The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Aoife Hennessy Department of Computing, Mathematics and Physics Waterford Institute of Technology Ireland

                   aoife.hennessy@gmail.com

Abstract We study the Euler-Seidel matrix of certain integer sequences, using the binomial transform and Hankel matrices. For moment sequences, we give an integral representation of the Euler-Seidel matrix. Links are drawn to Riordan arrays, orthogonal polynomials, and Christoffel-Darboux expressions.

1 Introduction

The purpose of this note is to investigate the close relationship that exists between the Euler-Seidel matrix [3, 4, 5, 6, 10] of an integer sequence, and the Hankel matrix [9] of that sequence. We do so in the context of sequences that have integral moment representations, though many of the results are valid in a more general context. While partly expository in nature, the note assumes a certain familiarity with generating functions, both ordinary and exponential, orthogonal polynomials [2, 8, 16] and Riordan arrays [12, 15] (again, both ordinary, where we use the notation (g, f), and exponential, where we use the notation [g, f]). Many interesting examples of sequences and Riordan arrays can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences, [13, 14]. Sequences are frequently referred to by their OEIS number. For instance, the binomial matrix B is \underline{A007318}.

The Euler-Seidel matrix of a sequence (a_{n})_{n≥0}, which we will denote by E = E_{a}, is defined to be the rectangular array (a_{n,k})_{n,k≥0} determined by the recurrence a_{0,k} = a_{k} (k ≥ 0) and

a_{n,k} = a_{n,k−1} + a_{n+1,k−1} (n ≥ 0, k ≥ 1).  (1)

The sequence (a_{0,k}), the first row of the matrix, is usually called the initial sequence, while the sequence (a_{n,0}), first column of the matrix, is called the final sequence. They are related by the binomial transform (or Euler transform, after Euler, who first proved this [7]). We recall that the binomial transform of a sequence a_{n} has general term b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}. Thus the first row and column of the matrix are determined from Eq. (1) as follows:

a_{n,0} = ∑_{k=0}^{n} \binom{n}{k}a_{0,k},  (2)

a_{0,n} = ∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}a_{k,0}.  (3)

In general, we have

a_{n,k} = ∑_{i=0}^{n} \binom{n}{i}a_{i+k,0} = ∑_{i=0}^{n} \binom{n}{i}a_{i+k}.  (4)

Example 1. We take a_{0,n} = a_{n} = C_{n} = \frac{1}{n+1} \binom{2n}{n}, the Catalan numbers. Thus the initial sequence, or first row, is the Catalan numbers, while the final sequence, or first column, will be the binomial transform of the Catalan numbers. We obtain the following matrix:

( \begin{matrix}1 & 1 & 2 & 5 & 14 & 42 & … \\ 2 & 3 & 7 & 19 & 56 & 174 & … \\ 5 & 10 & 26 & 75 & 230 & 735 & … \\ 15 & 36 & 101 & 305 & 965 & 3155 & … \\ 51 & 137 & 406 & 1270 & 4120 & 13726 & … \\ 188 & 543 & 1676 & 5390 & 17846 & 60398 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We now remark that the Catalan numbers C_{n} \underline{A000108} have the following moment representation:

C_{n} = \frac{1}{2π} ∫_{0}^{4} x^{n} \frac{\sqrt{x(4 − x)}}{x} dx.  (5)

For the rest of this note, we shall assume that all sequences discussed have a moment representation of the form

a_{n} = ∫_{R} x^{n}dμ_{a}

for a suitable measure dμ_{a}. The reader is directed to the Appendix for the link between the generating function of a sequence and the corresponding measure, when it exits.

We give some examples of such sequences.

Example 2. The aerated Catalan numbers.

We have seen (see Eq. (5)) that the Catalan numbers are a moment sequence. Similarly, the aerated Catalan numbers

1, 0, 1, 0, 2, 0, 5, 0, 14, …

can by represented by

C\frac{n}{2} \frac{1 + (−1)^{n}}{2} = \frac{1}{2π} ∫_{−2}^{2} x^{n}\sqrt{4 − x^{2}} dx.

This is the famous semi-circle distribution, of importance in random matrix theory.

Example 3. The factorial numbers n!.

We have the well-known integral representation of n! \underline{A000142}

n! = ∫_{0}^{∞} x^{n}e^{−x} dx.

