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Research article

A general method for solving linear matrix equations of elliptic biquaternions with applications

  • Received: 05 November 2019 Accepted: 21 February 2020 Published: 28 February 2020
  • MSC : 11R52, 15B33, 15A24

  • In this study, we obtain the real representations of elliptic biquaternion matrices. Afterwards, with the aid of these representations, we develop a general method to solve the linear matrix equations over the elliptic biquaternion algebra. Also we apply this method to the well known matrix equations X - AXB = C and AX - XB = C over the elliptic biquaternion algebra. Then, we give some illustrative numerical examples to show how the aforementioned method and its results work. Furthermore, we provide numerical algorithms for all the problems considered in this paper. Elliptic biquaternions are generalized form of complex quaternions and so real quaternions. This relation is valid for their matrices, as well. Thus, the obtained results extend, generalize and complement some known results from the literature.

    Citation: Kahraman Esen Özen. A general method for solving linear matrix equations of elliptic biquaternions with applications[J]. AIMS Mathematics, 2020, 5(3): 2211-2225. doi: 10.3934/math.2020146

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  • In this study, we obtain the real representations of elliptic biquaternion matrices. Afterwards, with the aid of these representations, we develop a general method to solve the linear matrix equations over the elliptic biquaternion algebra. Also we apply this method to the well known matrix equations X - AXB = C and AX - XB = C over the elliptic biquaternion algebra. Then, we give some illustrative numerical examples to show how the aforementioned method and its results work. Furthermore, we provide numerical algorithms for all the problems considered in this paper. Elliptic biquaternions are generalized form of complex quaternions and so real quaternions. This relation is valid for their matrices, as well. Thus, the obtained results extend, generalize and complement some known results from the literature.


    Let s=(sj)j0 be an integer sequence. For all m0, n1, the (m,n)-order Hankel matrix of s is

    Mm,n:=(sm+i+j)0i,jn1=(smsm+1sm+n1sm+1sm+2sm+nsm+n1sm+nsm+2n2.).

    The (m,n)-order Hankel determinant of s is Hm,n=detMm,n.

    Hankel determinants of automatic sequences have been widely studied, due to its application to the study of irrationality exponent of real numbers; see for example [1,3,5,6,7,9,15] and references therein. In 2016, Han [10] introduced the Hankel continued fraction which is a powerful tool for evaluating Hankel determinants. By using the Hankel continued fractions, Bugeaud, Han Wen and Yao [4] characterized the irrationality exponents of values of certain degree two Mahler functions at rational points. Recently, Guo, Han and Wu [8] fully characterized apwenian sequences, that is ±1 sequences whose Hankel determinants H0,n satisfying H0,n/2n11(mod2) for all n1.

    However, the Hankel determinants of other low complexity sequences, such as Sturmian sequences, are rarely known. Kamae, Tamura and Wen [11] explicitly evaluated the Hankel determinants of the Fibonacci word. Tamura [13] extended this result to infinite words generated by the substitutions aakb,ba (k1). In this paper, we study the Hankel determinants of the sequence generated by the substitution

    τ:1101,01.

    Denote by s=(sn)n0=limnτn(1) the fixed point of τ. Since τn+1(1)=τn(τ(1))=τn(1)τn(0)τn(1), the word τn(1) is a prefix of s. The first values of s can be obtained by finding τn(1). For example,

    s0s1s6=τ2(1)=τ(101)=1011101

    and

    s0s1s17=τ3(1)=τ(τ2(1))=10111011011011101.

    It follows from [14,Proposition 2.1] that s is a Sturmian sequence. See also the sequence A104521 in [12].

    We give the explicit values of Hankel determinants Hm,n for the sequence s for all m0 and n1. In Figure 1, we use the color at the point (m,n) to indicate the value of Hm,n. For example, if Hm,nHm,n, then points (m,n) and (m,n) will be marked by different colors. In particular, if Hm,n=0, then the point (m,n) is marked by white. Then we can see the distribution of first values of Hm,n from Figure 1 and the collection of all the points (m,n) with Hm,n=0 are the union of disjoint parallelograms. These parallelograms (together with their boundaries) are divided into parallelograms of three types, labelled by Uk,i, Vk,i and Tk,i where k0 and i1 (for detailed definitions, see Section 3).

    Figure 1.  Visualization of the Hankel determinants Hn,m (0n,m350).

    For all k0 and i1, Theorem 1.1 (resp. Theorem 1.2, Theorem 1.3) gives the exact value of Hm,n for all (m,n)Uk,i (resp. Vk,i, Tk,i); see Figure 2. Since for k0 and i1, the parallelograms Uk,i, Vk,i and Tk,i are disjoint and they tile the lattices in the first quadrant; see Proposition 3.1 in Section 3. Combing this nice property and Theorem 1.1, 1.2 and 1.3, we obtain the values of Hm,n for all m0 and n1.

