Constructing Exponential Riordan Arrays from Their A and Z Sequences

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Abstract We show how to construct an exponential Riordan array from a knowledge of its A and Z sequences. The effect of pre- and post-multiplication by the binomial matrix on the A and Z sequences is examined, as well as the effect of scaling the A and Z sequences. Examples are given, including a discussion of related Sheffer orthogonal polynomials.

1 Introduction

One of the most fundamental results concerning Riordan arrays is that they have a sequence characterization [13, 18]. This normally involves two sequences, called the A-sequence and the Z-sequence. For exponential Riordan arrays [9] (see Appendix), this characterization is equivalent to the fact that the production matrix [11] of an exponential array [g, f], with A-sequence A(t) and Z-sequence Z(t) has bivariate generating function

e^{zt}(Z(t) + A(t)z).

In this case we have

A(t) = f′(f̄(t)), Z(t) = \frac{g′(f̄(t))}{g(f̄(t))}.

Examples of exponential Riordan arrays and their production matrices may be found in the On-Line Encyclopedia of Integer Sequences [19, 20]. In that database, sequences are referred to by their A-numbers. For known sequences, we shall adopt this convention in this note.

A natural question to ask is the following. If we are given two suitable power series A(t) and Z(t), can we recover the corresponding exponential Riordan array [g(t), f(t)] whose A and Z sequences correspond to the given power series A(t) and Z(t)?

The next two simple results provide a means of doing this.

Lemma 1. For an exponential Riordan array [g(t), f(t)] with A-sequence A(t), we have

\frac{d}{dt}f̄(t) = \frac{1}{A(t)}.

Proof. By definition of the compositional inverse, we have

f(f̄(t)) = t.

Differentiating this with respect to t, we obtain

f′(f̄(t))\frac{d}{dt}f̄(t) = 1

or

\frac{d}{dt}f̄(t) = \frac{1}{f′(f̄(t))} = \frac{1}{A(t)}.

Lemma 2. For an exponential Riordan array [g(t), f(t)] with A-sequence A(t) and Z-sequence Z(t), we have

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)}.

Proof. We have

\frac{d}{dt} ln(g(f̄(t))) = \frac{g′(f̄(t))}{g(f̄(t))} \frac{d}{dt}f̄(t) = Z(t)\frac{1}{A(t)} = \frac{Z(t)}{A(t)}.

Thus if we can easily carry out the reversion from f̄(t) to f(t), a knowledge of A(t) and Z(t), along with the equations

\frac{d}{dt}f̄(t) = \frac{1}{A(t)},  \frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)}  (1)

will allow us to find f(t) and g(t). The steps to achieve this are as follows.

• Using the equation \frac{d}{dt} f̄(t) = \frac{1}{A(t)}, solve for f̄(t).

• Revert f̄(t) to get f(t).

• Sove the equation \frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} and take the exponential to get g(f̄(t)).

• Solve for g(t) by substituting f(t) in place of t in the last found expression. Constants of integration may be determined using such conditions as f̄(0) = f(0) = 0, and g(0) = 1. Example 3. We seek to find [g(t), f(t)] where

A(t) = \frac{1}{1 + t}, Z(t) = −\frac{1}{1 + t}.

We start by solving the equation

\frac{d}{dt}f̄(t) = 1 + t.

Since f̄(0) = 0, we find that

f̄(t) = t + \frac{t^{2}}{2} = t (1 + \frac{t}{2}) .

We revert this to get

f(t) = \sqrt{1 + 2t} − 1.

We now solve the equation

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = −1.

Thus we find that

ln(g(f̄(t))) = −t ⇒ g(f̄(t)) = e^{−t}.

Thus (since f̄(f(t)) = t) we get

g(t) = e^{−f(t)} = e^{1−\sqrt{1+2t}}.

Hence the exponential Riordan array with the given A and Z sequences is

[g, f] = [e^{1−\sqrt{1+2t}}, \sqrt{1 + 2t} − 1] .

We note that

[g, f]^{−1} = [e^{t}, t + \frac{t^{2}}{2} ]

which is the Pascal-like matrix \underline{A100862} [6].