Example 4. The aerated double factorials.

We recall that the double factorials \underline{A001147} are given by

(2n − 1)!! = ∏_{k=1}^{n}(2k − 1) = \frac{(2n)!}{n! 2^{n}}.

The aerated double factorials, which begin

1, 0, 3, 0, 5, 0, 15, 0, 105, 0, 945, …

have integral representation

\frac{1}{\sqrt{2π}} ∫_{−∞}^{∞} x^{n} e^{− \frac{x^{2}}{2}} dx.

The aerated double factorial numbers have exponential generating function e^{\frac{x^{2}}{2}} . (Note that the numbers 2^{n} · n! \underline{A000165} are also called double factorials).

2 The Euler-Seidel and Hankel matrix for moment sequences

We recall that for a sequence (a_{n})_{n≥0}, its Hankel matrix is the matrix H = H_{a} with general term a_{n+k}. Note that if a_{n} has o.g.f. A(x) then the bivariate generating function of H_{a} is given by

\frac{xA(x) − yA(y)}{x − y}.

If a_{n} has an exponential generating function G(x), then the n-th row (and n-th column) of H_{a} has exponential generating function given by

\frac{d^{n}}{dx^{n}} G(x).

Example 5. We have seen that the aerated double factorial numbers have e.g.f. e^{\frac{x^{2}}{2}} . Thus the n-th row of the Hankel matrix associated to them has e.g.f.

\frac{d^{n}}{dx^{n}} e^{\frac{x^{2}}{2}} = e^{\frac{x^{2}}{2}} ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2}(2k − 1)!!x^{n−2k}.

(We note that written as

∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2}(2k − 1)!! = e^{− \frac{x^{2}}{x}} \frac{d^{n}}{dx^{n}} e^{\frac{x^{2}}{x}} ,

this is a statement about scaled Hermite polynomials.)

Note that if

a_{n} = ∫ x^{n}dμ_{a}

then

a_{n+k} = ∫ x^{n+k}dμ_{a} = ∫ x^{n}x^{k}dμ_{a}.

The binomial matrix is the matrix B with general term \binom{n}{k}. The binomial transform of a sequence a_{n} is the sequence with general term

b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}.

In this case, the sequence b_{n} has o.g.f. given by

\frac{1}{1 − x}A (\frac{x}{1 − x}) .

The sequence (b_{n})_{n≥0} can be viewed as

B · (a_{n})^{t}.

Note that we have

b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}

= ∑_{k=0}^{n} \binom{n}{k} ∫ x^{k} dμ_{a}

= ∫ ∑_{k=0}^{n} \binom{n}{k}x^{k} dμ_{a}
= ∫ (1 + x)^{n} dμ_{a}.

In similar fashion, we have

a_{n} = ∫ (x − 1)^{n} dμ_{b}.

Proposition 6. We have

E_{a} = BH_{a}.  (6)

Proof. We have

BH_{a} = (\binom{n}{k}) · (a_{n+k}) .

The result follows from Eq. (4). We now let

b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k},

the binomial transform of a_{n}. In the sequel, we shall be interested in the product B^{−1}H_{b}. Example 7. Taking b_{n} = ∑_{k=0}^{n} \binom{n}{k}C_{k} \underline{A0007317}, the binomial transform of the Catalan numbers C_{n}, we obtain

H_{b} = ( \begin{matrix}1 & 2 & 5 & 15 & 51 & 188 & … \\ 2 & 5 & 15 & 51 & 188 & 731 & … \\ 5 & 15 & 51 & 188 & 731 & 2950 & … \\ 15 & 51 & 188 & 731 & 2950 & 12235 & … \\ 51 & 188 & 731 & 2950 & 12235 & 51822 & … \\ 188 & 731 & 2950 & 12235 & 51822 & 223191 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

Multiplying by B^{−1}, we obtain

B^{−1}H_{b} = ( \begin{matrix}1 & 2 & 5 & 15 & 51 & 188 & … \\ 1 & 3 & 10 & 36 & 137 & 543 & … \\ 2 & 7 & 26 & 101 & 406 & 1676 & … \\ 5 & 19 & 75 & 305 & 1270 & 5390 & … \\ 14 & 56 & 230 & 965 & 4120 & 17846 & … \\ 42 & 174 & 735 & 3155 & 13726 & 60398 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

which is the transpose of the Euler-Seidel matrix for C_{n}.

