Applied Umbral Calculus

Heinrich Niederhausen

Florida Atlantic University

Boca Raton, FL 33431

November 10, 2003

1 A Quick Introduction and First Examples

WORK IN PROGRESS!!!As a �rst approximation we may say that Umbral Calculus is about an isomorphism between

formal power series, linear functionals an a certain class of linear operators (shift invariant) onpolynomials.

The isomorphisms Examplepower series

f (t) =P

m�0 amtm

%. -&L j xm

m!

�= am

��! Sp (x) =

Pm�0 am

dm

dxmp (x)

functional linear operator

f (t) = eat

%. -&Evala j x

m

m!

�= am

m! ��! Eap (x) = p (x+ a)

evaluation shift operator

Common ground to the three concepts are special polynomial sequences, called She�er sequences. Apolynomial sequence (sm (x))m2N0 is a sequence of polynomials sm (x) 2 K [x] such that deg sm = m,s0 6= 0. It is convenient to de�ne sm = 0 for negative m. The coe�cient ring K is assumed to be anintegral domain1. For this introduction to Finite Operator Calculus it su�ces to choose K as R [!],the ring of real polynomials in some formal weight parameter !. A formal power series (t) 2 K [[x]]of order 1, i.e. , (0) = 0, 0 (0) invertible in K, will be called a delta series. We substitute thederivative operator D for t in a delta series, and obtain a shift-invariant linear operator2 Q on K [[x]]called a delta operator. The derivative D itself is a delta operator, and like D every delta operator

1Recap: A ring is an abelian group w.r.t. \addition", has an associative \multiplication" that is also (left andright) distributive over +. If the multiplication is commutative, then the ring is commutative. An integral domain isa commutative ring with a 1 and without zero divisors. A �eld is a commutative ring with a 1 in which the nonzeroelements form a multiplicative group. Fields are integral domains. K [x] is the vector space (over K) of polynomialsin x with coe�cients in K. A vector space over the ring S (the scalars) is an additive group (the vectors) with ascalar multiplication \compatible" with the addition.

2A linear operator Q : V ! V on a vector space V over S is a function satisfying Q (�u+ �v) = �Q (u) + �Q (v)for all u; v 2 V and �; � 2 S. We will write Qp instead of Q (p), and for functions p we write Qp (x) instead of(Qp) (x).

1

Q reduces degrees by one and has its null-space equal to the constant polynomials. The solutionsto the system of operator equations

Qsm (x) = sm�1(x)

are therefore only determined up to constants; every polynomial sequence (sm) solving that systemis a Q - She�er sequence. Initial values determine the constants, and if the initial values aresm (x) = �0;m, this special She�er sequence is called the Q - basic sequence.

Example 1 (a) Dxm=m! = xm�1= (m� 1)! (b) r�m�1+xm

�= (1� E�1)

�m�1+xm

�=�m�1+xm

���

m�1+x�1m

�=�m�2+xm�1

�All our operators will be formal power series in D, a degree reducing operator. This explains

the term \Finite Operator Calculus"; whenP

m�0 amDm is applied to a polynomial p of degree n,

then only the �rst n + 1 terms of the in�nite series matter. On any given p the operator acts as a�nite sum.Umbral Calculus can be used as a tool for solving recursions, if the exact solutions to such

recursions are She�er sequences. In that case, the recursion will lead to an operator equationinvolving the (unknown) delta operator Q, and the Q-basic sequence is expanded with the help ofthe Transfer Formulas (5) or (7). It is rare that an interesting combinatorial problem can be solvedby a basic sequence, but the following example is an exception.

Example 2 A staircase polygon (parallelogram polyominoe) can be viewed as a pair of lattice pathswith steps ! and ", starting and ending at a common point without intersections in between,thus forming a (staircase) polygon. In order to count the number c2n of staircase polygons withcircumference 2n, we actually enumerate truncated (by a line with slope �1) staircase polygonswhich have a gap i > 0 (from truncation) and call their number bm (i) if the remaining staircasecircumference is 2i+2m. Note that 2i is the smallest possible remaining circumference. The picturebelow explains the terminology and shows what we mean by the \handles" of the truncated polygon.

Truncated staircase polygon (gap i = 3, and m = i+ 10). . .

cut ! . . . � � � �. . . � � � � �

upper handle ! � � � � � �� . . . � � � � �

� � � . . . � � � �� � � � . . . � �� � � � � " lower handle

If we move the truncation line one unit to the right, we lose the old handles, and the gap changesaccording to their direction,

bm (i) = bm (i� 1)| {z }both handles inwards

+ bm�2 (i+ 1)| {z }both handles outwards

+ 2 bm�1 (i)| {z }handles parallel

2

(Conway, Delest, and Guttmann [?, 1997]). The initial values are b0 (i) = 1 for all i � 0, andbm (0) = 0 for m > 0. The above recursion can be interpreted as a system of di�erence equations,and from b0 (i) = 1 follows that the solution can be extended to a polynomial sequence (bm (x)).De�ne the operator B by linear extension of its action on the basis (bm), Bbm (x) := bm�1 (x) forall m. The recursion bm (x) = bm (x� 1) + bm�2 (x+ 1) + 2bm�1 (x) is equivalent to the operatoridentities

1 = E�1 + E1B2 + 2B

r = E1B2 + 2B

where r = 1 � E�1 is the backwards di�erence operator, a delta operator with basic sequence��x�1+mm

��m�0. The Transfer Theorem 24 shows that B is a delta operator, and its basic sequence is

our solution sequence (bm (x)) because of the initial values bm (0) = �0;m. By formulas (7) and (8)

bm (x) =

mXi=1

x [Bm]�E1B2 + 2B

�i 1x

�i� 1 + x

i

�=

mXj=0

�j

m� j

�22j�mx

m+ x

�m+ x

j

�X

bm (x) tm =

1

2� t+

r1

4� t!�x

([tm] f (t) stands for the coe�cient of tm in f (t)). The total number c2n of staircase polygons ofcircumference 2n equals the number of truncated staircase polygons of circumference 2n�2 and gap1,

c2n = bn�2 (1) =n�2Xj=1

�j

n� 2� j

�22j�n+2

n� 1

�n� 1j

�=1

n

�2n� 2n� 1

�(Levine [?, 1959]), the Catalan numbers.

Example 3 Let U1; : : : ; Un be identically and indepenently distributed random random variableswith common density f (y) = 1 for 0 � y � 1, and 0 else (uniform distributuion). Denote byU(1); U(2); : : : ; U(n) the order statistics of U1; : : : ; Un, i.e. U(1) is the smallest, and U(n) the largest ofthe n observations. Find � := Pr

�U(1) � a+ b; U(2) � a+ 2b; : : : ; U(i) � a+ ib; : : : ; x � U(n) � a+ nb

�,

where the parameters a and b satisfy the conditions 0 < a+ b < a+ nb < 1.Elementary probability theory says that for a+ nb � x � 1

pn (x) : = Pr�U(1) � a+ b; : : : ; U(n�1) � a+ (n� 1) b; x � U(n) � a+ nb

�=

Z x

a+nb

Z u(n�2)

a+(n�1)b

Z u(n�3)

a+(n�2)b� � �Z u(3)

a+2b

Z u(2)

a�b1du(1)du(2) � � � du(n�2)du(n�1)du(n).

Note that (pm) can be extended (in x as well as n) to a polynomial sequence satisfying Dpm (x) =pm�1 (x), and pm (a+mb) = 0 for all m > 0. From p1 (x) =

R xa+b1du(1) = x � a � b follows

p0 (x) =ddx(x� a� b) = 1. De�ne bm (x) = pm (a+ bm+ x) for all m � 0. Now bm (0) = �0;m and

E�bDbm (x) = E�bDpm (a+ bm+ x) = E

�bpm�1 (a+ bm+ x) = pm�1 (a+ b (m� 1) + x) = bm�1 (x)

which shows that (bm) is the basic sequence for the delta operator E�bD. The Transfer Theorem 24

says that bm (x) = x (x+ bm)m�1 =m! hence pn (x) = bn (x� a� bn) = (x� a� bn) (x� a)m�1 =m!and

� = pn (1) = (1� a� bn) (x� a)n�1 =n!

3

If necessary and possible, the initial conditions are formulated as a functional on K [x], andthe Functional Expansion Theorem 46 expands the She�er polynomials sm in terms of the basicsequence. We call this the \bottom up" approach: Starting with simple \building blocks" (incombinatorics they usually are binomial coe�cients) we construct the basic sequence, and fromthis more advanced material the exact solution is put together such that the initial conditions aresatis�ed. In the more commonly applied \top down" approach a generating function identity issolved, and the exact solution may be obtained by �nding coe�cients. We will show that in manyapplications Umbral Calculus delivers the generating function too; the more interesting cases maybe those where this is not the case (Theorem 24).

Remark 4 The Finite Operator Calculus development of Umbral Calculus began with a paper byMullin and Rota in 1970 [6], and became \popular" through th very readeble exposition \FiniteOperator Calculus" by Rota and his students Kahaner and Odlyzko in 1973 [9]. The paper \UmbralCalculus" by Roman and Rota [8, 1978] extended the theory, and put more emphasis on linearfunctionals. Several generalizations of concept have been created; especially important for my workis Jack Freeman's \Transforms of operators on K[x][[t]]" [2, 1985].

2 Sequences of Binomial Type

The basic sequences we have seen in the Introduction are also sequences of binomial type, sometimescalled sequences of convolution type We will use the three terms as synonyms (some people don't;see Di Bucchianico [1]). The justi�cation follows in Theorem 18.

De�nition 5 A polynomial sequence (bm) in K [x] is of binomial type provided that

bm (x+ y) =mXi=0

bi (x) bm�i (y) (1)

for all y 2 K and m 2m0.

Note that this convolution identity implies b0 (x) � 1, and bm (0) = 0 for all positive integersm. If we think of x, y, and t as formal variables, and de�ne the generating function b (x; t) =P

m�0 bm (x) tm, then the convolution is equivalent to b (x+ y; t) = b (x; t) b (y; t). Because b (x; t)

is not identical 0, there must exist a formal power series f (t) such that b (x; t) = f (t)x. Fromf (t) =

Pm�0 bm (1) t

m = 1+ b1 (1) t+ : : : follows that order (f) = 0. Hence the (formal) logarithm

� (t) := ln f (t) =Xm�1

(�1)m+1 (f (t)� 1)m

of f exists, and b (x; t) = ex�(t).

Exercise 6 Show that � (t) is of order 1, i.e. there exists a nonzero scalar c such that, � (t) = ct+higher order terms in t (Hint: Show that b1 (x) = cx).

