all j = 1; : : : ; s, yj (t) = tj (th(t)) = P1m m !sjm(th(t))m=s, and (tj (y))sg(tj (y)) = 1
1
1
2
1
1
1
1
1
=1
y; since tj (0) = 0 and (tj (y))s = s y + O(y ), we have: g(tj (y)) = sy +yO(y ) ; g(0) = 1s = hs? : Then, if tj (y) is a solution of y = th(t), we have the following condition: shs? = 1: (3.4) Let us now consider yj (t): by setting g(t) = hs? (1 + G(t)); with G(0) = 6 0; we obtain: 1 1 X X m =s m yj (t) = m!sjm (tsg(t))m=s = m!sjm hm=s s? t ((1 + G(t)) ) : 2
1
1
2
1
1
1
1
1
m=1
m=1
369
1
1
The previous expression and condition (3.4) prove the second part of the theorem. By the binomial theorem we also nd: 1 1=s ! X =s G(t)k ; (1 + G(t)) = k k and, since G(t) 2 F ; it follows that (1 + G(t)) =s is a formal power series belonging to F . This concludes our proof. p Example 3.2 Let us consider h(t) = t = 1 + t 2 F ; by solving the equation: p t = p1y+ y ; 1+t we obtain the following six functions: p p 1+i 3 t ?1 + i 3 t t ; f (t) = p3 ; ; f (t) = ? p3 f (t) = ? p3 1+t 2 1+t 2 1+t p p ?1 + i 3 t 1+i 3 t f (t) = t; f (t) = ; f ( t ) = ? : 2 2 By developing the rst three functions in a Mac Laurin series, we nd: f (t) = ?t + 31 t + O t ; p! p! p! p! 3 3 3 3 t +O t : 1 + i 1 ? i 1 + i 1 ? i ; f ( t ) = t ? t + O t t ? f (t) = 2 6 2 6 We have f (0) = f (0) = f (0) = 0; but their rst coecient is not a cubic root of unity. Conversely, the three polynomials f (t), f (t) and f (t) are the formal power series desired. 1
=0 1
1
2
0
3
2
3
3
3
3
1
2
3
4
5
4
4
2
2
3
6
1
1
3
3
7
7
4
3
7
3
4
5
6
We can extract the coecients f^kh s? from f^ h s? (y) by using a generalization of the Lagrange Inversion Formula (see [3]). This formula can be proven in the same way as the traditional one (see Goulden and Jackson [4]). Theorem 3.2 Let f (t) and h(t) = ts? g(t) 2 Fs? be two formal power series; the coecients f^kh s? of the function f^ h s? (y) de ned in formula (3.1) can be computed as follows: 8 > k=0 1 [tsk? ] f (t) k > 0: : ks g(t)k [ ;
1]
[ ;
1
[ ;
1]
[ ;
[ ;
1]
1
1]
0
1]
1
0
We are now ready to generalize Theorem 2.3 to the case h(t) 2 Fs? : Theorem 3.3 Let D = (d(t); h(t)) be a Riordan array with h(t) 2 Fs? and f (t) be a formal power series; then: s 1 X X 1 h s? n ^ (3.5) dn;k fk = s [t ]d(t) f (yj (t)) j k where yj (t)h(yj (t)) = th(t) for all j = 1; : : : ; s. Proof: The proof directly follows from Theorem 2.1 and formula (3.3). 1
1
[ ;
1]
=1
=0
370
4 Applications The following are some examples of combinatorial sums treated by Theorem 3.3: Example 4.1 Let us consider the Riordan array ! t 1 D = (1 ? t)p ; (1 ? t) 2
and the function f (t) = (1 ? t)p. By de nition, we nd: ! !k n + p ? 1 t 1 n dn;k = [t ] (1 ? t)p (1 ? t) = 2k + p ? 1 ; 2
2
and from Theorem 3.2:
! 0 (t) 1 p + 2 k f p h k ? ^ fk = 2k [t ] g(t)k = p + 2k 2k : [ ;1]
2
1
By solving the equation th(t) = yh(y), i.e. t =(1 ? t) = y =(1 ? y) , we nd: ?t : y (t) = t; y (t) = 1 ? 2t By making some evaluations, we get: ! p (1 ? t ) 1 f ( y ( t )) + f ( y ( t )) h = 2 1 + (1 ? 2t)p : f^ (th(t)) = 2 Finally, by applying Theorem 3.3, we obtain the following identity: ! 1 1 n+p?1 ! p X p + 2k = X dn;k f^kh = [tn]d(t)f^ h (th(t)) = 2 k + p ? 1 2 k p + 2 k k k ! 1 1 n = [t ] 2 1 + (1 ? 2t)p = 21 n; + 21 ?np (?2)n = 12 n; + p + nn ? 1 2n? : 2
1
2
2
2
1
[ ;1]
2
2
[ ;1]
[ ;1]
=0
=0
0
0
Example 4.2 Let us now consider the Riordan array
! n + k ? 2 or dn;k = n ? 2k D = 1 ? t; (1 ? t) and the function f (t) = 1=(1 ? t). By applying Theorem 3.2 to them, we get: ! 1 3 k ? 2 h f^k = ? 2k k ? 1 ; f^ h = 1; !
