A conjecture based on Somos-4 sequence and its extension

Linear Algebra and its Applications 436 (2012) 4285–4295

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A conjecture based on Somos-4 sequence and its extension Xiang-Ke Chang a,b,∗ , Xing-Biao Hu a a

LSEC, Institute of Computational Mathematics and Scientific Engineering Computing, AMSS, Chinese Academy of Sciences, P.O. Box 2719, Beijing 100190, PR China Graduate School of the Chinese Academy of Sciences, Beijing, PR China

b

ARTICLE INFO

ABSTRACT

Article history: Received 19 October 2011 Accepted 23 January 2012 Available online 21 February 2012

In two recent papers by Barry (2010) [29] and (2011) [30], it is conjectured that Somos-4 admits a solution expressed in terms of Hankel determinant with its elements satisfying a convolution recursion relation. In this paper, Barry’s conjecture on Somos-4 is firstly confirmed. Actually, we present a more generalized result. The proof is mainly based on new findings on properties for so-called Block–Hankel determinants. The method can also be used to prove another conjecture proposed by Michael Somos, which has been solved by Guoce Xin. © 2012 Elsevier Inc. All rights reserved.

Submitted by P. Šemrl AMS classification: 37J35 11B37 11B83 Keywords: Somos sequence Block–Hankel determinant Jacobi identity Conjecture

1. Introduction A Somos-k (k  4) sequence originally introduced by Michael Somos is a sequence of numbers defined by a recurrence as follows: Sn Sn−k

=

[ k/2] i=1

xi Sn−i Sn−k+i ,

(1)

where the xi are given integers. In recent years, such sequences have attracted a great deal of interest of researchers in number theory, algebraic combinatorics, statistical mechanics, as well as discrete integrable systems. ∗ Corresponding author at: LSEC, Institute of Computational Mathematics and Scientific Engineering Computing, AMSS, Chinese Academy of Sciences, P.O. Box 2719, Beijing 100190, PR China. E-mail addresses: [email protected] (X.-K. Chang), [email protected] (X.-B. Hu). 0024-3795/$ - see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2012.01.016

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For example, combinatorial interpretations for Somos-4 (A006720) and Somos-5 (A006721) have been found by Speyer [1] and for Somos-6 (A006722) and Somos-7 (A006723) by Carroll and Speyer [2]. For the case of k = 4, 5, 6, 7, if the coefficients are xi = 1 (1  i  [k/2]) and the initial values are given by Si = 1 (0  i  k − 1), then quadratic relation (1) can produce a sequence of integers [3–5], which is a property of what we call integrality. Such a property was first found by Michael Somos. It was realized that the deeper reason behind this lay in the fact that the recurrence (1) satisfies Laurent phenomenon, meaning that the iterates are polynomials in the four initial data, and their inverses, and these Laurent polynomials have integer coefficients. The Gale–Robinson conjecture[6,7] confirmed by Fomin and Zelevinsky [8] indicates that Somos-k (k = 4, 5, 6, 7) sequences exhibit the Laurent property, which is an essential feature of Fomin and Zelevinsky’s theory of cluster algebras [9,10]. For the case k = 4 and k = 5, Hone constructed explicit solutions to the bilinear recurrence (1) with complex numbers xi in terms of the Weierstrass sigma function in [11,12], respectively. He also indicated that Somos-4 can be thought of as an integrable symplectic map [13,14] by making a change of variables. But, unfortunately, it is not obvious from the form of the Weierstrass sigma function that sequences have the Laurent phenomenon. For more details, please consult [15–24]. Furthermore, there also exist several conjectures about Somos-4 or Somos-5. For instance, Michael Somos [25] observed that y(z ) given by y − y2 = z − z 3 yields the Somos-4 sequence, which has been proved by Xin [26]. Gosper and Schroeppel [27] made two near-addition formulas for Somos-4 and Somos-5, which has been solved by Ma [28]. More recently, Barry proposed another conjecture about Somos-4 in the paper [29,30], which is still open. We restate it as the following: Conjecture 1.1. Assume that an satisfies a convolution recursion relation: ⎧ ⎪ ⎪ if n = 0; 1, ⎪ ⎪ ⎨ an = α, if n = 1; ⎪ ⎪ ⎪  ⎪ n−2 ⎩ αa n−1 + β an−2 + γ i=0 ai an−2−i , if n > 1. Then the Hankel determinant Hn sequence.