This result is general. In order to prove this, we will use the follow lemma. Lemma 8.

x^{n}(1 + x)^{k} = ∑_{i=0}^{k} \binom{k}{i}x^{i+n} = ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j} ∑_{i=0}^{j+k} \binom{j + k}{i}x^{i}.  (7)

Proof. Since (1 + x)^{k} = ∑_{i=0}^{k} \binom{k}{i}x^{i} by the binomial theorem, we immediately have

x^{n}(1 + x)^{k} = x^{n} ∑_{i=0}^{k} \binom{k}{i}x^{i} = ∑_{i=0}^{k} \binom{k}{i}x^{i+n}.

But also, we have

∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j} ∑_{i=0}^{j+k} \binom{j + k}{i}x^{i} = ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j}(1 + x)^{j+k}

= (1 + x)^{k} ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j}(1 + x)^{j}
= (1 + x)^{k}x^{n}.

Proposition 9. The Euler-Seidel matrix of the sequence (a_{n})_{n≥0} is equal to the transpose of the matrix given by B^{−1}H_{b}, where H_{b} is the Hankel matrix of the binomial transform

b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}

of the initial sequence a_{n}. That is,

E_{a}^{t} = B^{−1}H_{b}.  (8)

Proof. The general element of

B^{−1}H = ((−1)^{n−k}\binom{n}{k}) · (b_{n+k})

is given by

∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j}b_{j+k}.

Now

∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j}b_{j+k} = ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j} ∑_{i=0}^{j+k} \binom{j + k}{i}a_{i}.

We thus wish to show that

∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j} ∑_{i=0}^{j+k} \binom{j + k}{i}a_{i} = ∑_{i=0}^{k} \binom{k}{i}a_{i+n}.

To this end, we assume that

a_{n} = ∫ x^{n}dμ_{a},

and observe that

∑_{i=0}^{k} \binom{k}{i}a_{i+n} = ∑_{i=0}^{k} \binom{k}{i} ∫ x^{i+n}dμ_{a}

= ∫ ∑_{i=0}^{k} \binom{k}{i}x^{i+n} dμ_{a}

= ∫ ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j} ∑_{i=0}^{j+k} \binom{j + k}{i}x^{i} dμ_{a}

= ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j} ∑_{i=0}^{j+k} \binom{j + k}{i}a_{i}.

The result now follows from Eq. (4).

Corollary 10.

E_{a} = (∫ x^{k}(1 + x)^{n}dμ_{a}) .  (9)

We are interested in characterising the main diagonal of the Euler-Seidel matrix of a_{n}, which by the above is the same as the main diagonal of B^{−1}H_{b}, where H is the Hankel matrix of b_{n}, the binomial transform of a_{n}.

Note that the diagonal is given by

a_{n,n} = ∑_{i=0}^{n} \binom{n}{i}a_{n+i}.

Example 11. We have seen that the diagonal of the Euler-Seidel matrix for the Catalan numbers C_{n} begins

1, 3, 26, 305, 4120, 60398, 934064, …

By the above, the general term of this sequence is

d_{n} = ∑_{i=0}^{n} \binom{n}{i}C_{n+i}.

Now consider the moment representation of the Catalan numbers given by

C_{n} = ∫ x^{n}dμ = \frac{1}{2π} ∫_{0}^{4} x^{n} \frac{\sqrt{x(4 − x)}}{x}dx.

We claim that

d_{n} = ∫ (x(1 + x))^{n}dμ = \frac{1}{2π} ∫_{0}^{4}(x(1 + x))^{n} \frac{\sqrt{x(4 − x)}}{x}dx.

This follows from the result above, or directly, since

∫ (x(1 + x))^{n}dμ = = ∫ (x + x^{2})^{n}dμ
= ∫ ∑_{i=0}^{n} \binom{n}{i}x^{2i}x^{n−i}dμ

= ∑_{i=0}^{n} \binom{n}{i} ∫ x^{n+i}dμ

= ∑_{i=0}^{n} \binom{n}{i}C_{n+i}.

Note that by the change of variable y = x(1 + x) we obtain in this case the alternative moment representation for d_{n} given by

d_{n} = \frac{1}{2π} ∫_{0}^{20} y^{n} \frac{\sqrt{2}(1 + \sqrt{1 + 4y})\sqrt{5\sqrt{1 + 4y} − 2y − 5}}{4y\sqrt{1 + 4y}} dy.