In like manner, we can show that

A(t) = \frac{1}{1 + 2t}, Z(t) = −\frac{1}{1 + 2t}

corresponds to the exponential Riordan array

[g, f] = [e^{\frac{1−\sqrt{1+4t}}{2}} , \frac{\sqrt{1 + 4t} − 1}{2}] ,

whose inverse

[g, f]^{−1} = [e^{t}, t + t^{2}]

is Pascal-like [6]. In general, if A(t) = −Z(t) = \frac{1}{1+rt}, then

[g, f] = [e^{\frac{1}{r} (1−\sqrt{1+2rt})}, \frac{1}{r}(\sqrt{1 + 2rt} − 1)] .

Then

[g, f]^{−1} = [e^{t}, t + r\frac{t^{2}}{2} ]

is a Pascal-type matrix.

2 Effect of the binomial transform

The next proposition shows the effect of changing Z(t) to Z(t) + 1 and to Z(t) + A(t), respectively. We recall that the binomial matrix B = [e^{t}, t].

Proposition 4. Let [g, f] be an exponential Riordan array with A and Z sequences A(t) and Z(t) respectively. Then the exponential Riordan array B · [g, f] has A and Z sequences A(t) and Z(t)+1 respectively, while the exponential Riordan array [g, f]·B has A and Z sequences A(t) and Z(t) + A(t) respectively.

Proof. Firstly, we let the exponential Riordan array [h, l] have A and Z sequences A(t) and Z(t) + 1 respectively. Then we have \frac{d}{dt} l̄(t) = \frac{1}{A(t)}, which implies that l(t) = f(t) (since l(0) = f(0) = 0). Now

\frac{d}{dt} ln(h(l̄(t))) = \frac{d}{dt} ln(h(f̄(t))) = \frac{Z(t) + 1}{A(t)} = \frac{Z(t)}{A(t)} + \frac{1}{A(t)}.

Thus

ln(h(f̄(t))) = ln(g(f̄(t))) + f̄(t) ⇒ h(f̄(t)) = g(f̄(t))e^{f̄(t)}.

We obtain that

h(t) = g(t)l^{t}

and so

[h(t), l(t)] = [e^{t}g(t), f(t)] = [e^{t}, t] · [g(t), f(t)] = B · [g(t), f(t)].

Secondly, we now assume that the exponential Riordan array [h, l] have A and Z sequences A(t) and Z(t) + A(t) respectively. As before, we see that l(t) = f(t). Also,

\frac{d}{dt} ln(h(l̄(t))) = \frac{d}{dt} ln(h(f̄(t))) = \frac{Z(t) + A(t)}{A(t)} = \frac{Z(t)}{A(t)} + 1.

Thus

ln(h(f̄(t))) = ln(g(f̄(t))) + t ⇒ h(f̄(t)) = g(f̄(t))e^{t}.

Now substituting f(t) for t gives us

h(t) = e^{f(t)}g(t).

Thus

[h, l] = [e^{f(t)}g(t), f(t)] = [g(t), f(t)] · [e^{t}, t] = [g(t), f(t)] · B.

We shall see examples of these results in the next section.

3 Effect of Scaling

In this section, we will assume that the exponential Riordan array with A and Z sequences A(t) and Z(t), respectively, is given by [g(t), f(t)]. We wish to characterize the exponential Riordan array [g^{∗}(t), f^{∗}(t)] whose A and Z sequences are A^{∗}(t) = rA(t) and Z^{∗}(t) = sZ(t) respectively.

Proposition 5. We have

[g^{∗}(t), f^{∗}(t)] = [g(rt)^{\frac{s}{r}} , rf(t)] .

Proof. We have

\frac{d}{dt}f̄^{∗}(t) = \frac{1}{rA} = \frac{1}{r} \frac{d}{dt}f̄(t).

Thus

f̄^{∗}(t) = \frac{1}{r}f̄(t) ⇒ f^{∗}(t) = rf(t).

Then

\frac{d}{dt} ln(g^{∗}(f̄^{∗}(t))) = \frac{sZ}{rA} = \frac{s}{r} \frac{d}{dt} ln(g(f̄(t))),

and so

ln(g^{∗}(f̄^{∗}(t))) = \frac{s}{r} ln(g(f̄(t))) = ln (g(f̄(t))^{\frac{s}{r}} ) .

Thus

g^{∗}(f̄^{∗}(t)) = g(f̄(t))^{\frac{s}{r}} ⇒ g^{∗}(\frac{1}{r}f̄(t)) = g(f̄(t))^{\frac{s}{r}} ⇒ g^{∗}(\frac{1}{r}t) = g(t)^{\frac{s}{r}} ,

or

g^{∗}(t) = g(rt)^{\frac{s}{r}} .