(2)

= det (ai+j )0i,jn−1 is a (α 2 γ 2 , γ 2 (β + γ )2 − α 2 γ 3 ) Somos-4

Here we use the notation (x1 , x2 , . . . , x[ k ] ) Somos-k sequence, which represents that the sequence 2

satisfies recursion relation (1). The main purpose of this paper is to show the following result: (p)

Theorem 1.2. For a fixed positive integer p, let an be computed via the following convolution recurrence:

a(np)

=

⎧ ⎪ ⎪ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1, (p) ⎪ ⎪ α an−1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ α a(p) n−1

(p)

+ β an−2 + γ +γ

if n

< 0;

if n

= 0;

[ n−p 2 ] (p) (p) i=0 api an−2−pi , if n

[ n−p 2 ] (p) (p) i=1 api an−2−pi ,

if n

= kp + 1, k  0 = kp + j, k  0,2  j  p.

Then we have (1)

(i) hn is a (α 2 γ 2 , γ 2 (β (2)

(ii) hn

+ γ )2 − α 2 γ 3 ) Somos-4 sequence. is a (α γ − α γ 3 (β + γ ), α 6 γ 4 (β + γ )(β + 2γ ) − α 8 γ 5 ) Somos-5 sequence. 6

3

4

(3)

X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

(iii) If β

(p)

+ γ = 0, then hn satisfies the relation: (p)

(p)

hn+p+3 h(np) (p)

where hn

4287



(p)

= det aip+j

(p)

= (α p γ )p+1 hn+p+2 hn+1 −

(α p γ )2p+1 (p) (p) hn+p+1 hn+2 , αp

0i,jn−1

.

Obviously, conclusion (i) confirms the conjecture due to Barry. Besides, conclusion (ii) indicates that a Block–Hankel determinant solution to Somos-5 recurrences is obtained while conclusion (iii) yields a solution for a special case of the Gale–Robinson recurrence. This paper is organized as follows. In Section 2, the so-called Block–Hankel determinants are investigated. As a result, several new results on Block–Hankel determinants are achieved. We present the proof of Theorem 1.2 in Section 3. Finally conclusions and discussions are given in Section 4.

2. Block–Hankel determinants and their properties In order to complete the proof of Theorem 1.2, it is necessary to investigate the properties of deter(p)

minant hn . As we know, Hankel determinants [31] have widely appeared in the theory of orthogonal polynomials, Padé approximation, continued fractions and combinatorial mathematics. (See [32–36] for more details.) In order to interpret a graph, Shingu and Kamioka [37] introduce the notion of Block–Hankel determinant, which is an extension of Hankel determinant. Now we give the definition of Block–Hankel determinant. (p,q) (p,q) ) ≡ det (ar +ip+jq )0i,jn−1 . An n × n Block–Hankel determinant (p  1, q  1) is defined by Hn (ar When p = 1 and q = 1,



(1,1)

(1,1) (1,1) ar +1 · · · ar +n−1

ar



(1,1) (1,1) (1,1)

a ar +2 · · · ar +n

r +1

, Hn (a(r 1,1) ) ≡

(4)

.. . . .. ..



. . . .





(1,1)

(1,1) (1,1)

a

ar +n · · · a r +n−1

r +2n−1

which is a conventional Hankel determinant. The following are two other examples:



(2,1) (2,1) (2,1)

ar +1 · · · ar +n−1

ar





(2,1) (2,1)

a(2,1)

a · · · a r +2 r +3 r +n+1

(2,1)

Hn (ar ) ≡



.. .. .. ..



. . . .





(2,1)

(2,1) (2,1)

a

r +2(n−1) ar +2n−1 · · · ar +3(n−1) and

Hn (a(r 4,2) )

≡

(4,2)

ar



a(4,2) r +4



..

.



(4,2)

a

r +4(n−1)

(4,2)

ar +2

(4,2)

(4,2)

(4,2)

· · · ar +2n+2 .. .. . .

(4,2)

· · · ar +6(n−1)

ar +4n−2



· · · ar +2(n−1)

ar +6

.. .

(5)

(4,2)







.









(6)

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X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

(p)

(p,1)

Obviously, hn in Theorem 1.2 have the same structure as Hn (a0 study of Block–Hankel determinants with q (p)

as Hn (ar

).