The above method of proof is easily generalised. Thus we have

Proposition 12. Let a_{n} be a sequence which can be represented as the sequence of moments of a measure:

a_{n} = ∫ x^{n}dμ_{a}.

Then the elements d_{n} of the main diagonal of the Euler-Seidel matrix have moment representation given by

d_{n} = ∫ (x(1 + x))^{n}dμ_{a}.

3 Examples: the Fibonacci and Jacobsthal cases

Example 13. We first of all look at the Fibonacci numbers, F_{n+1} = F (n + 1) \underline{A000045}. It is well known that the binomial transform of F (n + 1) is F (2n + 1) \underline{A122367}:

F (2n + 1) = ∑_{k=0}^{n} \binom{n}{k}F (k + 1).

Letting ϕ = \frac{1+\sqrt{5}}{2} and ϕ̄ = \frac{1}{ϕ} = \frac{1−\sqrt{5}}{2}, we have

F_{n+1} = \frac{1}{\sqrt{5}}(ϕ^{n+1} − ϕ̄^{n+1})
= \frac{1}{\sqrt{5}}(ϕϕ^{n} − ϕ̄ϕ̄^{n})
= \frac{1}{\sqrt{5}} ∫_{R} x^{n}(ϕδ_{ϕ} − ϕ̄δϕ̄) dx.

Then the general element of the Euler-Seidel matrix for F_{n+1} is given by

\frac{1}{\sqrt{5}} ∫_{R} x^{k}(1 + x)^{n}(ϕδ_{ϕ} − ϕ̄δϕ̄) dx = \frac{ϕ · ϕ^{k}(1 + ϕ)^{n} − ϕ̄ · ϕ̄^{k}(1 + ϕ̄)^{n}}{\sqrt{5}}
= \frac{ϕ^{k+1}(1 + ϕ)^{n} − ϕ̄^{k+1}(1 + ϕ̄)^{n}}{\sqrt{5}} ,

where δ_{a} represents the Dirac delta “function” at a:

∫_{R} f(x)δ_{a}dx =< δ_{a}, f >= f(a).

This gives us

E_{F_{n+1}} = ( \begin{matrix}1 & 1 & 2 & 3 & 5 & 8 & … \\ 2 & 3 & 5 & 8 & 13 & 21 & … \\ 5 & 8 & 13 & 21 & 34 & 55 & … \\ 13 & 21 & 34 & 55 & 89 & 144 & … \\ 34 & 55 & 89 & 144 & 233 & 377 & … \\ 89 & 144 & 233 & 377 & 610 & 987 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The diagonal of this matrix begins

1, 3, 13, 55, 233, …

This is F_{3n+1} \underline{A033887}, which by the above can be written

F_{3n+1} = \frac{ϕ^{n+1}(1 + ϕ)^{n} − ϕ̄^{n+1}(1 + ϕ̄)^{n}}{\sqrt{5}} .

Example 14. Our next example is based on the Jacobsthal numbers J_{n} \underline{A001045}, where

J_{n} = \frac{2^{n}}{3} − \frac{(−1)^{n}}{3} .

We have

J_{n+1} = \frac{2 · 2^{n}}{3} + \frac{(−1)^{n}}{3} = \frac{1}{3} ∫_{R} x^{n} (2δ_{2} + δ_{−1}) dx.

Then the Euler-Seidel matrix for J_{n+1} has general element given by

∫ x^{k}(1 + x)^{n} (2δ_{2} + δ_{−1}) dx = \frac{2 · 2^{k}(1 + 2)^{n} + (−1)^{k}(1 + (−1))^{n}}{3}
= \frac{2^{k+1}3^{n} + (−1)^{k} 0^{n}}{3} .

We have

E_{J_{n+1}} = ( \begin{matrix}1 & 1 & 3 & 5 & 11 & 21 & … \\ 2 & 4 & 8 & 16 & 32 & 64 & … \\ 6 & 12 & 24 & 48 & 96 & 192 & … \\ 18 & 36 & 72 & 144 & 288 & 576 & … \\ 54 & 108 & 216 & 432 & 864 & 1728 & … \\ 162 & 324 & 648 & 1296 & 2592 & 5184 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The diagonal sums, which begin

1, 4, 24, 144, 864, …

are \underline{A067411}, with general term

\frac{2^{n+1} 3^{n} + (−1)^{n} 0^{n}}{3} = \frac{2 · 6^{n} + 0^{n}}{3} .