Example 6. We let

A(t) = 1 + t, Z(t) = 1 + 2t.

We find that the corresponding exponential array is

[g, f] = [e^{2e^{t}−t−2}, e^{t} − 1] ,

which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 3 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 9 & 13 & 6 & 1 & 0 & 0 & ⋯ \\ 35 & 59 & 37 & 10 & 1 & 0 & ⋯ \\ 153 & 301 & 230 & 85 & 15 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

with production matrix which begins

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 2 & 2 & 1 & 0 & 0 & 0 & … \\ 0 & 4 & 3 & 1 & 0 & 0 & … \\ 0 & 0 & 6 & 4 & 1 & 0 & … \\ 0 & 0 & 0 & 8 & 5 & 1 & … \\ 0 & 0 & 0 & 0 & 10 & 6 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We now take

A^{∗}(t) = 3(1 + t), Z^{∗}(t) = 5(1 + 2t).

The corresponding exponential Riordan array is then given by

[g^{∗}(t), f^{∗}(t)] = [(e^{2e^{3t}−3t−2})^{\frac{5}{3}} , 3(e^{t} − 1)] .

This array begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 5 & 3 & 0 & 0 & 0 & 0 & ⋯ \\ 55 & 33 & 9 & 0 & 0 & 0 & ⋯ \\ 665 & 543 & 162 & 27 & 0 & 0 & ⋯ \\ 9895 & 9033 & 3573 & 702 & 81 & 0 & ⋯ \\ 165185 & 170103 & 76410 & 19575 & 2835 & 243 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

with production matrix which begins

( \begin{matrix}5 & 3 & 0 & 0 & 0 & 0 & … \\ 10 & 8 & 3 & 0 & 0 & 0 & … \\ 0 & 20 & 11 & 3 & 0 & 0 & … \\ 0 & 0 & 30 & 14 & 3 & 0 & … \\ 0 & 0 & 0 & 40 & 17 & 3 & … \\ 0 & 0 & 0 & 0 & 50 & 20 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

4 Further examples

Example 7. We take the Stirling number related choice of

A(t) = 1 + t, Z(t) = 1 + t.

From

\frac{d}{dt}f̄(t) = \frac{1}{1 + t},

we obtain

f̄(t) = ln(1 + t) ⇒ f(t) = e^{t} − 1.

Then from

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = 1

we obtain

ln(g(f̄(t))) = t ⇒ g(f̄(t)) = e^{t},

and hence

g(t) = e^{e^{t}−1}.

Thus we obtain

[g, f] = [e^{e^{t}−1}, e^{t} − 1] ,

which is \underline{A049020}. We have

[g, f] = S_{2} · B

where S_{2} is the matrix of Stirling numbers of the second kind (\underline{A048993}) and B is the binomial matrix (\underline{A007318}). The production array of [g, f] is given by

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 2 & 1 & 0 & 0 & 0 & … \\ 0 & 2 & 3 & 1 & 0 & 0 & … \\ 0 & 0 & 3 & 4 & 1 & 0 & … \\ 0 & 0 & 0 & 4 & 5 & 1 & … \\ 0 & 0 & 0 & 0 & 5 & 6 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Since this production matrix is tri-diagonal, the inverse matrix [g, f]^{−1} is the coefficient array of a family of orthogonal polynomials [4, 3]. The family in question is the family of Charlier polynomials, which has the Bell numbers (with e.g.f. e^{e^{t}−1}) as moments. The Charlier polynomials satisfy the three-term recurrence

P_{n}(t) = (t − n)P_{n−1}(t) − (n − 1)P_{n−2}(t),

with P_{0}(t) = 1, P_{1}(t) = t − 1.

Example 8. We take

A(t) = 1 + t Z(t) = 1 + t + t^{2}.

Again, we find that

f(t) = e^{t} − 1.

Then

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = \frac{1 + t + t^{2}}{1 + t} ,

and hence

ln(g(f̄(t))) = \frac{t^{2}}{2} + ln(1 + t).

Thus

g(f̄(t)) = e^{\frac{t^{2}}{2}} (1 + t),

and so

g(t) = e^{\frac{(e^{t}−1)^{2}}{2}} (1 + e^{t} − 1) = e^{t}e^{\frac{(e^{t}−1)^{2}}{2}} .