Example 7 (a) xm=m!with � (t) = (b)�m�1+xm�1

�with � (t) = (c) x (x+ bm)m�1 =m!

with � (t) =

4

The power series � (t) is fundamental for the Finite operator Calculus. It characterizes thesequences of binomial type. We therefore give a name to power series of order 1 in K [[t]].

De�nition 8 We say that � (t) 2 K [[t]] is a delta series i� � (t) is of order 1, and the coe�cientof t in � (t) has a reciprocal (a multiplicative inverse) in K.

We are now ready to state the theorem that will connect (in the next section) the sequences ofbinomial type withthe basic sequences.

Theorem 9 A polynomial sequence (bm) is of binomial type i� there exists a delta series � (t) suchthat

Pm�0 bm (x) t

m = ex�(t).

Proof. We have already shown the existence of a delta series � (t) for sequences of binomialtype. To see the other direction, let � (t) be a delta series, andX

m�0bm (x) t

m = ex�(t) = b (x; t)

for some polynomials sequence (bm). Hence b (x+ y; t) = e(x+y)�(t) = ex�(t)ey�(t) = b (x; t) b (y; t),which is equicalent to the convolution identity (1).

2.1 SomeProbability Distributions

It is shown in [1], Theorem 3.5.10, that for every sequence of binomial type (qn) there exists aweakly continuous convolution semigroup3 (�x)x�0 of probability measures on R such that

qm (x) =

Z 1�1

ym

m!d�x (y)

for x � 0 i� there exists an in�nitely divisible measure probability measure � on R with momentsequence (m!qm (1))m�0, i.e., m!qm (1) =

R1�1 y

m � (dy), andR1�1 y � (dy) 6= 0. In that case � = �1,

and m!qm (1) = E [Ym] if Y is a random variable with distribution �1.

Example 10 [1, Examples 3.5.15 ]Let �x be the point mass at x, �x (A) = 1 if x 2 A, and 0 else,for all (measurable) subsets of R.

1. The measures (�x)x�0 are a (weakly continuous) convolution semigroup, becauseRf d�x+y =R

f (w) �x+y (dw) = f (x+ y) =Rf (u+ y)�x (du) =

R Rf (u+ v)�x (du)�y (dv). We �nd

qm (x; �x) =R1�1 (y

m=m!) d�x (y) = xm=m!. Note that m!qm (1) = 1 =

R1�1 y

md �1 (y).

2. The Poisson semigroup �x = e�xP

k�0�xk=k!

��k leads to the basic polynomials

�m (x) =

Z 1�1(ym=m!) d�x (y) =

e�x

m!

Xk�0

xkkm

k!;

known as the exponential polynomials (see Dobinski's formula (Example ??) , �m (x) =Pmk=0 S (m; k)

xk

m!, where S (m:k) are the Stirling numbers of the second kind). In this case

�m (1) =Pm

k=0 S (m; k), the m-th Bell number.

3For measurable functions f holdsRfd�x+y =

R Rf (u+ v)�x (du)�y (dv), and for measurable sets A holds

�x+y (A) =R�x (A� v)�y (dv) =

R�y (A� u)�x (du)

5

3. The basic polynomials for the Gamma semigroup �x (dy) = e�y (yx�1=� (x)) dy are

gm (x) =

Z 10

ym

m!

yx�1

� (x)e�ydy =

�m�1+xm

�;

associated to the backwards di�erence operator r, as shown in the introduction.

2.1.1 Without measure theory

Every random variable Y with �nite moments and nonzero expectation de�nes a sequence of bino-mial type (bm) by polynomial extension of the values bm (k) := E [(Y1 + � � �+ Yk)m] =m! (k;m = 0; 1; : : : ),where Y1; Y2 : : : are i.i.d. random variables with the same distribution as Y (Exercise 11 ). Theconverse is not true, of course; for example, if b2 (1) = 0, the random variable Y would have thesecond moment equal to zero, and therefore all moments equal to zero (except E [Y 0]). We obtainthe sequence of binomial type (bm) from Y by extending bm (k) from the positive integers to all realnumbers. The sequence of binomial type has the generating function

Xm�0

bm (x) tm =

Xm�0

E [Y m] tm=m!

!x=MY (t)

x ; (2)

where MY (t) stands for the moment generating function of Y . For example, if Y has the bi-nomial distribution Pr (Y = k) =

�mk

�pk (1� p)m�k with parameters m and p, then MY (t) =P

m�0 tmPm

k=0 km�mk

�pk (1� p)m�k =m! = (1 + (et � 1) p)m and therefore bm (x;Bin (m; p)) =

Pk�0�mxk

�pk [tm] (et � 1)k

=Pm

k=0

�mxk

�pk k!

m!S (m; k).

Exercise 11 Let Y; Y1; Y2 : : : be i.i.d. random variables with �nite moments and nonzero expecta-tion. Show that there exists a unique basic sequence (bm) such that bm (k) := E [(Y1 + � � �+ Yk)m] =m!

for allm; k = 0; 1; 2; : : : . Hint:�P

n�0 bn (1) tn�k=P

i1;:::ik�0 (bi1 (1) � � � bik (1)) ti1+���+ik =

Pn�0 t

nP

ij�0; i1+���+ik=nEhYi11 ���Y

ikk

ii1!:::ik!

.

Exercise 12 Let Y be a geometric random variable, i.e. Pr (Y = k) = p (1� p)k, the probabilityof k failures before the �rst success in independent trials with success probability 0 < p < 1. Showthat bm (x) =

Pmk=0

�x+k�1k

�(1� p)k p�k k!

m!S (m; k) is the m-th basic polynomial obtained from this

distribution. Hint: You may want to use the identity (et � 1)k =P

m�kk!m!S (m; k) tm

2.1.2 Distributions of binomial type

Up to now we represented the moments of certain distributions by the values m!bm (k) of associatedpolynomials of binomial type. In the following some discrete distributions will be directly expressedby the polynomials.A distribution �i on [m] is of binomial type i�

�i = qi (�) qm�i (�) =qm (�+ �)

for some sequence (qm) of binomial type. It is easy to verify that the binomial and the hypergeo-metric distribution are of this type; however, it is interesting that both distributions are of binomialtype because of a more general construction.

6

A subclass of the distributions of binomial type is obtained by conditioning as follows. Let � (t)be the delta series such that

Pm�0 bm (x) t

m = ex�(t). Suppose that for some parameter space �the family of random variables X�, � 2 �, takes values in N0 such that for all i 2N0

Pr (X� = i) = pi��bi (�) (3)

for some p 2 R where � (p) converges, and � = e��(p). Furthermore, assume that the process X� isstationary with independent increments at � and , �; ; �+ 2 �,

Pr (X� = i and X�+ = i+ j) = Pr (X� = i) Pr (X = j) : (4)

Then the conditional distribution �i := Pr (X� = i jX�+� = m) is of binomial type on [m], because

�i = Pr (X� = i jX�+ = m) =Pr (X� = i) Pr (X = m� i)

Pr (X�+ = m)=pi��bi (�) p

m�i� bm�i ( )

pm��+ bm (a+ ):

Exercise 13 Suppose X� has the Poisson distribution Pr (X� = i) = e���i=i! for all i 2N0 and� 2 R+. Show by conditioning that the Binomial distribution is of binomial type.

Exercise 14 Take a sample of size m from a population of size � + , �; 2N0, which contains� marked and unmarked subjects. Show by conditioning on the binomial distribution that thehypergeometric distribution �i := Pr (i marked objects in the sample) is of binomial type. Hint: X�

counts the number of successes in � independent trials.

Problem 15 A positive distribution pi on [N ] is symmetric i�

pi = pN�i

for all i = 0; : : : ; N . Show that every positive symmetric distribution on [N ] is a distribution ofbinomial type,

pi =qi (1=2) qN�i(1=2)

qN (1)

for some sequence of binomial type (qn). Give details for the example p0 = pi for all i 2 [N ] (discreteuniform distribution).

3 Delta Operators

Every delta series � (t) has a compositional inverse ��1 (t), i.e., ����1 (t)

�= t = ��1 (� (t)). If

� (t) = ct+ : : : then ��1 (t) = 1ct+ : : : . That's why we required that in a delta series the coe�ent

[t] � (t) has a reciprocal. The inverse � (t)�1 is again a delta series.

Remark 16 In the ring of formal power series it it easy to confuse compositional and multiplicativeinverses. The latter are often written as � (t)�1, but I prefer 1=� (t) and 1

�(t), and call them recip-

rocals. Only power series of order 0 with constant terms that have a reciprocal, have a reciprocal.If � (t) = a+ : : : then 1=� (t) = 1

a+ : : : .

7

De�nition 17 Let Q be a linear operator on K [x]. We call Q a delta operator provided there is adelta series (t) such that Q = (D) (where D = d

dx). The polynomial sequence (qm) is the Q-basic

sequence i� Qqm = qm�1and qm (0) = �m;0 for all m � 0 (assume that q�1 � 0).

We will now establish the promised connection between basic sequences and sequences of bino-mial type.

Theorem 18 Let � (t) be a delta series. If (bm) is a polynomial sequence such thatP

m�0 bm (x) tm =

ex�(t) and B is the delta operator B = ��1 (D), then Bbm = bm�1 for all m � 0.

Proof. Let ��1 (t) =P

k�1 ktk. In this notationX

m�0��1 (D) bm (x) t

m = ��1 (D) ex�(t) =Xk�1

kDkex�(t) =

Xk�1

k� (t)k ex�(t)

= ��1 (� (t)) ex�(t) = tex�(t) =Xm�0

bm (x) tm+1 =

Xm�1

bm�1 (x) tm;

hence Bbm = bm�1.

Corollary 19 A polynomial sequence (bm) is of binomial type i� it is the�lnP

m�0 bm (1)Dm��1-

basic sequence.

Proof. Apply the above theorem, and Theorem 9.

Example 20 Some standard examples:

� (t) ex�(t) bm (x) ��1 (D) name of operatort ext xm

m!D derivavtive

ln (1 + t) (1 + t)x�xm

�eD � 1 = � forward di�erence

ln 11�t (1� t)�x

�m�1+xm

�1� e�D = r backwards di�erence operator

Example 21 The cumulant generating function of a random variable Y is the logarithm of itsmoment generating function, KY (t) = lnMY (t). In (2) we saw that

Pm�0 bm (x) t

m = MY (t)x if

bm (k) := E [(Y1 + � � �+ Yk)m] =m!. Hence (bm) is the KY (D)-basic sequence.

