The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences

1 2 Journal of Integer Sequences, Vol. 13 (2010), 3 Article 10.8.2 47 6 23 11

Paul Barry School of Science Waterford Institute of Technology Ireland [email protected]

Aoife Hennessy Department of Computing, Mathematics and Physics Waterford Institute of Technology Ireland [email protected]

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

1 On-Line Encyclopedia of Integer Sequences, [13, 14]. Sequences are frequently referred to by their OEIS number. For instance, the binomial matrix B is A007318. The Euler-Seidel matrix of a sequence (an)n 0, which we will denote by E = Ea, is defined ≥ to be the rectangular array (an,k)n,k 0 determined by the recurrence a0,k = ak (k 0) and ≥ ≥ an,k = an,k 1 + an+1,k 1 (n 0, k 1). (1) − − ≥ ≥ The sequence (a0,k), the first row of the matrix, is usually called the initial sequence, while the sequence (an,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 n n recall that the binomial transform of a sequence an has general term bn = k=0 k ak. Thus the first row and column of the matrix are determined from Eq. (1) as follows: P n n a = a , (2) n,0 k 0,k k=0 Xn n n k a ,n = ( 1) − ak, . (3) 0 k − 0 k X=0 In general, we have n n n n a = a = a . (4) n,k i i+k,0 i i+k i=0 i=0 X X 1 2n Example 1. We take a0,n = an = Cn = n+1 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: 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 . . .     ......   ......      We now remark that the Catalan numbers Cn A000108 have the following moment representation: 1 4 x(4 x) C = xn − dx. (5) n 2π x Z0 p For the rest of this note, we shall assume that all sequences discussed have a moment representation of the form n an = x dµa R Z for a suitable measure dµa. 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.

2 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 1+( 1)n 1 2 C n xn√ x2 dx. 2 − = 4 2 2π 2 − Z− 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! A000142

∞ n x n!= x e− dx. Z0 Example 4. The aerated double factorials. We recall that the double factorials A001147 are given by

n (2n)! (2n 1)!! = (2k 1) = . − − n! 2n k Y=1 The aerated double factorials, which begin

1, 0, 3, 0, 5, 0, 15, 0, 105, 0, 945,... have integral representation 2 1 ∞ n x x e− 2 dx. √2π Z−∞ x2 The aerated double factorial numbers have exponential generating function e 2 . (Note that the numbers 2n n! A000165 are also called double factorials). · 2 The Euler-Seidel and Hankel matrix for moment sequences

We recall that for a sequence (an)n 0, its Hankel matrix is the matrix H = Ha with general ≥ term an+k. Note that if an has o.g.f. A(x) then the bivariate generating function of Ha is given by xA(x) yA(y) − . x y − If an has an exponential generating function G(x), then the n-th row (and n-th column) of Ha has exponential generating function given by dn G(x). dxn

3 x2 Example 5. We have seen that the aerated double factorial numbers have e.g.f. e 2 . Thus the n-th row of the Hankel matrix associated to them has e.g.f. n 2 n x2 x2 ⌊ ⌋ d n n 2k e 2 = e 2 (2k 1)!!x − . dxn 2 − k X=0 (We note that written as

n 2 n ⌊ ⌋ n x2 d x2 (2k 1)!! = e− x e x , 2 − dxn k X=0 this is a statement about scaled Hermite polynomials.) Note that if n an = x dµa Z then n+k n k an+k = x dµa = x x dµa.

Z Z n The binomial matrix is the matrix B with general term k . The binomial transform of a sequence a is the sequence with general term n n n b = a . n k k k X=0 In this case, the sequence bn has o.g.f. given by 1 x A . 1 x 1 x − − The sequence (bn)n 0 can be viewed as ≥ t B (an) . · Note that we have n n b = a n k k k X=0 n n = xk dµ k a k X=0 Z n n = xk dµ k a k Z X=0 n = (1 + x) dµa. Z In similar fashion, we have n an = (x 1) dµb. − Z

4 Proposition 6. We have Ea = BHa. (6) Proof. We have n BHa = (an k) . k · + The result follows from Eq. (4). We now let n n b = a , n k k k X=0 1 the binomial transform of an. In the sequel, we shall be interested in the product B− Hb.