In this case, the production matrix is four-diagonal and begins

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 2 & 1 & 0 & 0 & 0 & … \\ 2 & 2 & 3 & 1 & 0 & 0 & … \\ 0 & 6 & 3 & 4 & 1 & 0 & … \\ 0 & 0 & 12 & 4 & 5 & 1 & … \\ 0 & 0 & 0 & 20 & 5 & 6 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The exponential Riordan array

[g, f] = [e^{t}e^{\frac{(e^{t}−1)^{2}}{2}} , e^{t} − 1]

begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 7 & 10 & 6 & 1 & 0 & 0 & ⋯ \\ 29 & 45 & 31 & 10 & 1 & 0 & ⋯ \\ 136 & 241 & 180 & 75 & 15 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The row sums of this array are the Dowling numbers \underline{A007405}.

We note that the exponential Riordan array

B^{−1} · [g, f] = [e^{−t}, t] · [g, f] = [e^{\frac{(e^{t}−1)^{2}}{2}} , e^{t} − 1]

has

A(t) = 1 + t Z(t) = t + t^{2}.

This array begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 0 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 1 & 0 & 0 & 0 & ⋯ \\ 3 & 4 & 3 & 1 & 0 & 0 & ⋯ \\ 10 & 19 & 13 & 6 & 1 & 0 & ⋯ \\ 45 & 91 & 75 & 35 & 10 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The first column of this array is \underline{A060311}, while its row sums are given by \underline{A004211}. The production matrix of this array begins

( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 1 & 0 & 0 & 0 & … \\ 2 & 2 & 2 & 1 & 0 & 0 & … \\ 0 & 6 & 3 & 3 & 1 & 0 & … \\ 0 & 0 & 12 & 4 & 4 & 1 & … \\ 0 & 0 & 0 & 20 & 5 & 5 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

where we see that the effect of the inverse binomial matrix is to subtract 1 from the diagonal.

In this example, we have Z(t) = 1 + t + t^{2} = A(t) + t^{2}. Thus the exponential Riordan array [g, f] is equal to the product

[h, l] · B

where the exponential Riordan array [h, l] has A and Z sequences of 1+t and t^{2}, respectively.

Example 9. We take

A(t) = 1 + t^{2}, Z(t) = 1 + t + t^{2}.

Then Thus

f(t) = tan(t).

Now

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = \frac{1 + t + t^{2}}{1 + t^{2}} = 1 + \frac{t}{1 + t^{2}} ,

and so

ln(g(f̄(t))) = ln \sqrt{1 + t^{2}} + t.

Thus

g(f̄(t)) = e^{t}\sqrt{1 + t^{2}} ⇒ g(t) = e^{tan(t)}\sqrt{1 + tan^{2}(t)} = \frac{e^{tan(t)}}{cos(t)}.

Thus the sought-for exponential Riordan array is given by

[g, f] = [e^{tan(t)} sec(t), tan(t)] .

This matrix begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 2 & 1 & 0 & 0 & 0 & ⋯ \\ 6 & 8 & 3 & 1 & 0 & 0 & ⋯ \\ 20 & 32 & 20 & 4 & 1 & 0 & ⋯ \\ 92 & 156 & 100 & 40 & 5 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

with production matrix that begins

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 1 & 0 & 0 & 0 & … \\ 2 & 4 & 1 & 1 & 0 & 0 & … \\ 0 & 6 & 9 & 1 & 1 & 0 & … \\ 0 & 0 & 12 & 16 & 1 & 1 & … \\ 0 & 0 & 0 & 20 & 25 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The first column is \underline{A009244}. We note that we have the following factorization

[g, f] = [e^{tan(t)} sec(t), tan(t)] = [sec(t), tan(t)] · B.