8

3.1 Transfer Formulas

If we cannot guess the B-basic sequence (bm) of some delta operator B it may be possible to connect(bm) with a known Q-basic sequence (qm), say. This is possible, because we need not expand deltaoperators in powers of D; a linear operator is a delta operator if it can be written as a delta seriesin terms of any other delta operator.

Theorem 22 (Transfer Theorem) Let Q be a delta operator with Q-basic sequence (qm)m2N0. IfQ = � (B) for some delta series � and linear operator B, then B is also a delta operator, and theB - basic sequence (bm)m2N0 has for all m 2 N0 the expansion

bm (x) =

mXi=0

cm;iqi (x) (5)

where cm;i is the coe�cient of tm in � (t)i. For the generating function of the basic sequence (bm)

holds Xm�0

bm (x) tm =

Xi�0qi (x)� (t)

i : (6)

The proof of this theorem is implicitly contained in [9]. The great applicability of this and thenext Transfer Theorem comes from the feature that we �nd bm (x) without explicitly knowing whatB is. The only information needed is Q = � (B).Proof. There exists a delta series (t) such that

Pm�0 qm (x) t

m = ex (t) and Q = �1 (D).Note that B is a delta operator because B can be expanded as a delta series in D, B = ��1 (Q) =��1 ( �1 (D)). HenceX

m�0bm (x) t

m = ex (�(t)) =Xi�0qi (x)� (t)

i =Xi�0qi (x)

Xm�i

�[tm]� (t)i

�tm

=Xm�0

tmmXi=0

qm�i (x)�[tm]� (t)m�i

�:

De�nition 23 We say that a linear operator T on K [x] is shift invariant i� TEa = EaT for allshift operators Ea. The ring of shift invariant operators is denoted by �.

Remember that Ea = eaD 2 K [[D]] for all a 2 K: If � 2 K [[t]] then � (D) eaD = eaD� (D)because the multiplication of power series is commutative. Hence all operators that are powerseries in D are shift invariant. It is shown in [9] that this are all of them, � = K [[D]]. Deltaoperators can be expanded as delta series � 2 � [[t]] if [t]� (t) is invertible; however this expansionis no longer unique and it often generates interesting identities. With the help of Lagrange-B�urmanninversion the expansion (5) generalizes to the following theorem.

Theorem 24 (Generalized Transfer Formula) Let Q be a delta operator with Q - basic se-quence (qm)m2N0. If Cm;i is the coe�cient of B

m in � (B)i, where � is the delta series Q = � (B) 2

9

� [[B]], then B is also a delta operator. The B - basic sequence (bm)m2N0 has for positive m theexpansion

bm (x) =

mXi=1

xCm;i1

xqi (x) : (7)

See [4] for the proof. The factor x in this formula will not cancel in general, because Cm;i is anoperator. For example, we can write the forward di�erence operator � : p (x) 7! p (x+ 1) � p (x)in terms of the backwards di�erence r,

� = E1 � 1 = E1�1� E�1

�= E1r 2 � [[r]] :

The �-basic sequence can be guessed as��

xm

��m�0. By 7

mXi=1

x�[rm]

�E1r

�i� 1x

�x

i

�= x

1

x+m

�x+m

m

�=

�m� 1 + x

m

�is the m-th basic polynomial which we used already in the staircase example 2 in the Introduction,Example 2.Unfortunately, Theorem 24 does not tell us the generating function

Pm�0 bm (x) t

m as Theorem22 does. However, in many applications the only operator occurring among the coe�cients of � areshift operators, and it may be possible to express E1 as (B), say, in terms of B using only scalarcoe�cients. In that case r = 1� 1= (B), and

Xm�0

bm (x) tm =

Xi�0

�i� 1 + x

i

��1� 1

(t)

�i= (t)x (8)

by Theorem 22. For example, in the introductory staircase example we found 1 = E�1+E1B2+2B,hence

E�1 =1

2

�1� 2B +

p1� 4B

�and

Xm�0

bm (x) tm =

1

2� t+

r1

4� t!�x

:

Example 25 (Three step vectors) A �-weighted (c; )-path has step set Sc; = f!; "; (c; )g,c 2 N1, 2 Z, where the (c; ) steps are weighted by �. Suppose D (m;n) is the number (generatingfunction) of �-weighted (c; )-paths to (m;n), under some restrictions. The recursion

D(m;n) = D(m;n� 1) +D(m� 1; n) + �D(m� c; n� ) (9)

must be solved on the appropriate recursion domain. If c is a positive integer, then this recursioncan be solved by a polynomial sequence (dm), say, except when c = 1 and � = �1. The associatedoperator B : dm ! dm�1 solves the equation

r = B + �E� Bc = � (B) :

10

We calculate

Cm;i = [Bm]� (B)i = [Bm]�B + �E� Bc

�i=�Bm�i

� �1 + �E� Bc�1

�i=

�Bm�i

� iXk=0

�i

k

��kE� kB(c�1)k =

�i

(m� i) = (c� 1)

��(m�i)=(c�1)E� (m�i)=(c�1)

From the seconed Expansion Theorem we obtain the B-basic sequence

bm(x j c; ) =mXi=1

xCm;i1

x

�i� 1 + x

i

�=

mXi=1

x

iCm;i

�i� 1 + xi� 1

�=

m�1Xi=0

x

m� iCm;m�i�m� i� 1 + xm� i� 1

�(10)

=m�1Xi=0

x

m� i

�m� ii= (c� 1)

��i=(c�1)E� i=(c�1)

�m� i� 1 + xm� i� 1

�

=

bm=ccXi=0

x

m� i (c� 1)

�m� i (c� 1)

i

��iE� i

�m� i (c� 1)� 1 + xm� i (c� 1)� 1

�

=

bm=ccXi=0

�m� (c� 1) i

i

��ix

x+m� ( + c� 1)i

�x+m� ( + c� 1)i

m� (c� 1)i

�for all m � 0.

Exercise 26 Suppose S = f(0; 1) ; (1; 0) ; (a1; b1) ; : : : ; (as; bs)g. Let Pr (m;n; k) := Pr (path reaches (m;n) in the k steps under some restrictions)and denote the s+ 2 step probabilities by

p10 : = Pr (step vector (1; 0) is selected)

p01 : = Pr (step vector (0; 1) is selected)

pajbj = Pr (step vector (aj; bj) is selected)

Hence

Pr (m;n; k) = p01 Pr (m;n� 1; k) + p10P (m� 1; n; k) +sXj=1

pajbj Pr (m� aj; n� bj; k � 1)

Suppose the s + 2 step probabilities can be parameterized by only 3 parameters �; �1 and �2 suchthat

p10 = ��1; p01 = ��2; pajbj = ��aj1 �

bj2 for all j = 1; : : : ; s:

If D (m;n; k) is the number of ways to reach (m;n) in k steps from the origin (under the samerestrictions). Verify that

Pr (m;n; k) = �k�m1 �n2D (m;n; k)

under this model. Let Pr (m;n) =P

k Pr (m;n; k) and �nd a recursion for sm (n) := p�m01 p

�n10 P (m;n; k).

Show: If S = f(0; 1) ; (1; 0) ; (a; b)g contains only three steps (s = 1) then

� =pa10p

b01

1� p10 � p01:

11

3.2 Polynomials of Abel-Type

We derived the basic sequence for r by noting that r = E1�. This procedure is a special case ofa more general result. If Q = EaB and

Pn�0 qm (x) t

m = ex (t) then for all m > 0

bm (x) =

mXi=1

x�[Bm]� (EaB)m�i

� 1xqi (x) = xE

am 1

xqm (x) =

x

x+ �mqm (x+ �m) : (11)

Therefore, if Q(a) := E�aQ then q

(a)m (x) = xqm (x+ am) = (x+ am) is the Q(a) - basic polynomial

according to (11). The Abel polynomials x (x+ am)m�1 =m! are the D(a)-basic polynomials (seeExample 3). The process of moving from Q to Q(a) generates many identities; we call this process\Abelization". If (qm) has the generating function e

x�(t) we will denote the generating function

of�q(a)m

�by ex�(a)(t). From Q = ��1 (D) and Q(a) := E�aQ follows Q(a) := e�aD��1 (D), thus

�(a) (t) =�e�at��1 (t)

��1(t). For this reason the generating function of Abel-type polynomials is

di�cult to obtain in general, even when the polynomials themselves are easily expanded.

4 She�er Polynomials

In the vector space K [x] every polynomial sequence is a basis; the basic sequences a special kind ofbases. They are useful to expand a much larger class of polynomial sequences, the She�er sequences.

De�nition 27 Let � (t) be a delta series, and � (t) 2 K [[t]] a power series of order 0. We say thatthe polynomial sequence (sm (x)) is a She�er sequence, provided that

Pm�0 sm (x) t

m = � (t) ex�(t).

We say that (sm) is associated to (bm), ifP

m�0 bm (x) tm = ex�(t). If B = ��1 (D), then we call

(sm) a B-She�er sequence.

She�er sequences are a generalization of Appel sequences (where � (t) = t). Every basic sequenceis also a She�er sequence (where � (t) = 1). From a recursion point of view, the She�er sequencesolve the same recursion as the basic sequence, but with di�erent initial values.

Theorem 28 Suppose (sm) is a She�er sequence. Then (sm) is a B-She�er sequence, i� Bsm =sm�1 for all m 2 N0.

Proof. Modify the proof of Theorem 18.The Binomial Theorem for She�er Sequences is a generalization of (1, and has many applications.

Theorem 29 (Binomial theorem for She�er Sequences) If (bm) is a sequence of binomialtype, and (sm) associated to (bm), then

sm (x+ y) =mXi=0

si (x) bm�i (y) (12)

for all x and y.

12

Proof. � (t) e(x+y)�(t) =�� (t) ex�(t)

�ey�(t)

This theorem allows to calculate (= expand in terms of the associated (bm)) (sm) in a recursivemanner, if initial values ym = sm (xm) are known for all m. Obviously s0 � y0 (which must bedi�erent from 0). Suppose sm (x) has already been expanded. Then

sm (x) = sm (xm + x� xm) =mXk=0

sk (xm) bm�k (x� xm) = ym +m�1Xk=0

sk (xm) bm�k (x� xm) :

If the initial conditions are less general, better expansion methods can be used. The easiestcase would be sm (x+ c) = �0;m, the next best after a basic sequence. Thus sm (x) = bm (x� c).Abelization yields another class of easy initial conditions. Note that (sm (x+ am+ c))m�0 is a Q(a)-

She�er sequence i� (sm) is a Q - She�er sequence. We write s[a]m (x) for sm (am+ x). In general, if

q(a)m (x) = xqm (x+ am) = (x+ am) is the Q(a)-basic polynomial, then

tm (x) := q(a)m (x� am) =

x� amx

qm (x)

is a She�er sequence for the delta operator EaQ(a) = Q. Note that sm (am) = �0;m. In symbols,�q(a)[�a]m

�is a Q - She�er sequence.