n n Example 7. Taking bn = k=0 k Ck A0007317, the binomial transform of the Catalan numbers Cn, we obtain P 1 2 5 15 51 188 . . . 2 5 15 51 188 731 . . .   5 15 51 188 731 2950 . . .  15 51 188 731 2950 12235 . . .  Hb =    51 188 731 2950 12235 51822 . . .     188 731 2950 12235 51822 223191 . . .     ......   ......      1 Multiplying by B− , we obtain

1 2 5 15 51 188 . . . 1 3 10 36 137 543 . . .   2 7 26 101 406 1676 . . . 1  5 19 75 305 1270 5390 . . .  B− Hb =   ,  14 56 230 965 4120 17846 . . .     42 174 735 3155 13726 60398 . . .     ......   ......      which is the transpose of the Euler-Seidel matrix for Cn. This result is general. In order to prove this, we will use the follow lemma. Lemma 8.

k n j+k n k k i+n n j n j + k i x (1 + x) = x = ( 1) − x . (7) i − j i i=0 j=0 i=0 X X X

5 k k k i Proof. Since (1 + x) = i=0 i x by the binomial theorem, we immediately have

P k k k k xn(1 + x)k = xn xi = xi+n. i i i=0 i=0 X X But also, we have

n j+k n n j n j + k i n j n j+k ( 1) − x = ( 1) − (1 + x) − j i − j j=0 i=0 j=0 X X X n k n j n j = (1+ x) ( 1) − (1 + x) − j j=0 X = (1+ x)kxn.

Proposition 9. The Euler-Seidel matrix of the sequence (an)n 0 is equal to the transpose of 1 ≥ the matrix given by B− Hb, where Hb is the Hankel matrix of the binomial transform n n b = a n k k k X=0 of the initial sequence an. That is,

t 1 Ea = B− Hb. (8) Proof. The general element of

1 n k n B− H = ( 1) − (bn k) − k · + is given by n n j n ( 1) − bj k. − j + j=0 X Now

n n j+k n j n n j n j + k ( 1) − bj k = ( 1) − ai. − j + − j i j=0 j=0 i=0 X X X We thus wish to show that

n j+k k n j n j + k k ( 1) − ai = ai n. − j i i + j=0 i=0 i=0 X X X To this end, we assume that n an = x dµa, Z 6 and observe that

k k k k a = xi+ndµ i i+n i a i=0 i=0 X X Z k k = xi+n dµ i a Z i=0 Xn j+k n j n j + k i = ( 1) − x dµa − j i j=0 i=0 Z X X n j+k n j n j + k = ( 1) − ai. − j i j=0 i=0 X X The result now follows from Eq. (4). Corollary 10. k n Ea = x (1 + x) dµa . (9) Z We are interested in characterising the main diagonal of the Euler-Seidel matrix of an, 1 which by the above is the same as the main diagonal of B− Hb, where H is the Hankel matrix of bn, the binomial transform of an. Note that the diagonal is given by

n n a = a . n,n i n+i i=0 X Example 11. We have seen that the diagonal of the Euler-Seidel matrix for the Catalan numbers Cn begins 1, 3, 26, 305, 4120, 60398, 934064,... By the above, the general term of this sequence is

n n d = C . n i n+i i=0 X Now consider the moment representation of the Catalan numbers given by

1 4 x(4 x) C = xndµ = xn − dx. n 2π x Z Z0 p We claim that

1 4 x(4 x) d = (x(1 + x))ndµ = (x(1 + x))n − dx. n 2π x Z Z0 p

7 This follows from the result above, or directly, since

(x(1 + x))ndµ = = (x + x2)ndµ

Z Zn n 2i n i = x x − dµ i i=0 Z X n n = xn+idµ i i=0 X Z n n = C . i n+i i=0 X Note that by the change of variable y = x(1 + x) we obtain in this case the alternative moment representation for dn given by

1 20 √2(1 + √1 + 4y) 5√1 + 4y 2y 5 d = yn − − dy. n 2π 4y√1 + 4y Z0 p The above method of proof is easily generalised. Thus we have