t
3
[ ;1]
[ ;1] 0
371
1
and, by solving the equation t =(1 ? t) = y =(1 ? y) for y; we obtain: p p1 ? 4t 3 t ? 1 ? (1 ? t ) 3 t ? 1 + (1 ? t ) 1 ? 4t : ; y ( t ) = y (t) = t; y (t) = 2t 2t Since the third expression is not a formal power series, we deduce that the correct solutions are y (t) and y (t). After some simplifying, we nd: p ^f h (th(t)) = 3 ? 2t + 1 ? 4t : 4(1 ? t) The [tn]d(t)f^ h (th(t)) is related to the Catalan numbers, de ned as Cn = [tn](1 ? p1 ?coecient 4t)=(2t), as follows: 1 n + k ? 2 ! 1 3k ? 2 ! 1 n + k ? 2 ! 1 3k ? 2 ! X X h dn; f^ ? n ? 2k 2k k ? 1 = n; ? n; ? k n ? 2k 2k k ? 1 = k ! p 1 ( ? 2 ) : 1 ? 4 t 2( n ? 1) 1 ? t ? 2 1 1 n ? n ? ? [ t ] = ? = ? 2 [t ] n; 2t 2 2n n ? 1 2 n; 2
1
3
2
1
2
3
3
2
2
2
[ ;1]
[ ;1]
[ ;1] 0 0
0
1
=1
=1
1
1
0
Therefore, for n > 0; the following identity holds: 1 n + k ? 2 ! 1 3k ? 2 ! X n ? 2k k k ? 1 = Cn? ? n; : k 1
(4.1)
1
=1
Example 4.3 In this example, we examine the following Riordan array: ! ! t n + 2 k ? 2 D = 1 ? t; (1 ? t) or dn;k = n ? 2k ; 4
and the same function f (t) as before. We easily obtain p ! 1 t ? 6t + 1 : t ? 3 ? 2(2 k ? 1) h h h f^k = ? 2k 2k ? 1 ; f^ = 1; f^ (th(t)) = 4(t ? 1) p This time, f^ h (th(t)) is related to Schroder's numbers Sn = [tn](1 + t ? t ? 6t + 1)=(4t): 1 n + 2k ? 2 ! 1 4k ? 2 ! 1 n + 2k ? 2 ! 1 4k ? 2 ! X X h ^ dn; f ? n ? 2k 2k 2k ? 1 = n ? 2k 2k 2k ? 1 = n; ? n; ? [ ;1]
[ ;1] 0
2
[ ;1]
[ ;1]
[ ;1] 0 0
k=1
2
0
p
1
k=1
p ? 3 + t ? 1 ? 6 t + t 1 ? 6t + t = ? S : 1 + t ? n? = ?[tn] = n; ? [t ] n; n? 4 4t 2
2
1
0
0
Therefore, the following identity holds: 1 n + 2k ? 2 ! 1 2(2k ? 1) ! X n ? 2k 2k 2k ? 1 = Sn? ? n; : k 1
=1
372
1
1
(4.2)
Example 4.4 This example is related to the Motzkin numbers de ned as Mn = [tn](1 ? t ? p 1 ? 2t ? 3t )=(2t ). We consider the Riordan array: ! n ? 3 k ? 1 D = (1; t(1 + t)) or dn;k = n ? 2k (?1)n 2
2
and the function f (t) = (1 + t). Among the solutions to the equation th(t) = yh(y); we p have to take y (t) = t and y (t) = (?1 ? t ? 1 ? 2t ? 3t )=2. Moreover, we note that h(t) = tf (t) and this yields: ! t f ( t ) t f ( t ) h f (y (t)) = y (t) ; f^ (th(t)) = 2 1 + y (t) : 1
2
2
2
2
Since f^kh = ?