The notation Q(a); q(a)m and s

[a]m will be used in the following.

Example 30 (Three step vectors, continued) Consider the f!; "; (c; )g-paths as in Example25, and assume that c, a and � are positive integers. Find the number of (c; )-path to (m;n)which stay weakly above the line y = ax, where a is a positive integer. Formula4

(2;�3)-paths above y = 2x" n �& �6 �" �! � �5 �" �" �4 �" �" �3 �" �2 �" �1 �" �0 �" �

0 1 2 3 ! m

" n 1 6 22 54 �6 1 5 15 23 �5 1 4 9 �4 1 3 4 �3 1 2 �2 1 1 �1 1 �0 1 �

0 1 2 3 ! mSample path to (3; 6) The number of (2;�3)-paths above y = 2x

The path counts are determined by the recursion D(m;n) = D(m;n� 1) +D(m� 1; n) + �D(m�c; n � ) as in (9), and the initial values D (m; am� 1) = 0. The polynomial extension dm (x)satis�es the conditions, and is therefore a She�er sequence associated to the basic sequence (10)

bm (x j c; ) =bm=ccXi=0

�m� (c� 1) i

i

��ix

x+m� ( + c� 1)i

�x+m� ( + c� 1)i

m� (c� 1)i

�:

4Pbn=cc

k=0

�m+1�k( �1)

k

��m+n�k( +c�1)

n�ck�

m�na+1m+1�k( �1)

13

The She�er polynomial sm (x) = dm (x+ am� 1) are associated to b(a)m (x j c; ), and from sm (0) =�0;m follows sm (x) = b

(a)m (x j c; ) = xbm (x+ am j c; ) = (x+ am). Hence for n � am holds

D (m;n) = dm (n) = b(a)m (n+ 1� am j c; ) = (n+ 1� am) bm (n+ 1 j c; ) = (n+ 1)

=

bm=ccXi=0

�m� (c� 1) i

i

��i (n+ 1� am)

n+ 1 +m� ( + c� 1)i

�n+ 1 +m� ( + c� 1)i

m� (c� 1)i

�:

For f!; "g-paths make = 0, hence D (m;n) = (n+1�am)n+1+m

�n+1+mm

�. For Dyck paths choose a = 1

and m = n.

4.1 Linear Recursions

Solving one-dimensional linear recursions is a traditional topic of enumerative combinatorics. Weshow in this section how She�er sequences can be applied to approach such recursions. If weapply Umbral Calculus to �nd an explicit expression for a recursively de�ned sequence of numbers�0; �1; : : : , �0 6= 0, solving �n =

Pnj=1 �j�n�j+ n, we typically think of a She�er sequence (sn) such

that sn (0) = �n for all n 2 N0. The initial values are enforced by the sequence s0 (1) ; s1 (1) ; : : : ,and the expansion of the solution follows from the Binomial Theorem for She�er sequences

sn (1) =nXk=0

sk (0) bn�k (1) ;

where (bn) is the corresponding basic sequence.

Example 31 (Fibonacci numbers) The Fibonacci numbers Fn can be de�ned by F0 = F1 =1 and Fn = Fn�1 + Fn�2 for all n � 2. The usual generating function approach [?, Chpt. 4]immediately shows that

Pn�0 Fnt

n = 1= (1� t� t2); we want to show a di�erent way to �nd anexpression for the numbers Fn. Let (sn) be a polynomial sequence such that

sn (x)� sn�1(x)� sn�2 (x) = sn (x+ 1)

and s0 (x) � 1, sn (1) = 0 for all n 2 N1. The numbers sn (0) and Fn follow the same recursion andsatisfy the same initial values; they are equal. Let (bn (x)) be a basic sequence such that b0 (1) � 1,b1 (1) = b2 (1) = �1;and bn (1) = 0 for all n � 2. We can obtain the corresponding delta series� (t) from

Pn�0 bn (x) t

n = ex�(t) =�P

n�0 bn (1) tn�x= (1� t� t2)x. Hence 1 =

Pn�0 sn (1) t

n =�Pn�0 sn (0) t

n�e�(t) =

�Pn�0 Fnt

n�(1� t� t2) and therefore

Pn�0 Fnt

n = (1� t� t2)�1, From

14

here on the calculations are the same as with standard generation function methods:

Fn = [tn]�1

t2 + t� 1 (13)

= [tn]

1=p5

t+ 12+ 1

2

p5� 1=

p5

t+ 12� 1

2

p5

!=

1p5�12+ 1

2

p5� ��12

p5� 1

2

�n � 1p5�12� 1

2

p5� �

12

p5� 1

2

�n=

2n+1p5�p5� 1

�n+1 � 2n+1p5��p5� 1

�n+1 = 2n+1��p5� 1

�n+1 � 2n+1 �p5� 1�n+1p5 (�4)n+1

=

�1 +

p5�n+1 � �1�p5�n+12n+1

p5

(see also Identity (??))

The general problem of solving one dimensional linear recursions is addressed in the followingproposition.

Proposition 32 Let ` be a positive integer. Suppose the numbers �n solve for n � ` the linearrecursion

�n =nXj=1

�j�n�j + n

where �1; �2; : : : and `; `+1; : : : are sequences of given constants. The initial values �0; �1; : : : ; �`�1are also known. Then

�n =nXk=`

k�tn�k

� 1

1�P1

j=1 �jtj

and Xn�0

�ntn =

P1k=` kt

k

1�P

j�1 �jtj

where k = �k �P

j�1 �j�k�j for k = 0; : : : ; `� 1.

Remark 33 Suppose the recursion is homogeneous (all n's are zero). If there are only �nitelymany factors �1; : : : ; �d, the generating function

Pn�0 �nt

n is rational. By Stanley's Theorem4.1.1 [10]

�n =

kXi=1

pi (n) �ni

where the 1=�i's are the k distinct roots of the polynomial

1�dXj=1

�jtj =

kYi=1

(1� �it)mi

and each pi (n) is a polynomial (in n) of degree less than the multiplicity mi of root 1=�i.

15

Proof. Write the recursion as

�n �nXj=1

�j�n�j = n

for n � `, and de�ne the sequence (bn) of binomial type in terms of the initial values b0 (1) = 1,and bj (1) = ��j for j > 1. Let b1 (1) = ��1 if �1 6= 0, and b1 (1) = 1 else (note that b1 (1) mustbe di�erent from 0 for any basic sequence). Thus

Pn�0 bn (x) t

n = ex�(t) =�P

n�0 bn (1) tn�x=�

1 + �0;�1t�P

n�1 �ntn�x. We think of �n as the value sn (0) of some yet undeterminded She�er

sequence (sn) associated to (bn). By the Binomial Theorem for She�er Sequences (12)

n + �0;�1�n�1 = �n + �0;�1�n�1 �nXj=1

�j�n�j =nXj=0

sn�j (0) bj (1) = sn (1)

for n � `. For 0 � n < ` we must de�ne sn (1) = �n + �0;�1�n�1 �Pn

j=1 �j�n�j, validating the

recursion sn (1) =Pn

j=0 sn�j (0) bj (1) for all n � 0. FromP

n�0 sn (x) tn =

�Pn�0 sn (0) t

n�ex�(t)

we �ndP

n�0 sn (1) tn =

�Pn�0 sn (0) t

n�e�(t), thus X

n�0�nt

n

! 1 + �0;�1t�

Xn�1

�ntn

!=

`�1Xn=0

�n + �0;�1�n�1 �

nXj=1

�j�n�j

!tn +

Xn�`

( n + �0;�1�n�1) tn

=`�1Xn=0

�n �

nXj=1

�j�n�j

!tn +

Xn�`

ntn + �0;�1t

Xn�0

�ntn

The sequences (��n) and�[tn] 1

1�P1j=1 �jt

j

�are called orthogonal (if the latter can be explicitly

determined), and the sequences (�n) and ( n) are an inverse pair. For examples of inverse pairs seeSection 4.2.

Example 34 (Derangement Numbers) The derangement numbers dn denote the number of per-mutations � of [n] that are derangements, i.e., �i 6= i for all i = 1; : : : ; n. They follow the recursiondn = ndn�1+(�1)n,with initial value d0 = 1. In the notation of Proposition 32, �n := dn=n!, ` = 1,�1 = 1, �1 = 1, and n = (�1)

n =n!. Thus

dn = n!�n = n!

1 +

nXk=1

(�1)k

k!

!��tn�k

� 1

1� a1t

�= n!

nXk=1

(�1)k

k!:

If we apply proposition to this problem, we need to search for a basic sequence (bn) such thatPnj=0 bj+n;n�n�j = (�1)

n for n � 1, thus

Example 35 (Bernoulli Numbers) The Bernoulli number Bn solve the system of equations �n;0 =Pnk=0

�n+1k+1

�1n+1Bn�k for n � 0. Dividing by n! shows that the numbers Bn=n! can be calculated

from the linear recursion Bn=n! = �Pn

k=11

(k+1)!Bn�k= (n� k)!. Applying the proposition with

ak = �1= (k + 1)! for k � ` = 1 gives the exponential generating functionXn�0

Bnn!tn =

B01 +

Pj�1 t

j= (j + 1)!=

t

t+P

j�1 tj+1= (j + 1)!

=t

t+ (et � 1� t) =t

et � 1 :

The generating function is not rational; it can be easily derived by other approaches.