Thus we can say that the exponential Riordan array [sec(t), tan(t)], which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 0 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 0 & 1 & 0 & 0 & 0 & ⋯ \\ 0 & 5 & 0 & 1 & 0 & 0 & ⋯ \\ 5 & 0 & 14 & 0 & 1 & 0 & ⋯ \\ 0 & 61 & 0 & 30 & 0 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

has A sequence defined by 1 + t^{2} and Z sequence defined by t. Thus its production matrix is given by

( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 0 & 1 & 0 & 0 & 0 & … \\ 0 & 4 & 0 & 1 & 0 & 0 & … \\ 0 & 0 & 9 & 0 & 1 & 0 & … \\ 0 & 0 & 0 & 16 & 0 & 1 & … \\ 0 & 0 & 0 & 0 & 25 & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We can infer from this that the inverse array

[sec(t), tan(t)]^{−1} = [\frac{1}{\sqrt{1 + t^{2}}} , tan^{−1}(t)]

is the coefficient array of the family of orthogonal polynomials

P_{n}(t) = tP_{n−1}(t) − (n − 1)^{2}P_{n−2}(t),

with P_{0}(t) = 1 and P_{1}(t) = t.

Example 10. In this example, we let

A(t) = 1 + t, Z(t) = \frac{1}{1 − t}.

As before, we get f(t) = e^{t} − 1. Now

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = \frac{1}{1 − t^{2}} ,

and hence

ln(g(f̄(t))) = \frac{1}{2} ln (\frac{1 + t}{1 − t}) .

We infer that

g(t) = \sqrt{\frac{e^{t}}{2 − e^{t}}} .

The function g(t) generates the sequence \underline{A014307} which begins

1, 1, 2, 7, 35, 226, 1787, 16717, 180560, 2211181, 30273047, ….

It has many combinatorial interpretations [7, 15, 17].

The exponential Riordan array

[g, f] = [\sqrt{\frac{e^{t}}{2 − e^{t}}} , e^{t} − 1]

begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 7 & 10 & 6 & 1 & 0 & 0 & ⋯ \\ 35 & 45 & 31 & 10 & 1 & 0 & ⋯ \\ 226 & 271 & 180 & 75 & 15 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

with production matrix that begins

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 2 & 1 & 0 & 0 & 0 & … \\ 2 & 2 & 3 & 1 & 0 & 0 & … \\ 6 & 6 & 3 & 4 & 1 & 0 & … \\ 24 & 24 & 12 & 4 & 5 & 1 & … \\ 120 & 120 & 60 & 20 & 5 & 6 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

In general, the exponential Riordan array with

A(t) = 1 + t, Z(t) = \frac{r}{1 − t},

is given by

[g, f] = [(\frac{e^{t}}{2 − e^{t}} )^{r/2} , e^{t} − 1] .

Example 11. For this example, we take

A(t) = e^{−t}, Z(t) = e^{t}.

Then

\frac{d}{dt}f̄(t) = \frac{1}{A(t)} = \frac{1}{e^{−t}} = e^{t},

and so we get

f̄(t) = e^{t} + C = e^{t} − 1

since f̄(0) = 0. Thus

f(t) = ln(1 + t).

Now

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = \frac{e^{t}}{e^{−t}} = e^{2t},

and so

ln(g(f̄(t))) = \frac{e^{2t}}{2} − \frac{1}{2} ⇒ g(f̄(t)) = e^{\frac{1}{2} (e^{2t}−1)}.

Substituting f(t) for t we get

g(t) = e^{\frac{1}{2} (e^{2 ln(1+t)}−1)} = e^{t+ \frac{t^{2}}{2}} .

Thus

[g, f] = [e^{t+ \frac{t^{2}}{2}} , ln(1 + t)] .

We note that if we have

A(t) = Z(t) = e^{−t},

then we obtain

[g, f] = [1 + t, ln(1 + t)].

Interestingly, this last exponential Riordan array has a production matrix that is equal the ordinary Riordan array

(\frac{1 + 2t}{1 + t}, \frac{t}{1 + t})

with its first row removed.

5 Orthogonal polynomials

When Z(t) = α + βt and A(t) = 1 + γt + δt^{2}, the production matrix of the corresponding exponential Riordan array [g, f] is tri-diagonal, beginning as follows.

( \begin{matrix}α & 1 & 0 & 0 & 0 & 0 & … \\ β & α + γ & 1 & 0 & 0 & 0 & … \\ 0 & 2(β + δ) & α + 2γ & 1 & 0 & 0 & … \\ 0 & 0 & 3(β + 2δ) & α + 3γ & 1 & 0 & … \\ 0 & 0 & 0 & 4(β + 3δ) & α + 4γ & 1 & … \\ 0 & 0 & 0 & 0 & 5(β + 4δ) & α + 5γ & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

As a consequence, [g, f]^{−1} is the coefficient array of the family of orthogonal polynomials P_{n}(t) defined by the three-term recurrence [8, 12, 21]

P_{n}(t) = (t − (α + (n − 1)γ))P_{n−1}(t) − (n − 1)(β + (n − 2)δ)P_{n−2}(t),

with P_{0}(t) = 1 and P_{1}(t) = x − α. These are precisely the Sheffer orthogonal polynomials [1, 13].