Proposition 12. Let an be a sequence which can be represented as the sequence of moments of a measure: n an = x dµa. Z Then the elements dn of the main diagonal of the Euler-Seidel matrix have moment representation given by n dn = (x(1 + x)) dµa. Z 3 Examples: the Fibonacci and Jacobsthal cases

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

n n F (2n +1) = F (k + 1). k k X=0 1+√5 ¯ 1 1 √5 Letting φ = 2 and φ = φ = −2 , we have

1 n+1 n+1 Fn+1 = (φ φ¯ ) √5 − 1 = (φφn φ¯φ¯n) √5 − 1 n ¯ = x (φδφ φδφ¯) dx. √5 R − Z 8 Then the general element of the Euler-Seidel matrix for Fn+1 is given by k n ¯ ¯k ¯ n 1 k n ¯ φ φ (1 + φ) φ φ (1 + φ) x (1 + x) (φδφ φδφ¯) dx = · − · √5 R − √5 Z φk+1(1 + φ)n φ¯k+1(1 + φ¯)n = − , √5 where δa represents the Dirac delta “function” at a:

f(x)δadx =< δa,f >= f(a). R Z This gives us 112358 . . . 2 3 5 8 13 21 . . .   5 8 13 21 34 55 . . . E  13 21 34 55 89 144 . . .  Fn+1 =   .  34 55 89 144 233 377 . . .     89 144 233 377 610 987 . . .     ......   ......    The diagonal of this matrix begins  1, 3, 13, 55, 233,...

This is F3n+1 A033887, which by the above can be written φn+1(1 + φ)n φ¯n+1(1 + φ¯)n F3n+1 = − . √5

Example 14. Our next example is based on the Jacobsthal numbers Jn A001045, where 2n ( 1)n Jn = − . 3 − 3 We have n n 2 2 ( 1) 1 n Jn+1 = · + − = x (2δ2 + δ 1) dx. 3 3 3 R − Z Then the Euler-Seidel matrix for Jn+1 has general element given by k n k n k n 2 2 (1+2) +( 1) (1+( 1)) x (1 + x) (2δ2 + δ 1) dx = · − − − 3 Z 2k+13n +( 1)k 0n = − . 3 We have 1 1 3 5 11 21 . . . 2 4 8 16 32 64 . . .   6 12 24 48 96 192 . . . E  18 36 72 144 288 576 . . .  Jn+1 =   .  54 108 216 432 864 1728 . . .     162 324 648 1296 2592 5184 . . .     ......   ......      9 The diagonal sums, which begin

1, 4, 24, 144, 864,... are A067411, with general term

2n+1 3n +( 1)n 0n 2 6n + 0n − = · . 3 3 4 Hankel matrices, Riordan arrays and orthogonal polynomials

From the last section, we have Ea = BHa and t 1 Ea = B− Hb where bn is the binomial transform of an. The second equation shows us that

1 t Ea = (B− Hb) t 1 t = Hb(B− ) t 1 = Hb(B )− , since Hb is symmetric. Thus we obtain

t 1 BHa = Hb(B )− , which implies that t Hb = BHaB . (10)

Since det(B) = 1, we deduce the well-known result that the Hankel transform of bn is equal to that of an. We can also use this result to relate the LDU decomposition of Hb [11, 17] to that of Ha. Thus we have

t Hb = BHaB t t = B LaDaLa B · · t = (BLa)Da(BLa) .

One consequence of this is that the coeﬃcient triangle of the polynomials orthogonal with respect to dµb is given by 1 1 La− B− , 1 where La− is the coeﬃcient array of the polynomials orthogonal with respect to dµa.