(p,1)

= 1, namely, Hn (ar

). Thus, we concentrate on the ). For simplicity, we abbreviate this (p)

In the sequel, we investigate some properties of Block–Hankel determinants Hn (ar ) with the elements in (3). The proofs of the following lemmas can be completed by the same steps:

• Step 1: Perform column operations recursively. • Step 2: If necessary, decompose the determinant obtained by Step 1 to two parts. • Step 3: Perform row operations recursively in order to get a determinant in a simple expression. Lemma 2.1. For n (p)

Hn (a0

(p)

Jn (a0

(p)

) = (α p γ )n−1 Hn−1 (a−1 ) + (β + γ )(α p γ )n−2 Jn−1 (a0 ),

where

(p)

 1, we have

)≡

(p) (p)

(p)

Ap−1 a0 a1

(p) (p) (p)

A2p−1 ap ap+1

. .. ..

.

. . .

(p) ( p) ( p)

Anp−1 a (n−1)p a(n−1)p+1

(p)

···

an−2

··· .. .

an+p−2

(p)

.. .

(p)

· · · an+(n−1)p−2



















(7)

(8)

and (p)

Hn (a−1 )

(p)

= −(α p γ )n−2 Kn−1 (a−1 ),

where

(p) Kn (a−1 )

≡

(p) (p)

(p) a−1 a0

Ap−1

(p) (p) (p)

A2p−1 ap−1 ap

. .. ..

.

. . .

(p) (p)

Anp−1 a(p) (n−1)p−1 a(n−1)p

(9)

(p)

···

an−3

··· .. .

an+p−3

(p)

.. .

(p)

· · · an+(n−1)p−3











.







(10)

(p)

Here Akp−1 is defined by (p)

Akp−1

(p)

= akp−1 −

k −1 i=1

(p) (p) a(k+1−i)p , (p)

Aip−1

ap

(p)

Ap−1

(p)

= ap−1 .

(11)

Proof. Here we only give the proof of Eq. (7). By making the appropriate substitutions, Eq. (9) can be obtained in a similar manner without Step 2. Step 1: We rewrite



(p) (p) (p)



a0

a1 · · · an−1





(p) (p) (p)

ap ap+1 · · · an+p−1

(p)



(12) Hn (a0 ) =

.. .. .. ..



.



. . .



(p)

(p) (p)

a

(n−1)p a(n−1)p+1 · · · an+(n−1)p−1

X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

4289

by using the recursion relation (3) to obtain Eq. (7). We first subtract the (ip + 1)th column multiplied (p)

= 1, 2, . . . , [ n−p 3 ]. Next, if n = kp + 2, k = 1, 2, . . ., subtract the (n − 2)th column multiplied by (β + γ ) from the nth column. And then, subtracting the (n − 1)th column multiplied by α , we have

by γ an−3−ip from the nth column for i

 (p) H n a0

=



a(p)

0

a(pp)





a(p) 2p



..



.



(p)

a(n−1)p







(p) (p) (p)

an+p−2 γ ap an−3



2 

(p) (p) (p) an+2p−2 γ api an+2p−3−pi

. i=1

.. ..



. .

n −1



(p) (p) (p) an+(n−1)p−2 γ api an+(n−1)p−3−pi

(p)

··· ···

an−2

··· .. . ···

0

(13)

i=1

Applying a similar procedure to the (n − 1)th, . . ., 2nd columns, we obtain



(p)

a0 0

(p) (p) (p) (p)

(β + γ )ap−1 + γ ap a−1

ap

2

 (p) (p) (p)

a(p)  (β + γ )a2p−1 + γ γ api a2p−1−pi

2p (p) Hn a0 =

i =1

.. ..

. .



n −1

(p) (p) (p) (p)

a api a(n−1)p−1−pi

(n−1)p (β + γ )a(n−1)p−1 + γ

i =1





(p) (p)

γ a p a n− 3



2

 (p) (p) ··· γ api an+2p−3−pi

. i =1



.. ..

. .



n −1

(p) (p) ··· γ api an+(n−1)p−3−pi



··· ···

0

i =1

(14)

(p)

Step 2: Decompose Hn (a0

a(p)

0

a(pp)





a(p) ˜ n (a(0p) ) =

2p H

..



.



(p)

a(n−1)p

) into two parts, namely, 0

(p)

(β + γ )ap−1 (p)

··· ···

(β + γ )a2p−1

···

.. .