4 Hankel matrices, Riordan arrays and orthogonal polynomials

From the last section, we have

E_{a} = BH_{a}

and

E_{a}^{t} = B^{−1}H_{b}

where b_{n} is the binomial transform of a_{n}. The second equation shows us that

E_{a} = (B^{−1}H_{b})^{t}
= H_{b}^{t}(B^{−1})^{t}
= H_{b}(B^{t})^{−1},

since H_{b} is symmetric. Thus we obtain

BH_{a} = H_{b}(B^{t})^{−1},

which implies that

H_{b} = BH_{a}B^{t}.  (10)

Since det(B) = 1, we deduce the well-known result that the Hankel transform of b_{n} is equal to that of a_{n}. We can also use this result to relate the LDU decomposition of H_{b} [11, 17] to that of H_{a}. Thus we have

H_{b} = BH_{a}B^{t}
= B · L_{a}D_{a}L_{a}^{t} · B^{t}
= (BL_{a})D_{a}(BL_{a})^{t}.

One consequence of this is that the coefficient triangle of the polynomials orthogonal with respect to dμ_{b} is given by

L_{a}^{−1}B^{−1},

where L_{a}^{−1} is the coefficient array of the polynomials orthogonal with respect to dμ_{a}.

Example 15. We take the example of the Catalan numbers a_{n} = C_{n} and their binomial transform b_{n} = ∑_{k=0}^{n} C_{k}. It is well known that the Hankel transform of C_{n} is the all 1’s sequence, which implies that D_{a} is the identity matrix. Thus in this case,

H_{a} = L_{a}L_{a}^{t}

where

L_{a} = L_{C_{n}} = (c(x), xc(x)^{2}) (\underline{A039599})

with

L_{a}^{−1} = (\frac{1}{1 + x}, \frac{x}{(1 + x)^{2}} ) (\underline{A129818})

with general term (−1)^{n−k}\binom{n+k}{2k}, where we have used the notation of Riordan arrays. The polynomials

P_{n}(x) = ∑_{k=0}^{n}(−1)^{n−k}\binom{n + k}{2k}x^{k}

are thus a family of polynomials orthogonal on [0, 4] with respect to the density function

\frac{1}{2π} ^{\frac{\sqrt{x(4−x)}}{x}} [18]. It is known that the bi-variate generating function of the inverse of the n-th principal minor of H_{a} is given by the Christoffel-Darboux quotient

\frac{P_{n+1}(x)P_{n}(y) − P_{n+1}(y)P_{n}(x)}{x − y}.

We deduce that the orthogonal polynomials defined by b_{n} have coefficient matrix

L_{b}^{−1} = L_{a}^{−1}B^{−1}
= (\frac{1}{1 + x}, \frac{x}{(1 + x)^{2}} ) · (\frac{1}{1 + x}, \frac{x}{1 + x})
= (\frac{1 + x}{1 + 3x + x^{2}} , \frac{x}{1 + 3x + x^{2}} ) .

It turns out that these polynomials Q_{n}(x) are given simply by

Q_{n}(x) = P_{n}(x − 1).

Thus H_{b}^{−1} has n-th principal minor generated by

\frac{Q_{n+1}(x)Q_{n}(y) − Q_{n+1}(y)Q_{n}(x)}{x − y}.

In similar manner, we can deduce that the Euler-Seidel matrix E_{a} = E_{C_{n}} is such that the n-th principal minor of E_{a}^{−1} is generated by

\frac{P_{n+1}(x)P_{n}(y − 1) − P_{n+1}(y − 1)P_{n}(x)}{x − y} = \frac{P_{n+1}(x)Q_{n}(y) − Q_{n+1}(y)P_{n}(x)}{x − y}.

Example 16. For the aerated double factorials, we have

H_{a} = L_{a}D_{a}L_{a}^{t}

where

L_{a} = [e^{\frac{x^{2}}{2}} , x], D_{a} = diag(n!).