10 Example 15. We take the example of the Catalan numbers an = Cn and their binomial n transform bn = k=0 Ck. It is well known that the Hankel transform of Cn is the all 1’s sequence, which implies that Da is the identity matrix. Thus in this case, P t Ha = LaLa where 2 La = LCn =(c(x), xc(x) ) (A039599) with 1 1 x L− = , (A129818) a 1+ x (1 + x)2 n k n+k with general term ( 1) − 2k , where we have used the notation of Riordan arrays. The polynomials − n n k n + k k Pn(x)= ( 1) − x − 2k k X=0 are thus a family of polynomials orthogonal on [0, 4] with respect to the density function √x(4 x) 1 − 2π x [18]. It is known that the bi-variate generating function of the inverse of the n-th principal minor of Ha is given by the Christoﬀel-Darboux quotient

Pn (x)Pn(y) Pn (y)Pn(x) +1 − +1 . x y − We deduce that the orthogonal polynomials deﬁned by bn have coeﬃcient matrix

1 1 1 Lb− = L−a B− 1 x 1 x = , , 1+ x (1 + x)2 · 1+ x 1+ x 1+ x x = , . 1 + 3x + x2 1 + 3x + x2

It turns out that these polynomials Qn(x) are given simply by

Qn(x)= Pn(x 1). − 1 Thus Hb− has n-th principal minor generated by

Qn (x)Qn(y) Qn (y)Qn(x) +1 − +1 . x y −

In similar manner, we can deduce that the Euler-Seidel matrix Ea = ECn is such that the 1 n-th principal minor of Ea− is generated by

Pn (x)Pn(y 1) Pn (y 1)Pn(x) Pn (x)Qn(y) Qn (y)Pn(x) +1 − − +1 − = +1 − +1 . x y x y − −

11 Example 16. For the aerated double factorials, we have

t Ha = LaDaLa where x2 La = [e 2 , x], Da = diag(n!). The associated orthogonal polynomials (which are scaled Hermite polynomials) have coeﬃ- cient matrix x2 1 2 La− = [e− , x], and we have n ⌊ 2 ⌋ n k n 2k Pn(x)= (2k 1)!!( 1) x − , 2k − − k X=0 or equivalently,

n n−k n k n + k 1+( 1) − k Pn(x)= Bessel∗ , k ( 1) 2 − x , 2 − 2 k X=0 where (2n k)! n + k Bessel∗(n, k)= − = (2k 1)!!, k!(n k)!2n k 2k − − − (see [1]). With Qn(x)= Pn(x 1) − 1 we again have that the Euler-Seidel matrix Ea is such the n-th principal minor of Ea− is generated by

Pn (x)Pn(y 1) Pn (y 1)Pn(x) Pn (x)Qn(y) Qn (y)Pn(x) +1 − − +1 − = +1 − +1 . x y x y − − 5 Appendix: The Stieltjes transform of a measure

The Stieltjes transform of a measure µ on R is a function Gµ deﬁned on C R by \ 1 Gµ(z)= µ(t). R z t Z − If f is a bounded continuous function on R, we have

f(x)µ(x)= lim f(x) Gµ(x + iy)dx. R − y 0+ R ℑ Z → Z If µ has compact support, then Gµ is holomorphic at inﬁnity and for large z,

∞ a G (z)= n , µ zn+1 n=0 X 12 n where an = R t µ(t) are the moments of the measure. If µ(t)= dψ(t)= ψ′(t)dt then

R 1 t ψ(t) ψ(t0)= lim Gµ(x + iy)dx. − −π y 0+ ℑ → Zt0 n If now g(x) is the generating function of a sequence an, with g(x)= n∞=0 anx , then we can deﬁne 1 1 ∞ a P G(z)= g = n . z z zn+1 n=0 X By this means, under the right circumstances we can retrieve the density function for the measure that deﬁnes the elements an as moments.

1 √1 4z Example 17. We let g(z)= − 2z− be the g.f. of the Catalan numbers. Then

1 1 1 x 4 G(z)= g = 1 − . z z 2 − x r ! Then √2 x2 + y2 x2 8x + y2 + 16 x2 + 4x y2 Gµ(x + iy)= − − − , ℑ − qp p 4 x2 + y2 and so we obtain p

1 √2 x2 + y2 x2 8x + y2 + 16 x2 + 4x y2 ψ′(x) = lim − − − −π y 0+ − qp p 4 x2 + y2  →   1 x(4 x) p = − .  π x 2 p 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.

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

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

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