16

Problem 36 A Fibonacci-like sequence �0; �1; : : : solves a recurrence of the form

�n = u�n�1 + v�n�2 + n

for n � 1, with initial values �0, �1, and given nonhomogeneous terms 1; 2; : : : . Find an expres-sion fo �n and its generating function. As examples, solve

1. the Fibonacci recursion Fn = Fn�1 + Fn�2 for all n � 2, F0 = F1 = 1

2. the Lucas recursion Ln = Ln�1 + Ln�2 for all n � 2, L0 = 2 and L1 = 1

3. the recursion �n = (a� 1)�n�1 + a�n�1 + n� 1

Problem 37 (Tiling rectangles with squares) S. Heubach [3] found recursions for the numberT km;n of all tilings of an n�m rectangle 1� 1 and 2� 2 tiles, where exactly k tiles of size 2� 2 areused. She explicitly calculated

T k2;n =

�n� kk

�T k3;n = 2k

�n� kk

�;

and for n� 4-rectangles found the recursion

T k4;n+1 = Tk4;n + 3T

k�14;n�1 + T

k�24;n�1 + 2

kXr=2

T k�r4;n�r

for all n > 4, and the initial values T k4;4 = 1 if k is even, and 0 else. Find T k4;n andP

k�0 Tk4;k+jt

k..For n� 5-rectangles Heubach found the recursion

T k5;n+1 (x+ 1 j 5) = T k5;n (x j 5) + 4T k�15;n�1 (x j 5) + 3T k�25;n�1 (x+ 1 j 5) + 2k+1Xr=3

FrTk+1�r5;n+1�r (x j 5)

where Fr denotes the r-th Fibonacci number, F0 = F1 = 1, and the initial values Tk5;k (0 j 5) = 3k=2 if

k is even, and 0 else. FindP

k�0 Tk5;k+jt

k.For more on generating functions of related tiling problems see Merlini, Sprugnoli, and Verri

([5], http://dsiII.dsi.uni�.it/~merlini/tiling.ps, ps-�le).

Linear recursions of the form

�n =nXj=1

�n;j�n�j + n

for all n � ` are certainly more challenging. Here��i;j�is a given in�nite triangular coe�cient

matrix. The study of inverse pairs solves one particular case as follows. Let ` be a positive integer,and let (bn (x)) be a sequence of binomial type with generating function

Pn�0 bn (x) t

n = ex�(t).Suppose the numbers �n solve for n � ` the linear recursion

nXj=0

bj (c+ �n)�n�j = n

17

where c,�, and `; `+1; : : : are given constants. The initial values �0; �1; : : : ; �`�1 are also known.Then

�n =

nXk=0

k��n;n�k

where k =Pk

j=0 bj (k)�k�j for k = 0; : : : ; `� 1;and ��n;k = c+n�c+(n�k)�bk(�c� (n� k)�). See Section

4.2 for more details. However, this inversion does not apply for linear recurrences of �nite order,where

Pdj=0 bj (c+ �n)�n�j = n for a �xed order d. In this case we must require that for all j > d

holds bj (c+ �n) = 0 and all n, thus bj cannot be a polynomial. We can avoid any reference tosequences of binoial type if we formulate the inverse pair relationship as follows.

Proposition 38 Let ` be a positive integer, and � (t) a power seris of order 0, with nonzero linearterm. Suppose the numbers �n solve for n � ` the linear recursion

nXj=0

an�j;j�j = n

where `; `+1; : : : is a sequence of given constants, and jan;j= (n+ j) = [tn]� (t)j. The initial values

�0; �1; : : : ; �`�1 are also known. Then

�n =nXk=0

k�tn�k

�� (t)�n

where k =Pk

j=0 ak�j;j�j for k = 0; : : : ; `� 1.

Proof. De�ne that bn (j) = jan;j= (n+ j), thusXn�0

bn (j) tn =

Xn�0

jan;jtn= (n+ j) = � (t)j = ej ln�(t)

which shows that (bn (x)) is a sequence of binomial type. We think of �n as the value sn (0) of someyet to determine She�er sequence (sn) associated to

�xx�nbn (x� n)

�. By the Binomial Theorem

n =

nXj=0

sj (0) an�j;j =

nXj=0

sj (0)n

n� n+ j bn�j (n� n+ j) = sn (n)

for n � `. For 0 � n < ` we must de�ne sn (n) =Pn

j=0 �jnjbn�j (j), validating the recursion

sn (n) =Pn

j=0 sj (0)njbn�j (j) for all n � 0. The polynomial sequence pn (x) := sn (n+ x) is She�er

with associated basic polynomials bn (x). The Binomial Theorem says that for all n � 0

�n = sn (0) = pn (�n) =nXj=0

pj (0) bn�j (�n) = nnXj=0

sj (j) an�j;�n=j

=nXk=0

sk (k)�tn�k

�� (t)�n :

18

4.2 Inverse Relations

A pair of number sequences (�n) and���n�is called orthogonal i�

�0;n =

nXk=0

�k��n�k

for all n 2 N0, and �0 = ��0 = 1. Suppose (�n) is another number sequences with �0 6= 0, and de�ne

n :=

nXk=0

�k�n�k (14)

for all n 2 N0. It is easy to verify that

�n =nXk=0

k��n�k: (15)

The identities (14) are called inverse relations (Riordan [7, Chpt. 2]). If an inverse pair of relationsis given, only one of the two must be shown; the other will hold automatically. Constructing inversepairs hinges on �nding the sequence

���n�orthogonal to (�n). In an Umbral Calculus interpretation

of inverse relations we view (�n) as the values of a basic sequence (bn) evaluated at some argumentc 6= 0. Such a basic sequence must exist for every (�n) and every c 6= 0. Then ��n = bn (�c) becauseof the Binomial Theorem,

�0;n = bn (0) =nXk=0

bk (c) bn�k (�c) =nXk=0

�k��n�k:

The numbers �n can now be seen as the values sn (0) = �n of some She�er sequence (sn) associatedto (bn). By the Binomial Theorem

nXk=0

�k�n�k =nXk=0

sk (0) bn�k (c) = sn (c) = n:

The inverse relation (15) is another instance of the Binomial Theorem,

�n = sn (0) =nXk=0

sk (c) bn�k (�c) =nXk=0

k��n�k:

Such an inverse relation is the special case ` = 1 of the linear recurrence �n =Pn

j=1 �j�n�j + nfor all n � ` (see Proposition 32).

Example 39 The orthogonal pair

�n = c (c+ �n)n�1 =n!; ��n = c (c� �n)

n�1 (�1)n =n!

occurs with � = 1 as No. 1 in Riordan's [7, Table 2.2] Table 3.1 of Abel inverse relations. This pairis de�ned with the help of the Abel polynomials bn(x) = x (x+ �n)n�1 =n!, the basic polynomialsfor E��D. Hence �n = bn(c), and ��n = bn (�c).

19

The orthogonal pair No. 2

�n = (c+ �n)n=n! ��n =

�c2 � �2n

�(c� �n)n�2 (�1)n =n!

in the same table use the same basic polynomials but we can view (�k) as an evaluation of theShe�er sequence ((x+ �n)n=n!) for E��D at c. It is easy to see that ��k = bk (�c)� �bk�1 (�� c).This pair is orthogonal because for n > 0

nXk=0

�k��n�k =

nXk=0

(c+ �k)k

k!(bn�k (�c)� �bn�k�1 (�� c))

=(c� c+ �n)n

n!� �(c+ �� c+ � (n� 1))

n�1

(n� 1)! = 0:

In a more demanding version of inverse relations the sequence (�n) is replaced by a triangularmatrix

��n;k

�n�0; k=0:::n. The inverse relations are now de�ned by

n =nXk=0

�k�n;n�k

�n =nXk=0

k��n;n�k

for all n 2 N0, where��n;k

�and

���n;k

�are orthogonal matrices in the sense of

�n;j =nXk=j

�k;k�j��n;n�k

(Gould class of inverse relations). Finding the orthogonal pair is the main di�culty; if there existsa basic sequence (bn) such that

�n;k = bk (c+ �n)

(k � n) then Umbral Calculus tells us that

��n;k =c+ n�

c+ (n� k)�bk(�c� (n� k)�) (16)

because

nXk=j

�k;k�j��n;n�k =

nXk=j

bk�j (c+ �k)c+ n�

c+ k�bn�k(�c� k�)

=

n�jXk=0

bk (c+ j� + �k)�c� n�

�c� n� + (n� j � k)�bn�j�k(�c� n�+ (n� j � k)�)

= bn�j(c+ j�� c� n� + (n� j)�) = �n;j:

by (??). See also Proposition 38.

20

Exercise 40 Class 2 in Riordan's Table [7, Table 2.2] of Gould classes of inverse relations is thepair

�n;k =

��p� (q � 1)n� 1

k

�and ��n;k =

�p+ (q � 1) (n� k)

k

�+ q

�p+ (q � 1) (n� k)

k � 1

�(In Riordan's notation �n;n�k = An;k and ��n;n�k = (�1)

n�k Bn;k). Show: This pair is orthogonal.

The following exercises are based on the special properties��kk

�= �

��k�1k�1�and (�k)k

k!= � (�k)

k�1

(k�1)! .

Exercise 41 Show:

�n;k =

�c� � (n� k)

k

�and

��n;k =

��n� ck

�� �

��n� c� 1k � 1

�is an orthogonal pair. This is class 1 in Riordan's Table [7, Table 2.2] of Gould classes of inverserelations. Hint5

Exercise 42 Show:

�n;k =

�c� 2�n+ �k

k

�and

��n;k =2�n� c

� (2n� k)� c

�� (2n� k)� c

k

�� � 2�n� �� 1� c

� (2n� k)� 1� c

�� (2n� k)� 1� c

k � 1

�is an orthogonal pair.

Exercise 43 Show:

�n;k = (c+ � (n� k))k =k! and ��n;k = (�1)

k (c+ � (n� k)) (c+ �n)k�1 =k!is an orthogonal pair. For � = 1 this is No. 3 in Riordan's [7, Table 3.1] table of Abel inverserelations (In Riordan's notation �n;n�k = An;k= (n� k)! and ��n;n�k = (�1)

n�k Bn;k= (n� k)!).Exercise 44 Show:

�n;k =(c� �k + 2�n)k

k!and

��n;k = (�1)k (c+ 2�n� �k)n�k�2

(n� k)!�(2�n+ c)2 + �k (�� 4�n� 2c)

�is an orthogonal pair. For � = 1 and c = 0 this is No. 5 in Riordan's [7, Table 3.1] table of Abelinverse relations.

Exercise 45 Show: If (bn (x))n�0 is a sequence of binomial type and a; u are two arbitrary butnonzero constants, then

n =nXk=0

�kbn�k (un)()m+nXk=0

bn+m�k (am)�k =n+mXk=0

kam� u (n+m)

am� uk bn+m�k(am� uk)

5Note that

��n;n�k = (�1)n+k

��c� (�� 1)n� k

n� k

�+ (�� 1)

�c� 1� (�� 1)n� k

n� k � 1

��:

21

5 The Particular Functional

We are looking for the particular She�er sequence (sm) which satis�es some given \initial" condi-tions. We assume that these conditions can be expressed as values of a particular functional L,so that hL j smi is known for all m. There may be many such functionals. With every choice ofparticular functional L comes a unique shift invariant (particular) operator

�L :=Xm�0

hL j bmiBm: (17)

If hL j 1i 6= 0 then � (L) is invertible. It is shown in [8] that the operator �L is invariant underthe choice of the delta operator B and its basic sequence (bm). For example, if L = Evala it isconvenient to choose the pair D and (xm=m!) to show that

�Evala =Xm�0

am

m!Dm = eaD = Ea; (18)

the shift operator by a. It is also easy to show that a functional de�ned as hL j smi =Evala jBksm

�for all m � 0 and given a and k, has the associated operator �L = EaBk.Now we are ready to state the Functional Expansion Theorem [?].