..

.

(p)

(β + γ )a(n−1)p−1 · · ·





(p) (p)

γ ap an−3



2 

(p) (p) γ api an+2p−3−pi



i=1

..



.

n −1



(p) (p) γ api an+(n−1)p−3−pi

0

i=1

and

(p)

ˆ n (a0 H

)=



a(p)

0

a(pp)





a(p) 2p



..



.



(p)

a(n−1)p

0

(p) (p)

γ ap a−1

γ

γ

2  i=1

n −1 i=1

(p) (p)

api a2p−1−pi

.. .

(p) (p)

api a(n−1)p−1−pi

··· ··· ··· ..

.

···





(p) (p)

γ ap an−3



2 

(p) (p) γ api an+2p−3−pi

. i=1

..



.

n −1



(p) (p) γ api an+(n−1)p−3−pi

0

i=1

(15)

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X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

(p)

(p) ) and Hˆ n (a0 ). for i = 2, . . . , k − 1. Then the

˜ n (a0 Step 3: Lastly, we perform the same row operations to both the determinants H (p)

(p)

For k = 3, 4, . . . , we subtract the ith multiplied by a(k−i+1)p /ap result in (7) obviously holds.  (p)

(p)

In addition, Hn (a−1 ) and Hn (a0 Lemma 2.2. For n (p)

Hn (a0

Ln (ap

 1, we have (p)

) = γ n−1 Hn−1 (ap−1 ) − βγ n−2 Ln−1 (a(pp) ),

where

(p)

) respectively have another expression.

)≡

(p) (p)

(p)

Bp−1 ap ap+1

(p) (p) (p)

B2p−1 a2p a2p+1

. .. ..

.

. . .

(p) ( p) (p)

Bnp−1 anp anp+1

(16)



· · · an+p−2

(p)



(p)

· · · an+2p−2



.. ..

.

.

(17)



(p)

· · · an+np−2

and (p)

Hn (a−1 )

(p)

= −γ n−2 Ln−1 (ap−1 ),

where

(p) Ln (ap−1 )

(p)

Here Bkp−1 (k (p)

Bkp−1

≡

(p) (p)

(p)

Bp−1 ap−1 ap

(p) (p) (p)

B2p−1 a2p−1 a2p

. .. ..

.

. . .

(p) p) (p)

Bnp−1 a(np −1 anp

(18)



· · · an+p−3

(p)



(p)

· · · an+2p−3

. .. ..

.

.

(19)



(p)

· · · an+np−3

 1) is defined by (p)

= akp−1 −

k −1 i=1

(p) (p) a(k−i)p , (p)

Bip−1

a0

(p)

Bp−1

(p)

= ap−1 .

(20)

Proof. Step 1: We start by performing column operations. Firstly, subtract the (ip

+ 1)th column = = kp + 2, k = 1, 2, . . ., subtract the (n − 2)th column multiplied by β from the nth column. And then, subtracting the (n − 1)th (p) column multiplied by α and adding the (n − 2)th column multiplied by γ a0 , we obtain (p) multiplied by an−3−ip from the nth column for i

(p)

Hn (a0

)=



a(p)

0

a(pp)





a(p) 2p



..



.



(p)

a(n−1)p

(p)

0

3 0, 1, . . . , [ n− ]. Next, if n p

(p) (p)

β ap−1 + γ a0 ap−1 (p)

β a2p−1 + γ

1  i=0

(p) (p)

api a2p−1−pi

.. .

(p)

β a(n−1)p−1 + γ

n −2 i=0

··· ··· ··· ..

(p) (p)

api a(n−1)p−1−pi

.

···

Next, following Step 2 and 3, we can easily obtain the result in (16).





(p) (p)

γ a0 an+p−3



1 

(p) (p) γ api an+2p−3−pi

. i=0

..



.

n −2



(p) (p) γ api an+(n−1)p−3−pi

0

i=0

X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

4291

The proof of Eq. (18) is similar, the details of which is omitted here.  (p)

Employing the similar procedure in the proof of Lemma 2.1 to Hn (ar Step 2, we also have the following lemma. Lemma 2.3. For n Hn (a(r p) )

 1, we have (p)

= α r (α p γ )n−1 Hn−1 (ar −1 ), 1  r  p − 1. (p)

At last, we analyze the term Hn (ap Lemma 2.4. For n Hn (a(pp) )

Jn (ap

(21) (p)

) and another expression of Hn (ap−1 ).