Example 12. We take the case of

A(t) = 1 + t + t^{2}, Z(t) = 1 + t.

We have

\frac{d}{dt}f̄(t) = \frac{1}{1 + t + t^{2}} .

Choosing the constant of integration so that f̄(0) = 0, we get

f̄(t) = \frac{2}{\sqrt{3}} tan^{−1} (\frac{2t + 1}{\sqrt{3}}) − \frac{π}{3\sqrt{3}}.

Thus

f(t) = \frac{\sqrt{3}}{2} tan (\frac{\sqrt{3}t}{2} + \frac{π}{6} ) − \frac{1}{2}

= \frac{2 sin (\frac{\sqrt{3}t}{2})}{\sqrt{3} cos (\frac{\sqrt{3}t}{2}) − sin (\frac{\sqrt{3}t}{2})}

= \frac{2 tan (\frac{\sqrt{3}t}{2})}{\sqrt{3} − tan (\frac{\sqrt{3}t}{2})}.

We now have

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = \frac{1 + t}{1 + t + t^{2}} ,

and hence

ln(g(f̄(t))) = \frac{1}{\sqrt{3}} tan^{−1} (\frac{2t + 1}{\sqrt{3}}) + \frac{1}{2} ln(1 + t + t^{2}) − \frac{π}{6\sqrt{3}}.

From this we infer that

g(t) = \frac{\sqrt{3}e^{\frac{x}{2}}}{\sqrt{3} cos (\frac{\sqrt{3}t}{2}) − sin (\frac{\sqrt{3}t}{2})}.

The function g(t) generates the sequence \underline{A049774}, which counts the number of permutations of n elements not containing the consecutive pattern 123.

The sought-for matrix is thus

[g, f] = [\frac{\sqrt{3}e^{\frac{x}{2}}}{\sqrt{3} cos (\frac{\sqrt{3}t}{2}) − sin (\frac{\sqrt{3}t}{2})}, \frac{2 sin (\frac{\sqrt{3}t}{2})}{\sqrt{3} cos (\frac{\sqrt{3}t}{2}) − sin (\frac{\sqrt{3}t}{2})}] .

This exponential Riordan array is \underline{A182822}, which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 5 & 12 & 6 & 1 & 0 & 0 & ⋯ \\ 17 & 53 & 39 & 10 & 1 & 0 & ⋯ \\ 70 & 279 & 260 & 95 & 15 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

with production matrix that begins

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 2 & 1 & 0 & 0 & 0 & … \\ 0 & 4 & 3 & 1 & 0 & 0 & … \\ 0 & 0 & 9 & 4 & 1 & 0 & … \\ 0 & 0 & 0 & 16 & 5 & 1 & … \\ 0 & 0 & 0 & 0 & 25 & 6 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Example 13. We change the previous example slightly by taking

A(t) = 1 + 2t + t^{2} = (1 + t)^{2}, Z(t) = 1 + t.

Then we have

\frac{d}{dt}f̄(t) = \frac{1}{(1 + t)^{2}} ⇒ f̄(t) = −\frac{1}{1 + t} + 1 = \frac{t}{1 + t}.

This means that

f(t) = \frac{t}{1 − t}.

Now we have

\frac{d}{dt} ln(g(f̄(t))) = \frac{Z(t)}{A(t)} = \frac{1}{1 + t},

and hence

ln(g(f̄(t))) = ln(1 + t) ⇒ g(f̄(t)) = 1 + t.

This implies that

g(t) = 1 + f(t) = 1 + \frac{t}{1 − t} = \frac{1}{1 − t}.

Thus

[g, f] = [\frac{1}{1 − t}, \frac{t}{1 − t}] .

Thus [g, f]^{−1} is the coefficient array of the Laguerre polynomials [5].