Theorem 46 Suppose (sm)m2N0 is a B-She�er sequence and L a functional such that hL j 1i 6= 0.The polynomials sm (x) can be expanded in terms of the B-basic sequence (bm)m2N0 as

sm (x) =

mXk=0

hL j ski��1L bm�k (x) :

They have the generating function

1Xm=0

sm (x) tm =

P1k=0 hL j ski tkP1j=0 hL j bji tj

1Xm=0

bm (x) tm:

The Binomial Theorem for She�er Sequences [9]

sm (x+ a) =mXk=0

sk (a) bm�k (x) (19)

is a special case of the Functional Expansion Theorem if we choose L = Evala,

sm (x+ a) = Easm (x) = E

a

mXk=0

hEvala j ski��1Evalabm�k (x) = Ea

mXk=0

hEvala j skiE�abm�k (x)

=mXk=0

sk (a) bm�k (x) :

22

Example 47 (Tiling a rectangle with 2� 2 squares) Denote by pm (k) the number of tilingsof a 4 � m + k rectangle with m squares of size 2 � 2. Obviously p0 (k) = 1 for all k � 0, andpm (0) = 1 for even k � 0, and 0 else.

Tiling with three 2x2 squares

4 � � �3 � � �2 � � � � �1 � � � � �

1 2 3 4 5 6 7

Heubach [?] found the recursion

pk (x+ 1) = pk (x) + 3pk�1 (x) + pk�2 (x) + 2

kXr=2

pk�r (x) :

If B is the delta operator mapping pm into pm�1, the above recursion is equivalent to the operatoridentity

E1 = 1 + 3B +B2 + 2Xr�2

Br =1 + 2B �B31�B :

By (8) the B-basic sequence (bm) has the generating functionXm�0

bm (x) tm =

�1 + 2t� t31� t

�xWe choose L = Eval0 as the particular functional and obtain from the Functional Expansion Theo-rem 46

1Xm=0

pm (x) tm =

P1k=0 hL j pki tkP1j=0 hL j bji tj

�1 + 2t� t31� t

�x=

1

1� t2

�1 + 2t� t31� t

�x:

Merlini, Sprugnoli, and Verri [5, 2000] developed an algorithm transforming tiling problems likethis into a regular grammar, and \automatically" produce the generating function from the tilingproblem with the help of a computer algebra package.

5.1 Example: Area under elevated Schr�oder paths

The three di�erent step vectors on a Schr�oder path are the horizontal step (w; 0) of length w � 2,and the diagonal steps (1; 1) and (1;�1). The (w; 0)-steps get the (multiplicative) weight !. Thepaths start at the origin, end on the x-axis, and never go below the x - axis. Elevated Schr�oderpaths stay strictly above the x - axis, with the exception of the �rst and last step. Sulanke [?],[?]found the following recursion for the total weighted area �m under all elevated Schr�oder paths from(0; 0) to (m+ 2; 0).

�m =

8>>>><>>>>:2m if m is even and 0 � m � w � 10 if m is odd and 0 < m � w � 1

2w + ! (w + 1) if m = w and w is even! (w + 1) if m = w and w is odd

4bm�2 + 2!bm�w � !2bm�2w if m > w:

23

The factors �j are therefore �1 = 0; �2 = 4; �w = 2!, and �2w = �!2.

Elevated Schr�oder paths (w = 2, step vectors !!;%;&)" n 9 = area �& weight = 12 �% � �&1 �% � � � �&0 �% � � � � � �

0 1 2 3 4 5 m

" n 8 = area weight = !2 �!! �&1 �% � � � �&0 �% � � � � � �

0 1 2 3 4 5 m" n 7 �& �& 11 �% � �% � �&0 �% � � � � � �

0 1 2 3 4 5 m

" n 6 �& !1 �!! �% � �&0 �% � � � � � �

0 1 2 3 4 5 m" n 9 �& !1 �% � �!! �&0 �% � � � � � �

0 1 2 3 4 5 m

" n 5 !2

1 �!! �!! �&0 �% � � � � � �

0 1 2 3 4 5 m

�4 = 16 + 20! + 5!2

For 0 � m < w we are given the initial values �m = (1 + (�1)m) 2m�1, and also �w = (1 + (�1)w) 2w�1+! (w + 1). Hence

Xm�0

�mtm =

P`�1k=0

��k �

Pkj=2 �j�k�j

�tk

1 +P

j�2 �jtj

=

Pw�1k=0 (�k � 4�k�2) tk + (�w � 4�w�2 � 2!�0) tw

(1� !tw)2 � 4t2

=1 + (! (w + 1)� 2!) tw

(1� !tw)2 � 4t2;

�m =

m=wXj=0

�m� j (w � 1) + 1

j

�2m�1�jw (m+ 1)�1 + (�1)m�jw�!jm� j (w � 1) + 1 :

If the horizontal step vector (w; 0) has even length, w = 2v, then this sum simpli�es

�2m =

m=vXj=0

4n�jv!j�2m� j (2v � 1)

j

�2m+ 1

2m+ 1� 2vj :

For \ordinary" elevated Schr�oder paths (w = 2) holds

�2m =1

2

�1 +

p(1 + !)

�2m+1+1

2

�1�

p(1 + !)

�2m+1:

This (generating) function is familiar: The numbers �m of non-sel�ntersecting f(1; 0); (�1; 0); (0; 1)g -paths starting at (0; 0) and having lengthm (Stanley [?]) are enumerated by 1

2

�1 +

p(1 + !)

�m+1+

12

�1�

p(1 + !)

�m+1, when the weight ! is given to the runs (maximal subwalks) of ! - steps.

24

They solve the recursion �m = 2�m�1 + !�m�2.

�k and �k when ! = 1k = 0 1 2 3 4 5 6�k = 1 3 7 17 41 99 239�k = 1 0 7 0 41 0 239

6 Applying the Functional Expansion Theorem

None of the above examples needed the power of the Functional Expansion Theorem 46, becausethe binomial formula su�ces when explicit initial values are given. In [?] we investigated She�ersequences (sm) satisfying initial conditions of a di�erent form

sm(x) = tm(x) for all m = 0; 1; : : : ; `� 1Z 1

0

sm(y + cm)dy = 0 for all m = `; `+ 1; : : :

or

sm(x) = tm(x) for m = 0; : : : ; `� 1

sm(cm) =m�1Xi=0

(�1)i ci+1sm�i�1(cm) for all m � `

where t0 (x) ; : : : t`�1 (x) are given (initial) polynomials. Another common feature of the aboveexamples is that the existence of polynomial solutions to the recursion and initial values is ratherobvious. The following example di�ers on both counts.

6.1 Example: Privileged Access

Suppose a lattice walker can choose from the in�nite number of steps in the step set S :=f(1; 1) ; (1;�1)g [ f(i; 0) j i 2 N1g. The horizontal steps (i; 0) are weighted with �i. Such weightsare useful if we want to study the corresponding random walk with geometric probabilities for thehorizontal steps. We also require that the & - step keeps the walk in the �rst quadrant; it cannotbe chosen when the path has reached the x-axis. However, after leaving the origin the walker hasspecial access to the boundary y = �1 by selecting the �tting access step vector. In this examplewe let f(0;�n) jn 2 N1g be the set of special access steps, and we give them the weight �. Thewalker can vertically \jump down" to the boundary from any position. It is implied that the pathmust leave the boundary in the next step.

Sample path to (13; 1) with access steps f(0;�n) j n > 0gn2 �# �&1 � 99K �% ... �% �0 � ! �# �% ... � 99K �%�1 � �% � � � � �% � � � � � � �

0 1 2 3 4 5 6 7 8 9 10 11 12 ! m

25

The (weighted) numbers A (m;n) of such paths from the origin to (m;n) have no polynomialextension such that am (n) = A (m;n) on the support of A for any polynomial sequence (am (x)).

n A (m;n)3 12 1 3�1 1 2� 5�2 + ��+ 2 + �0 1 � 2�2 + ��+ 1 + � 4�3 + 3��2 + (3 + 3�+ �2)�+ 2�+ �2

�1 � (1 + �) ��2�2 + (�+ 1) (2 + �)

���4�3 + (5 + 3�)�2 + (6 + 4�+ �2)�+ 3 + 3�+ �2

�0 1 2 3 ! m

However the transformation M :=�1 �11 0

�maps the given problem into an equivalent walk where

such an extension exists. The steps&,%, (i; 0) are mapped into (2; 1) ; (0; 1) and (i; i) ; i 2 N1, thelatter taken with weight �i. The transformed special access steps (0;�n) are the horizontal steps(n; 0), n > 0, with weight �. Let D (m;n) be the weighted number of transformed paths to (m;n),D (m;n) = A (n; n�m). We denote the polynomial extension of D (m;n) by dm (n)

dm (n) = D (m;n) for all n � m� 1:

The polynomial extension dm (n) (special access values in boxes)n 1 3� 5�2 + ��+ 2 + � 4�3 + 3��2 + (3 + 3�+ �2)�+ 2�+ �2

2 1 2� 2�2 + ��+ 1 + � 0�3 + 2��2 + � (0 + 2�+ �2) + 2�+ �2

1 1 � 0�2 + ��+ 0 + � ��3+��2+���1+�+�2

�+2�+�2

0 1 0� ��2+��� 1+� 0�3+0�2+��0+ 0�+�2

�+2�+�2

�1 1 �� ��2+��� 2+� ...