k?2 k?1
2
2
[ ;1]
2
2
2
; by some simplifying, we obtain: p 1 n ? 3k ? 1 ! (?1)n 3k ? 2 ! X 1 ? 2t ? 3t = ? 1 M 1 ? t ? n ? = n; ? n; ? [t ] n; 2k 2k ? 1 4t 2 n? n ? 2k k [ ;1]
3 2
2
2
0
0
0
2
=1
2
and the following identity is proved: 1 n ? 3k ? 1 ! 1 3k ? 2 ! X = (?1)nMn? : k 2 k ? 1 n ? 2 k k
(4.3)
2
=1
Example 4.5 Let us examine the Riordan array D having h(t) 2 F and the function f (t) 1
de ned by:
1 1 p 1 1 ! D = t ln 1 ? t ; t t ln 1 ? t ; f (t) = 1t ln 1 ?1 t : We can note that: 1 p k (p + 2k)! " n + p # 1 1 p 1 1 k k n p n = (n + p)! p + 2k ; dn;k = [t ] t ln 1 ? t t t ln 1 ? t = [t ] ln 1 ? t h i where mn is a Stirling number of the rst kind which satis es: 1 m X 1 !n n; ln 1 ? t = m t n n! m 2
2
2
+2
+
=0
(see Graham, Knuth and Patashnik [5]). Moreover, by some simplifying, we have f^ h = f and, for k > 0 : 0 1 0 1 1 f^kh = 21k [t k? ] gf((tt))k = 21k [t k] B @ k? k? CA : (4.4) (1 ? t) t ln ?t t ln ?t [ ;1] 0
[ ;1]
2
1
2
1
1
1
373
2
1
1
1
2
1
0
It is not very easy to extract the previous coecient and we do it by examining one term at a time. By di erentiating, we obtain the following for m > 1: 1 1 ?m 1 1 ?m 1 d k ? k = k [t ] dt t ln 1 ? t = [t ] t ln 1 ? t 1 1 ?m 1 1 m ? 1 m ? 1 1 k k = k [t ] t ln 1 ? t ? k [t ] 1 ? t t ln 1 ? t ; and thus: 1 1 ?m m ? k ? 1 1 1 ?m 1 k = m ? 1 [tk ] t ln 1 ? t : [t ] 1 ? t t ln 1 ? t We use this result to transform the rst term of (4.4) into an expression involving the second one. We then examine the generating function of the Stirling polynomials (see, e.g., [5], p. 258): 1 1 x X 1 j (x + j )tj ; ln = x t 1?t j and, by setting x = ?2k + 1 in it, we immediately deduce that: f^kh = 21k [ k(1) ? (1 ? 2k) k (1)] = k (1): By solving the equation ln (1 ? t) = ln (1 ? y); we obtain solutions y (t) = t and y (t) = t=(t ? 1) and this yields: f^ h (th(t)) = ? ln(1t ? t) + ln(12? t) : Then, by applying Theorem 3.3, we have: 1 1 p 1 1 1 (p + 2k )! " n + p # X t 1 1 n k (1) = [t ] t ln 1 ? t t ln 1 ? t ? 2 t ln 1 ? t = k (n + p)! p + 2k # # " " ( p + 1)! ( p + 1)! n + p + 1 n + p = (n + p + 1)! p + 1 ? 2(n + p)! p + 1 : +1
+1
1
+1
=0
[ ;1]
2
2
2
2
2
1
2
[ ;1]
2
=0
References [1] M. Bicknell and V. E. Hoggatt. Unit determinants in generalized Pascal triangles. Fibonacci Quarterly, 11:131{144, 1973. [2] L. Carlitz. Generalized Stirling and related numbers. Rivista di Matematica della Universita di Parma, serie 4, 4:79{99, 1978. [3] C. Corsani, D. Merlini, and R. Sprugnoli. Pseudo-inversion of combinatorial sums. Proc. of the 7th FPSAC, Paris, pages 117{125, 1995. 374
[4] I. P. Goulden and D. M. Jackson. Combinatorial Enumeration. John Wiley & S., 1983. [5] L. R. Graham, D. E. Knuth, and O. Patashnik. Concrete Mathematics. Addison-Wesley, 1989. [6] P. Henrici. Applied and Computational Complex Analysis, I. Wiley, New York, 1988. [7] V. E. Hoggatt and M. Bicknell-Johnson. Numerator polynomial coecient arrays for Catalan and related sequence convolution triangles. Fibonacci Quarterly, 15:30{34, 1977. [8] E. Jabotinsky. Sur la representation de la composition de fonctions par un produit de matrices. Application a l'iteration de ex et de ex ? 1. Comptes Rendus Hebdomadaires des Sciences de L'Academie des Sciences, 224:323{324, 1947. [9] E. Jabotinsky. Representation of functions by matrices. Application to Faber polynomials. Proceedings of the American Mathematical Society, 4:546{553, 1953. [10] D. G. Rogers. Pascal triangles, Catalan numbers and renewal arrays. Discrete Mathematics, 22:301{310, 1978. [11] S. Roman. The Umbral Calculus. Academic Press, Orlando, 1984. [12] S. Roman and G. C. Rota. The umbral calculus. Advances in Mathematics, 27:95{188, 1978. [13] L. V. Shapiro, S. Getu, W.-J. Woan, and L. Woodson. The Riordan group. Discrete Applied Mathematics, 34:229{239, 1991. [14] R. Sprugnoli. Riordan arrays and combinatorial sums. Discrete Mathematics, 132:267{ 290, 1994. [15] R. Sprugnoli. Riordan arrays and the Abel-Gould identity. Discrete Mathematics, 142:213{233, 1995. [16] H. S. Wilf. Generatingfunctionology. Academic Press, 1990. [17] D. Zeilberger. A holonomic system approach to special functions identities. Journal of Computational and Applied Mathematics, 32:321{368, 1990.
375