 1, we have (p)

= α p (α p γ )n−1 Hn−1 (ap−1 ) + α p (β + γ )(α p γ )n−2 Jn−1 (a(pp) ),

where

(p)

)(1  r  p − 1) without

)≡

(p)

A2p−1



(p)

A3p−1



..



.

(p)

A

(p)

(p)

ap

(p)

a2p a2p+1

··· .. .

(p) (p) (n+1)p−1 anp anp+1

···

.. .

.. .





(p) an+2p−2



..



.



(p) an+np−2

(p)

· · · an+p−2

ap+1

(p)

(22)

(23)

and (p)

Hn (ap−1 )

(p)

= −α p (α p γ )n−2 Kn−1 (ap−1 ),

where

(p) Kn (ap−1 )

≡

(p)

A2p−1



(p)

A3p−1



..



.

(p)

A

(p)

(p)

(p)

(p)

.. .

.. .

(24)





(p) an+2p−3



. ..



.



(p) an+np−3

(p)

ap−1 ap

· · · an+p−3

a2p−1 a2p

··· .. .

(p) (p) (n+1)p−1 anp−1 anp

···

(25)

Proof. Likewise, we just give the detailed proof of Eq. (22), as Eq. (24) can be proved in a similar way without Step 2. Step 1: We start by performing column operations. Firstly, subtract the (ip + 1)th column multiplied (p)

= 1, 2, . . . , [ n−p 3 ]. Next, if n = kp + 2, k = 1, 2, . . ., subtract the (n − 2)th column multiplied by (β + γ ). And then, subtracting the (n − 1)th column multiplied by α , we get by an−3−ip from the nth column for i

Hn (a(pp) )

=

(p) (p) (p)

a(pp) (β + γ )ap−1 + γ ap a−1



2 

(p) (p) (p)

a(p) (β + γ )a2p−1 + γ γ api a2p−1−pi 2p



i=1

.. ..



. .



n 

(p) ( p) (p) (p)

a api a(n−1)p−1−pi

(n−1)p (β + γ )a(n−1)p−1 + γ i=1





2 

(p) (p) ··· γ api an+2p−3−pi



i=1

. ..

..

. .



n 

(p) (p) ··· γ api an+(n−1)p−3−pi

(p) (p)

···

Then the result in (22) can be obtained by following Step 2 and Step 3. 

γ ap an−3

i=1

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X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

3. The proof of Theorem 1.2 In this section we give the proof of Theorem 1.2. The following derivations hold for any positive integer p as long as the expressions make sense.

Proof. We begin with four determinant identities. Define (p)

D1

≡ Kn (a−1 ),

D2

≡ Jn (a0 ),

D3

≡ Hn (a−1 ),

(p)

(p)



1 0

0

(p) (p) (p)

Bp−1 ap−1 ap

(p) (p) (p) D4 ≡

B2p−1 a2p−1 ap+1

. .. ..

.

. . .

(p) (p) (p)

B np−1 anp−1 anp







(p) an+p−3



(p) an+2p−3

.

..



.



(p) an+np−3

···

0

··· ··· .. . ···

Employing the Jacobi determinant identity [38,39], we have four relations: ⎞

⎛ Di Di ⎝

1 n 1 n

⎞

⎛

⎠ = Di ⎝

1 1

⎞

⎛

⎠ Di ⎝

n n

⎞

⎛

⎠ − Di ⎝

1 n

⎞

⎛

⎠ Di ⎝

n 1

⎠,

i

= 1, 2, 3, 4,

which are equivalent to (p)

(p)

Kn (a−1 )Hn−2 (ap−1 ) (p)

Jn (a0

(p)

(p)

(p)

(p)

(p)

= Kn−1 (a−1 )Hn−1 (ap−1 ) − Kn−1 (ap−1 )Hn−1 (a−1 ), (p)

(p)

(26)

(p)

(p)

)Hn−2 (ap ) = Jn−1 (a0 )Hn−1 (ap ) − Jn−1 (ap )Hn−1 (a0 ),

(27)

(p) (p) (p) (p) Hn (a−1 )Hn−2 (a(pp) ) = Hn−1 (a−1 )Hn−1 (a(pp) ) − Hn−1 (a0 )Hn−1 (ap−1 ), (p) (p) (p) Ln−1 (a(pp) )Hn−2 (ap−1 ) = Ln−2 (a(pp) )Hn−1 (ap−1 ) + Ln−1 (ap−1 )Hn−2 (a(pp) ),

respectively.