0 1 2 3 ! m(the bold face values are extensions outside the support of A)

The polynomials are solutions to the system of di�erence equations

dm (x)� dm (x� 1) = dm�2 (x� 1) +Xj�1

�jdm�j (x� j)

and must satisfy the initial conditions d0 (x) = 1, d1 (0) = 0, and the transformed access condition

dm (m� 1) =Xi�1�dm�i (m� 1) (20)

for all m > 1. Denote by B the linear operator that maps dm (x) into dm�1 (x). The recursion for(dm) shows that

r = E�1B2 +Xj�1

E�j�jBj = E�1B2 +E�1�B

1� E�1�B (21)

26

The basic sequence (bm (x)) can be expanded according to the transform formula (7)

bm (x) =

mXi=1

x [Bm]

�E�1B2 +

E�1�B

1 + E�1�B

�i1

x

�i+ x� 1

i

�

=

mXi=1

iXj=0

�i

j

��m� 2j � 1m� i� j

��m�2j

x

i

�x+ i�m+ j � 1

i� 1

�for m > 1, because

[Bm]

�E�1B2 +

E�1�B

1 + E�1�B

�i= E�i

�Bm�i

��B +

�

1� E�1�B

�i=�Bm�i

� iXj=0

�i

j

�Bi�jE�i�j

�1� E�1�B

��j=

iXj=0

�i

j

��m+ 2j � 2i� 1m+ j � 2i

�E�m�j+i�m+2j�2i =

iXj=0

�i

j

��m� 2j � 1m� i� j

�E�m+j�m�2j:

The operator identity (21) can be \simpli�ed" to the �nite recursion

r = �E�1B�2� E�1

�+ E�1B2

�1� E�1�B

�:

After solving for E1 we obtain from (8)

Xm�0

bm (x) tm =

1

2

�t2 + 2�t+ 1

�+

r1

4(t2 + 2�t+ 1)2 � �t (1 + t2)

!xXm�0

b(1)m (x) tm =

0@1� 2�t�q(1� 2�t)2 � 4t2 (�t� 1)2

2t2 (1� �t)

1Ax

:

It is convenient to switch to the polynomials d[1]n (x) = dn (n+ x) with delta operator B(1) = E

�1B

and basic polynomials b(1)m = xbm (x+m) = (x+m) (see Remark ??). In terms of

�d[1]m

�the privilege

access condition (20) becomes d[1]m (�1) =

Pi�1 �d

[1]m�i (i� 1). We de�ne the particular functional

L by linear extension ofL j d[1]m

�=Eval�1 j d[1]m

��Xi�1�Evali�1 jBi(1)d[1]m

�(22)

and obtain the particular operator

�L = E�1 �

Xi�1�Ei�1Bi(1) = E

�1�1�

�E1B(1)1� E1B(1)

�:

Note thatDL j d[1]0

E= 1,

DL j d[1]1

E= d

[1]1 (�1) � � = ��, and

DL j d[1]m

E= 0 for all m > 1. By the

27

Functional Expansion Theorem 46,

dm (x) = d[1]m (x�m) = � (L)�1�b(1)m (x�m)� �b(1)m�1 (x�m)

�=

1� E1B(1)1� (1 + �)E1B(1)

�b(1)m (x�m+ 1)� �b(1)m�1 (x�m+ 1)

�=

�1 +

�E1B(1)1� (1 + �)E1B(1)

��b(1)m (x�m+ 1)� �b(1)m�1 (x�m+ 1)

�= b(1)m (x�m+ 1)� �b(1)m�1 (x�m+ 1) +

�2

1 + �b(1)m�1 (x�m+ 1)

+�mXl=1

(1 + �)l�1 b(1)m�l (x+ l �m+ 1)� �

mXl=1

(1 + �)l�2 �b(1)m�l (x+ l �m)

= b(1)m (x�m+ 1)� �

1 + �b(1)m�1 (x�m+ 1)

+�mXl=1

(1 + �)l�1�b(1)m�l (x+ l �m+ 1)�

�

1 + �b(1)m�l (x+ l �m)

�= (x�m+ 1)

�bm (x+ 1)

x+ 1� �bm�1 (x)(1 + �)x

�+�

mXl=1

(1 + �)l�1�x+ l �m+ 1

x+ 1bm�l (x+ 1)�

� (x+ l �m)(1 + �)x

bm�l (x)

�

7 Bivariate Umbral Calculus

For the remaining part of the paper we will focus on the bivariate case; generalizations of thefollowing theorems to higher dimensions are straight forward. More on multivariate umbral calculuscan be found in [?, 1979] and [?], [?]. A bivariate She�er sequence (pm;n (x; y))m;n2N0 has a generatingfunction of the form X

m;n�0pm;n (x; y) s

mtn = � (s; t) ex�1(s;t)+y�2(s;t)

where � (s; t) 2 K [[s; t]] has order 0, i.e., � (0; 0) 6= 0, and �1 (s; t) ; �2 (s; t) is a pair of delta seriesin K [[s; t]], which means that �1 (s; t) =s and �2 (s; t) =t are both power series of order 0 in K [[s; t]].If � (s; t) = 1 the resulting She�er sequence is a bivariate basic sequence. For every pair of deltaseries �1 (s; t) ; �2 (s; t) there exists a compositional inverse 1 (s; t) ; 2 (s; t), also a delta pair, suchthat

�1 ( 1 (s; t) ; 2 (s; t)) = s and �2 ( 1 (s; t) ; 2 (s; t)) = t:

Let D1 := @=@x and D2 := @=@y. Using the above notation it can be shown that

1 (D1;D2) pm;n (x; y) = pm�1;n (x; y) and 2 (D1;D2) pm;n (x; y) = pm;n�1 (x; y) :

We call 1 (D1;D2) ; 2 (D1;D2) a pair of delta operators, and (pm;n)the associated She�er sequence.Of course, all operators in K [[D1;D2]] commute. Again we will state two transfer theorems.

28

Theorem 48 (Bivariate Transfer Theorem I) Let Q1; Q2 be a delta operator with Q1; Q2 -basic sequence(qm;n (x; y))m;n2N0. If Q1 = �1 (B1; B2) and Q2 = �2 (B1; B2) for some pair �1; �2 of delta series andlinear operator B1; B2, then B1; B2 is also a pair of delta operators, and the B1; B2-basic sequence(bm;n)m2N0 has for all m;n 2 N0 the expansion

bm;n (x; y) =

mXi=0

nXj=0

cm;n;i;jqi;j (x; y) (23)

where cm;n;i;j is the coe�cient of Bm1 B

n2 in �1 (B1; B2)

i �2 (B1; B2)j. For the generating function of

the basic sequence (bm;n) holdsXm;n�0

bm;n (x; y) smtn =

Xi;j�0

qi;j (x; y)�1 (s; t)i �2 (s; t)

j : (24)

The proof is analogous to the proof of Theorem 22.The Jacobian J (�1; �2) of a pair �1 (s; t) ; �2 (s; t) of power series is de�ned as

J (�1; �2) (s; t) :=

���� @P1=@s @P2=@s@P1=@t @P2=@t

���� :Theorem 49 (Bivariate Transfer Theorem II) Let Q1; Q2 be a delta operator with Q1; Q2 -basic sequence(qm;n (x; y))m;n2N0. If B1 = �1 (Q1; Q2)Q1 and B2 = �2 (Q1; Q2)Q2 for some pair �1; �2 of power se-ries of order 0, then B1; B2 is also a pair of delta operators, and the B1; B2-basic sequence (bm;n)m2N0has for all m;n 2 N0 the expansion

bm;n (x; y) = �1 (Q1; Q2)�m�1 �2 (Q1; Q2)

�n�1 J (Q1�1; Q2�2) (Q1; Q2) qm;n (x; y) (25)

Proof. Suppose Qi = ��1i (D1;D2) and Bi = ��1i (D1;D2) for some pairs of delta series �1; �2

and �1; �2. For notational simplicity we de�ne

!i (s; t) := ��1i (�1 (s; t) ; �2 (s; t)) :

HenceXm;n�0

bm;n (x; y) smtn = ex�1(s;t)+y�2(s;t) =

Xm;n�0

qm;n (x; y)!1 (s; t)m !2 (s; t)

n

=Xm;n�0

qm;n (x; y)X

i�m;j�nsitj

�sitj�!1 (s; t)

m !2 (s; t)n

=X

i�0;j�0sitj

X0�m�i;0�n�j

�sitj�!1 (s; t)

m !2 (s; t)n qm;n (x; y)

=X

i�0;j�0sitj

Xm�0;n�0

�sitj�!1 (s; t)

i�m !2 (s; t)j�nQm1 Q

n2qi;j (x; y) :

29

By Lagrange - B�urmann inversion�sitj�!1 (s; t)

i�m !2 (s; t)j�n =

�sm�i�1tn�j�1

�!�11 (s; t)�i�1 !�12 (s; t)�j�1 J

�!�11 ; !

�12

�(s; t)

= [smtn]

�!�11 (s; t)

s

��i�1�!�12 (s; t)

t

��j�1J�!�11 ; !

�12

�(s; t) :

Noting that!�11 (s; t) = s�1 (s; t) and !�12 (s; t) = t�2 (s; t)

�nishes the proof.

Example 50 Suppose we have the pair of operator identities

r1 = E�11 B1

�2 = E�11 B2 + !B1E�11 B2

We apply the above theorem with B1 = �1 and B2 = �2=�E�11 + !B1E

�11

�= �2 (1 + �1) = (1 + !�1).

Hence �1 (s; t) = 1, �2 (s; t) = (1 + s) = (1 + !s), and

bm;n (x; y) = �1 (�1;�2)�m�1 �2 (�1;�2)

�n�1 J (s�1; t�2) (�1;�2)

�x

m

��y

n

�=

�1 + �1

1 + !�1

��n�1 ����� 1 �2@�2@�1

0 1+�11+!�1

������x

m

��y

n

�= E�n1 (1 + !�1)

n

�x

m

��y

n

�=1Xi=0

�n

i

�!i�x� nm� i

��y

n

�:

For comparison we apply the �rst Transfer Theorem to the same problem, with �1 (B1; B2) = B1and

�2 (B1; B2) =B2 (1 + !B1)

E11=B2 (1 + !B1)

1 + �1

=B2 (1 + !B1)

1 +B1:

Hence the B - basic sequence has the generating function isXm;n�0

bm;n (x; y) smtn =

Xm;n�0

�x

m

��y

n

�sm�t (1 + !s)

1 + s

�n= (1 + s)x

�1 + t

1 + !s

1 + s

�y:

30

If we apply the bivariate transfer formula 24 we get the expression

bm;n (x; y) =mXi=0

nXj=0

[Bm1 B

n2 ]B

i1

�B2 (1 + !B1)

1 +B1

�j!�x

i

��y

j

�

=

�y

n

� mXi=0

��Bm�i1

��1 +

(! � 1)B11 +B1

�n��x

i

�

=

�y

n

� mXi=0

�Bm�i1

� nXk=0

�n

k

��(! � 1)B11 +B1

�k!�x

i

�=

�y

n

� mXi=0

nXk=0

�n

k

�(! � 1)k

��Bm�i�k1

�(1 +B1)

�k��x

i

�=

�y

n

� 1Xk=0

�n

k

�(! � 1)k

�x� km� k

�:

The two solution we found for bm;n (x; y) re ect the well-known hypergeometric function identity�x

m

�2F1

��m;�n�x ; 1� !