We finish by noting that the simple addition of t to A(t) has allowed us to go from the relatively complicated exponential Riordan array

[\frac{\sqrt{3}e^{\frac{x}{2}}}{\sqrt{3} cos (\frac{\sqrt{3}t}{2}) − sin (\frac{\sqrt{3}t}{2})}, \frac{2 sin (\frac{\sqrt{3}t}{2})}{\sqrt{3} cos (\frac{\sqrt{3}t}{2}) − sin (\frac{\sqrt{3}t}{2})}]

to the simple exponential Riordan array

[\frac{1}{1 − t}, \frac{t}{1 − t}] .

6 Appendix: exponential Riordan arrays

The exponential Riordan group [6, 9, 11], is a set of infinite lower-triangular integer matrices, where each matrix is defined by a pair of generating functions g(t) = g_{0} + g_{1}t + g_{2}t^{2} + ⋯ and f(t) = f_{1}t + f_{2}t^{2} + ⋯ where g_{0} ≠ 0 and f_{1} ≠ 0. We usually assume that

g_{0} = f_{1} = 1.

The associated matrix is the matrix whose i-th column has exponential generating function g(t)f(t)^{i}/i! (the first column being indexed by 0). The matrix corresponding to the pair f, g is denoted by [g, f]. The group law is given by

[g, f] · [h, l] = [g(h ∘ f), l ∘ f].

The identity for this law is I = [1, t] and the inverse of [g, f] is [g, f]^{−1} = [1/(g ∘ f̄), f̄] where f̄ is the compositional inverse of f.

If M is the matrix [g, f], and u = (u_{n})_{n≥0} is an integer sequence with exponential generating function U (t), then the sequence Mu has exponential generating function g(t)U(f(t)). Thus the row sums of the array [g, f] have exponential generating function given by g(t)e^{f(t)} since the sequence 1, 1, 1, … has exponential generating function e^{t}.

As an element of the group of exponential Riordan arrays, the binomial matrix B with (n, k)-th element \binom{n}{k} is given by B = [e^{t}, t]. By the above, the exponential generating function of its row sums is given by e^{t}e^{t} = e^{2t}, as expected (e^{2t} is the e.g.f. of 2^{n}).

To each exponential Riordan array L = [g, f] is associated [10, 11] a matrix P called its production matrix, which has bivariate g.f. given by

e^{zt}(Z(t) + A(t)z)

where

A(t) = f′(f̄(t)), Z(t) = \frac{g′(f̄(t))}{g(f̄(t))}.

We have

P = L^{−1}L̄

where L̄ [16, 22] is the matrix L with its top row removed.

The ordinary Riordan group is described in [18].

References

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[6] P. Barry, On a family of generalized Pascal triangles defined by exponential Riordan array, J. Integer Seq., 10 (2007), Article 07.3.5.

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[8] T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach, 1978.

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http://www.combinatorics.net/ppt2004/Louis%20W.%20Shapiro/shapiro.htm.

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http://arxiv.org/abs/math/0702638v1, February 22 2007.

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[14] A. Hennessy and P. Barry, Generalized Stirling numbers, exponential Riordan arrays, and orthogonal polynomials, J. Integer Seq., 14 (2011), Article 11.8.2.

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[16] P. Peart and W.-J. Woan, Generating functions via Hankel and Stieltjes matrices, J. Integer Seq., 3 (2000), Article 00.2.1.

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http://arxiv.org/abs/1301.6327.

[18] L. W. Shapiro, S. Getu, W.-J. Woan, and L. C. Woodson, The Riordan group, Discr. Appl. Math., 34 (1991), 229–239.

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cally at http://oeis.org, 2013.

[20] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, Notices Amer. Math. Soc., 50 (2003), 912–915.

[21] G. Szegö, Orthogonal Polynomials, 4th ed., Amer. Math. Soc., 1975.

[22] H. S. Wall, Analytic Theory of Continued Fractions, AMS Chelsea Publishing, 2000.

2010 Mathematics Subject Classification: Primary 11C20; Secondary 11B83, 15B36, 33C45. Keywords: exponential Riordan array, A sequence, Z sequence, production matrix, orthogonal polynomial.

(Concerned with sequences \underline{A004211}, \underline{A007318}, \underline{A007405}, \underline{A009244}, \underline{A014307}, \underline{A048993}, \underline{A049020}, \underline{A049774}, \underline{A060311}, \underline{A100862}, and \underline{A182822}.)

Received October 8 2013; revised version received December 30 2013. Published in Journal of Integer Sequences, January 6 2014.

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