The associated orthogonal polynomials (which are scaled Hermite polynomials) have coefficient matrix

L_{a}^{−1} = [e^{− \frac{x^{2}}{2}} , x],

and we have

P_{n}(x) = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}(2k − 1)!!(−1)^{k}x^{n−2k},

or equivalently,

P_{n}(x) = ∑_{k=0}^{n} Bessel^{∗} (\frac{n + k}{2} , k) (−1)^{\frac{n−k}{2}} \frac{1 + (−1)^{n−k}}{2} x^{k},

where

Bessel^{∗}(n, k) = \frac{(2n − k)!}{k!(n − k)!2^{n−k}} = \binom{n + k}{2k}(2k − 1)!!,

(see [1]). With

Q_{n}(x) = P_{n}(x − 1)

we again have that the Euler-Seidel matrix E_{a} is such the n-th principal minor of E_{a}^{−1} is generated by

\frac{P_{n+1}(x)P_{n}(y − 1) − P_{n+1}(y − 1)P_{n}(x)}{x − y} = \frac{P_{n+1}(x)Q_{n}(y) − Q_{n+1}(y)P_{n}(x)}{x − y}.

5 Appendix: The Stieltjes transform of a measure

The Stieltjes transform of a measure μ on ℝ is a function G_{μ} defined on ℂ \ ℝ by

G_{μ}(z) = ∫_{ℝ} \frac{1}{z − t}μ(t).

If f is a bounded continuous function on ℝ, we have

∫_{ℝ} f(x)μ(x) = − lim_{y→0^{+}} ∫_{ℝ} f(x)ℑG_{μ}(x + iy)dx.

If μ has compact support, then G_{μ} is holomorphic at infinity and for large z,

G_{μ}(z) = ∑_{n=0}^{∞} \frac{a_{n}}{z^{n+1}} ,

where a_{n} = ∫_{ℝ} t^{n}μ(t) are the moments of the measure. If μ(t) = dψ(t) = ψ′(t)dt then

ψ(t) − ψ(t_{0}) = −\frac{1}{π} lim_{y→0^{+}} ∫_{t_{0}}^{t} ℑG_{μ}(x + iy)dx.

If now g(x) is the generating function of a sequence a_{n}, with g(x) = ∑_{n=0}^{∞} a_{n}x^{n}, then we can define

G(z) = \frac{1}{z}g (\frac{1}{z}) = ∑_{n=0}^{∞} \frac{a_{n}}{z^{n+1}} .

By this means, under the right circumstances we can retrieve the density function for the measure that defines the elements a_{n} as moments.

Example 17. We let g(z) = \frac{1−\sqrt{1−4z}}{2z} be the g.f. of the Catalan numbers. Then

G(z) = \frac{1}{z}g (\frac{1}{z}) = \frac{1}{2} (1 − \sqrt{\frac{x − 4}{x}}) .

Then

ℑG_{μ}(x + iy) = −\frac{\sqrt{2}\sqrt{\sqrt{x^{2} + y^{2}}\sqrt{x^{2} − 8x + y^{2} + 16} − x^{2} + 4x − y^{2}}}{4\sqrt{x^{2} + y^{2}}} ,

and so we obtain

ψ′(x) = −\frac{1}{π} lim_{y→0^{+}} \{−\frac{\sqrt{2}\sqrt{\sqrt{x^{2} + y^{2}}\sqrt{x^{2} − 8x + y^{2} + 16} − x^{2} + 4x − y^{2}}}{4\sqrt{x^{2} + y^{2}}} \}

= \frac{1}{2π} \frac{\sqrt{x(4 − x)}}{x}.

6 Acknowledgements

The author would like to thank the anonymous referees for their careful reading and cogent suggestions which have hopefully led to clearer paper.

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2010 Mathematics Subject Classification: Primary 11B83; Secondary 11B39, 11C20, 15B05, 15B36, 42C05. Keywords: Euler-Seidel matrix, Hankel matrix, integer sequence, moment sequence, orthogonal polynomial, Riordan array.

(Concerned with sequences \underline{A000045}, \underline{A000108}, \underline{A000142}, \underline{A000165}, \underline{A000731}, \underline{A001045}, \underline{A001147}, \underline{A007317}, \underline{A007318}, \underline{A033887}, \underline{A039599}, \underline{A067411}, \underline{A122367}, and \underline{A129818}.)

Received April 27 2010; revised version received July 21 2010. Published in Journal of Integer Sequences, August 2 2010.

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