(p)

(p)

(28) (29)

(p)

Actually, from Lemmas 2.1–2.4, we can conclude that terms Kn (a−1 ), Kn (ap−1 ), Jn (a0 (p)

(p)

), Jn (ap ),

(p) (p) (p) (p) Ln−1 (ap ), Ln−1 (ap−1 ) and Hn (ap−1 ) are all linear combinations of the variables Hn (a−1 ), Hn (a0 ) and (p)

Hn (ap

). When these substitutions are applied to relations (26)–(29), it results in four equations (for (p) (p) (p) (p) (p) (p) simplicity, here we shall abbreviate Hn (a−1 ), Hn (a0 ) and Hn (ap ) to en , hn and gn , respectively): (p)

(p)

(p)

(p)

(p)

en+1 hn−p−1

= (a(pp) γ )p en(p) hn−p − γ (a(pp) γ )2p−2 en−1 hn−p+1 ,

(30)

(p) (p) hn+1 gn−2

(p) (a(pp) γ )h(np) gn−1

(31)

=

(p) − γ hn−1 gn(p) ,

X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

(p)

en(p) gn−2

(p)

⎡ (p) en(p) gn−2

p −1

(p)

= en−1 gn−1 − =

1

p −1 

⎣γ

β

i=1

p −1 

−

i=1

(p)

ai

i=1

(p)

ai

(p)

(p)



(p)

n−2−i

(ap γ ) (p)



(p)



⎤ (p) h(np) hn−p−1 ⎦ .

(p)

(33)

from the equations above. From Eqs. (32) and (33), we

 p−1  (p) (p) (p) [ai (a(pp) γ )n−1−i ]hn−1 hn−p

γ = 1+ β −

(p)

(32)

(a(pp) γ )n−1−i hn−1 hn−p

Now, we proceed to eliminate en and gn immediately obtain: (p) (p) en−1 gn−1

(p)

[ai (a(pp) γ )n−1−i ]hn−1 hn−p ,

i=1

p−1 1  (p) (p) n−2−i (p) (p) [ai (ap γ ) ]hn hn−p−1 .

β

4293

(34)

i=1

(p)

Additionally, according to relations (33) and (34), multiplying gn−1 on both sides of Eq. (30), we get (p) en(p) gn−1

 (p)

p−2

= γ (ap γ )

γ 1+ β p −1

1

 p−1  (p) (p) (p) [ai (a(pp) γ )n−1−i ]hn−1 hn−p+1 i=1

(p)

(p)

(p)

− (a(pp) γ )−p [ai (a(pp) γ )n−1−i ]hn+1 hn−p−1 . β i=1 (p)

(35)

(p)

In order to eliminate en and gn , we need another two equations. One is obtained by multiplying

(p)

gn

on both sides of Eq. (30), namely, (p)

(p)

en+1 gn(p) hn−p−1

(p)

(p)

(p)

= (a(pp) γ )p en(p) gn(p) hn−p − γ (a(pp) γ )2p−2 en−1 gn(p) hn−p+1 ,

(36)

(p)

the other is obtained by multiplying en−1 on both sides of Eq. (31), namely, (p)

(p)

(p)

hn+1 en−1 gn−2 (p) (p)

(p)

(p)

(p)

(p)

= (a(pp) γ )h(np) en−1 gn−1 − γ hn−1 en−1 gn(p) . (p)

(p)

(37)

(p) (p)

(p)

Note that en gn−2 , en−1 gn−1 and en gn−1 can be expressed in terms of hn in Eqs. (33), (34) and (p) (p) (35), therefore, according to combining Eqs. (36) and (37) and eliminating en−1 gn , we obtain (p)

(p)

(p)

γ (β + γ )(a(pp) )2p−2 h(np) hn−1 hn−p+2 hn−p−1 (p)

(p)

(p)

(p)

(p)

(p)

(p) 2

−hn+2 hn−1 hn−p hn−p−1 + (a(pp) γ )p+1 hn+1 hn−1 hn−p (p)

(p)

(p)

(p)

= γ (β + γ )(a(pp) )2p−2 hn+1 hn−2 hn−p+1 hn−p (p)

(p)

(p)

2

(p)

(p)

−hn+1 h(np) hn−p+1 hn−p−2 + (a(pp) γ )p+1 h(np) hn−p+1 hn−p−1 .