�=

�x� nm

�2F1

��m;�n

1� n�m� x ;!�:

The Functional Expansion Theorem 46 generalizes as expected, with �L de�ned as

�L :=Xm;n�0

hL j bm;niBm1 Bn2 (26)

for any B1; B2 - basic sequence (bm;n).

Theorem 51 Suppose (sm;n)m;n2N0 is a B1B2 - She�er sequence and L a linear functional onK [[s; t]] such that hL j 1i 6= 0. The polynomials sm;n (x; y) can be expanded in terms of the B1B2 -basic sequence (bm;n) as

sm;n (x) =

mXj=0

nXk=0

hL j sj;ki��1L bm�j;n�k (x; y) :

They have the generating function

Xm;n�0

sm;n (x; y) smtn =

Pj;k�0 hL j sj;ki sjtkPj;k�0 hL j bj;ki sjtk

Xm;n�0

bm;n (x; y) smtn:

7.1 Bi-indexed Umbral Calculus

Three variable recursions like pm;n (�) = pm;n (� � 1)+pm�1;n (�)+pm;n�1 (�) are not part of the abovebivariate theory. Very basic combinatorial objects, like multinomial coe�cients, follow recursionsof this type. An obvious approach to a polynomial theory for this type of recursions equates the

31

variable x and y in a bivariate polynomial sequence to give (pm;n (�))m;n2N0 . Abusing the notationwe we write pm;n (�) = pm;n (�; �), and we introduce the diagonalization operator � : K [x; y]! K [�]

�p (x; y) = p (�) :

Suppose (sm;n) is a She�er sequence with generating function � (s; t) ex�1(s;t)+y�2(s;t). By K [[s; t]] -

linear extension we de�ne

�Xm;n�0

sm;n (x; y) smtn :=

Xm;n�0

�sm;n (x; y) smtn = � (s; t) e�(�1(s;t)+�2(s;t)):

The following lemma is easy to verify.

Lemma 52 For all k 2 N0 holds� (D1 +D2)k = Dk�

A bi-indexed She�er sequence (sm;n (�)) is a double sequence of univariate polynomials such thatdeg sm;n (�) = m+ n for all m;n 2 N0, s0;0 (�) 6= 0, and

� (s; t) e��(s;t)

where � (s; t) is of order 0, and � (s; t) = �1 (s; t) + �2 (s; t) for some pair of bivariate delta series.Obviously this pair is not unique. If (sm;n (x; y)) is a bivariate Q1; Q2 - She�er sequence then(�sm;n (x; y)) is a bi-indexed She�er sequence, and

� (Q1 +Q2) sm;n (x; y) = � (sm�1;n (x; y) + sm;n�1 (x; y)) :

We say that (sm;n (�)) is a Q - She�er sequence i� Qsm;n (�) = sm�1;n (�) + sm;n�1 (�), and call Qa delta operator associated to (sm;n (�)). Trivial examples of bi-indexed She�er polynomials areproducts of two univariate She�er polynomials. The following example shows another importantclass.

Example 53 Let be a univariate delta series, and let (qm)m2N0 be the ordinary univariate Q =

�1 (D) - basic sequence. De�ne pm;n (�) :=�m+nm

�qm+n (�) for all m;n 2 N0. Hence

Xm;n2N0

pm;n (�) smtn =

Xk2N0

qk (�)

kXm=0

�k

m

�smtk�m = e� (s+t)

If we de�ne�1 (s; t) = (s+ t)� (t) and �2 (s; t) = (t)

then �1 (s; t) ; �2 (s; t) is a delta pair, and �1 (s; t)+�2 (s; t) = (t). This shows that��m+nm

�qm+n (�)

�m;n2N0

is a bi-indexed basic sequence. It is a Q - basic sequence because

Qpm;n (�) =

�m+ n

m

�qm+n�1 (�) = pm�1;n (�) + pm;n�1 (�) :

Note that Q� = � (Q1 +Q2) if

Q1 = �1 (D1 +D2)� �1 (D2) and Q2 =

�1 (D2) (27)

32

We can check this by applying Lemma 52:� (Q1 +Q2) = � �1 (D1 +D2) = �1 (D). The Q1; Q2 -

basic polynomials diagonalize to pm;n (�) = �qm;n (x; y), andXm;n�0

qm;n (x; y) smtn = ex (s+t)+(y�x) (t) (28)

qm;n (x; y) =nXi=0

�m+ n� i

m

�qm+n�i (x) qi (y � x) : (29)

Properties of the bivariate behind the bi-indexed She�er sequences can be used to solve certain re-cursions. For example, if (qm;n (�)) is a Q - basic sequence, then fn;m (�) := (m+ 1) qm+1;n (�) =� is aQ - She�er polynomial, and so is gn;m (�) := (n+ 1) qm;n+1 (�) =�, and the shifted linear combination

sm;n (�) := qm;n (� � �)� afm�1;n (� � �)� bgm;n�1 (� � �) = (� � � � am� bn)qm;n (� � �)� � � (30)

for some given constants a and b. ThisQ - She�er sequence satis�es the initial condition sm;n (� + am+ bn) =�0;m�0;n. If (rm;n (�)) is any Q - She�er sequence, and Q� = � (Q1 +Q2), then (rm;n (� + am+ bn))is a Q(a;b) - She�er sequence, where Q(a;b)� = �

�E�a1 E

�a2 Q1 + E

�b1 E

�b2 Q2

�. The bi-indexed polyno-

mial sequence�� qm;n(�+am+bn)

�+am+bn

�m;n2N0

is theQ(a;b) - basic sequence, the Abelization of (qm;n (�)).More

about multi-indexed UC can be found in [?, 1980] and Watanabe [?, 1986].

7.2 Example: A Slow Walk

Suppose the lattice walker can choose at every tick of the clock one of the steps f!; ";%;�g, where�= (0; 0) is the \pause step". Give weight ! to the% - step vector. Let be the number of weightedpaths from (0; 0) to (m;n) in k steps (seconds) under the restriction

D (i; j; l) = 0

if (time) l 6 ai + bj where a and b are positive integers, and (i; j) 6= (0; 0). In other words, thelattice point (i; j) can not be reached in a short time l; the walk is slow.

D (n;m; k) for a = 2 and b = 1; ! = 1

4

3

2

1

1"%�!

4 6

3 5

2 4

01 031"%�! 02

4 6 8

3 5 7

02 04 0611 03 0510 02 04

4 6 8

03 05 0712 04 0621 03 0510 12 04

04 06 08 01013 05 07 0932 04 06 0831 43 05 0710 22 04 06

k = 0 k = 1 k = 2 k = 3 k = 4The subscripts show the values of ai+ bj. The levels k = ai+ bj are in bold.

The solution to the di�erence equation

D (n;m; k)�D (n;m; k � 1) = D (n� 1;m; k � 1)+D (n;m� 1; k � 1)+!D (n� 1;m� 1; k � 1)

can be extended to a polynomial sequence (dm;n (�)) because D (0; 0; k) = 1 for all k � 0. Of course,D (m;n; k) = 0 if any of the three parameters is negative. In order to calculate the nonzero values

33

of D (m;n; k) the only additional initial condition is D (m;n; am+ bn) = �m;0�n;0. The recursionand operator equation for the polynomial extension are

dn;m (� + 1) = dn;m (�) + dn�1;m (�) + dn;m�1 (�) + !dn�1;m�1 (�)

�� = � (B1 +B2 + !B1B2)

where Bdn;m (�) = dn;m (� � 1) + dn�1;m (� � 1) and B� = � (B1 +B2). In addition to d0;0 (�) = 1the solution also must have the initial values dn;m (an+ bm) = �(n;m);(0;0). It follows from Lemma52 that �� = � (E11E

12 � 1), hence it is su�cient to determine B such that

B1 +B2 + !B1B2 = E11E12 � 1 = E11

�E12 � 1

�+ E11 � 1

E�11 (B1 +B2 + !B1B2) = r1 +�2:

We decide to split this sum of operators into r1 = E�11 B1 and �2 = E�11 B2 + !B1E�11 B2,

because we found already the B1; B2 - basic sequence (bm;n (x; y)) in Example 50, bm;n (x; y) =�yn

�Pmi=0

�ni

��x�nm�i�!i. The numbers U (m;n; k) of unrestricted paths to (m;n) in k steps have initial

values U (m;n; 0) = 0 except for U (0; 0; 0) = 1. Hence U (m;n; k) = bm;n (k; k) on the support ofU . The slow walk extensions have initial values dn;m (an+ bm) = �(n;m);(0;0). We apply formula (30)and get

D (m;n; k) = dn;m (k) =k � an� bm

k

�k

n

� mXi=0

�n

i

��k � nm� i

�!i

on the support of D.

References

[1] Di Bucchianico,A. (1997) Probabilistic and Analytic Aspects of the Umbral Calculus. CWITract, Amsterdam, ISBN 90 6196 471 7.

[2] Freeman, J.M. (1985). Transforms of operators on K[x][[t]], Congr. Numerantium 48, 115-132.

[3] Heubach, S. (1999) Tiling an n-by-m area with squares of size up to k-by-k (m � 5). Congr.Numerantium 140, 43-64, www.calstatela.edu/faculty/sheubac/CGTC30.PDF.

[4] Niederhausen, H. A Formula for Explicit Solutions of Certain Linear Re-cursions on Polynomial Sequences, Congr. Numerantium 49 (1985) 87-98,www.math.fau.edu/Niederhausen/HTML/Papers/Formula4explicitSolutions.ps

[5] Merlini, D., Sprugnoli, R. and Verri, M. C. (2000). Strip tiling and regular grammars, Th.Computer Sc., 242, 109-124

[6] Mullin, R., Rota, G. C. (1970). On the Foundations of Combinatorial Theory III: Theory ofBinomial Enumeration, in B. Harris (ed.). Graph Theory and Its Applications, Academic Press,New York, pp. 167-213.

[7] Riordan, J. (1968). Conbinatorial Identities, Wiley, New York.

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[8] Roman, S.M., and Rota,G.-C. (1978). The Umbral Calculus, Adv. in Math. 27, 95 - 188.

[9] Rota, G.-C., Kahaner, D., and Odlyzko, A. (1973). On the Foundations of CombinatorialTheory VIII: Finite operator calculus, J. Math. Anal. Appl. 42, 684-760.

[10] Stanley, R.P. (1986) Enumerative Combinatorics, Vol. I, Wadsworth & Brooks/Cole, Monterey,California.

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