(38)

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X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

(p) (p)

(p)

(p)

Dividing by hn hn−1 hn−p+1 hn−p on both sides and rearranging the relation above, for any positive integer p, we have (p)

γ (β + γ )(ap )2p−2 (p)

(p) −2 h(p) p n−p+2+i hn−p−1+i

+

= γ (β + γ )(ap ) where

n

m ti

means an empty sum, when m

γ (β + γ )(α p γ )2p−2

(p)

(p)

(p) p+1 (p) (p) γ ) hn hn−p−1 (p) (p) hn−1 hn−p (p) (p) (p) (p) (p) hn+2 hn−p−1 −(ap γ )p+1 hn+1 hn−p (p) (p) hn hn−p+1

+

(39)

,

> n. This yields a quantity independent of n, namely,

(p)

p −2

hn−p+2+i hn−p−1+i

i=0

hn−p+1+i hn−p+i

(p)

(p)

hn+1 hn−p−2 −(ap

(p) (p) i=0 hn−p+1+i hn−p+i ( p ) (p) p−2 2p−2  hn−p+3+i hn−p+i (p) (p) i=0 hn−p+2+i hn−p+1+i

(p)

(p)

+

(p)

hn+1 hn−p−2

(p) (p)

− (α p γ )p+1 hn hn−p−1 (p)

(p)

hn−1 hn−p

= C (p). (40)

As a result, (i) and (ii) in Theorem 1.2 can be obtained by putting p where C (p) can be determined by initial values. When β + γ

= 1 and p = 2, respectively, (p)

(p)

(p)

(p)

(p)

(p)

= 0, from Lemmas 2.1, 2.3 and 2.4, we conclude that Hn (a−1 ), Hn (ap−1 ) and Hn (ap ) (p)

all can be linearly expressed in terms of Hn (a0 ). Then, substituting Hn (a−1 ), Hn (ap−1 ) and Hn (ap into Eq. (28), we immediately obtain conclusion (iii). 

)

4. Conclusion and discussions In this paper, we have presented Block–Hankel determinant solutions to a series of specific Somos recurrences. The main results are summarized as follows: (i) Conjecture 1.1 is confirmed. (ii) A Block–Hankel determinant solution to Somos-5 is constructed. (iii) Block–Hankel determinant solutions to other specified Somos recurrences are also given. (p)

Noting that in our main result the elements an of the Block–Hankel determinant solutions to Somos recurrences satisfy recursion relation (3), it is natural to ask whether there exist Block–Hankel determinant solutions, whose elements satisfy other recursion relations. These problems are still needed to be considered in the future. Actually, as for Somos-5 sequence, we have the following conjecture: (2)

Conjecture 4.1. Let an be computed via the following convolution recurrence: ⎧ ⎪ if n < 0; 0, ⎪ ⎪ ⎨ (2) 1 , if n = 0; an = ⎪ ⎪ n−2 ⎪  [ ] (2) (2) (2) (2) ⎩ α an−1 + β an−2 + γ i=20 a2i an−2−2i if n  1. (2)

(41)

(2)

= det (a2i+j )0i,jn−1 is a (α 6 γ 3 + α 2 γ 2 (β + γ )[α 2 γ + (β + γ )2 ], −α 8 γ 5 − α 4 γ 3 (β + γ )[3β(β + γ )2 + γ 2 (β + γ ) + α 2 β(β − γ ))] Somos-5 sequence. Then hn

Numerical experiments indicate that the above conjecture holds, however, a rigorous proof still remains open. Additionally, as for Somos-4 sequence, there exists another conjecture made by Michael Somos [25]. In [26], Xin solved it by use of continued fraction method. It is remarked that we can also give an alternative proof to this conjecture by similar determinant technique in this paper. Here we omit the details.

X.-K. Chang, X.-B. Hu / Linear Algebra and its Applications 436 (2012) 4285–4295

4295

Acknowledgments We would like to thank Yi He for her helpful discussions. This work was partially supported by the National Natural Science Foundation of China (Grant No. 11071241) and the knowledge innovation program of LSEC and the Institute of Computational Mathematics, AMSS, CAS. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39]

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