    Figure 2.  Theorem 1.1 (1) and (2) give the values of Hm,n for (m,n) on the upper edge (marked by red dots) and the lower edge (marked by blue dots) of the parallelogram Uk,i, respectively. Theorem 1.1 (3) give the values of Hm,n for (m,n) marked by black and white circles in the parallelogram Uk,i. Theorem 1.2 and 1.3 are illustrated in the same way.

    To state our results, we need four (technically defined) integer sequences (fn)n0, (αi)i1, (βi)i1 and (γi)i1; see Section 3 for their definitions. These four sequences are used to locate the corners of parallelograms Uk,i, Vk,i and Tk,i. Our main results are the following.

    Theorem 1.1. Let k0 and Uk=i1Uk,i. For all (m,n)Uk,

    1. when n=f2k+31, Hm,n=(1)k+1(1)f2k+52Φk+1(m+n)f2k+12;

    2. when n=f2k, Hm,n=(1)k+1(1)f2k+212f2k+12;

    3. when f2k<n<f2k+31, if m+n=αif2k+2+1 or αi for some i1, then

    Hm,n=(1)(f2k+3n)k(1)(f2k+31n)(f2k+32n)2f2k+12;

    otherwise Hm,n=0.

    Theorem 1.2. Let k0 and Vk=i1Vk,i. For all (m,n)Vk,

    1. when n=f2k+21, Hm,n=(1)f2k+2+f2k+132f2k+12;

    2. when n=f2k, Hm,n=(1)k+1(1)f2k+212f2k+12;

    3. when f2k<n<f2k+21, if m+n=βi+1 or βi+f2k+1 for some i1, then

    Hm,n=(1)(f2k+2n)(k+1)(1)(f2k+21n)(f2k+22n)2f2k+12;

    otherwise Hm,n=0.

    Theorem 1.3. Let k0 and Tk=i1Tk,i. For all (m,n)Tk,

    1. when n=f2k+21, Hm,n=(1)f2k12f2k;

    2. when n=f2k+1, Hm,n=(1)Φk(m+n)f2k+121f2k;

    3. when f2k<n<f2k+21, if m+n=γif2k+1 or γi for some i1, then

    Hm,n=(1)(f2k+21n)k(1)(f2k+21n)(f2k+22n)2(1)f2k12f2k;

    otherwise Hm,n=0.

    Sketch of proofs of main results

    The computational result indicates that the collection of points (m,n) such that the Hankel determinant Hm,n=0 is covered by disjoint parallelograms of three different types, denoted by U,,V, and T,; see Figure 1. This inspires us to calculate the Hankel determinant Hm,n in each parallelogram (according to the location of (m,n)). Then connect the values of Hankel determinants in different parallelograms.

    Step 1 Locate the parallelograms.

    We first find that the second coordinates of the corners of those parallelograms can be expressed in terms of an integer sequence (fn)n0 (see Figure 2 for example). Then we see that the first coordinates of corners of three types of parallelograms are determined by three integer sequences (αi)i1, (βi)i1 and (γi)i1 introduced in Section 3. Then we characterize the parallelograms (observed in Figure 1) by (3.1). Proposition 3.1 showed that parallelograms defined by (3.1) tile the first quadrant.

    Step 2 Calculate Hm,n for (m,n) inside the parallelograms Uk,, Vk, and Tk,.

    Lemma 4.1 show that Hm,n=0 for all (m,n) which are not on the boundary of those parallelograms. Now the white part in Figure 1 is clear.

    Step 3 Reduction on the boundary of Uk,, Vk, and Tk,.

    ● By Lemma 4.2 and Lemma 4.6 (resp. Lemma 4.8, Lemma 4.10), calculating Hm,n for all (m,n) on the boundary of Uk, (resp. Vk,, Tk,) is reduced to calculate the determinant Hm,n for only one point (m,n) on its boundary. See Figure 3.

    Figure 3.  Illustration for Lemma 4.2, 4.6, 4.8 and 4.10.

    ● Lemma 4.3 connects the values of Hm,n for (m,n) on the lower edge of Uk, and Vk,. Lemma 4.4 builds similar connections for the values of Hm,n for (m,n) on the lower edge of Tk,i and Tk,i+1. See Figure 4.

    Figure 4.  Illustration for Lemma 4.3, 4.4, 5.1 and 5.2.

    Therefore, to obtain the values of Hm,n for all (m,n) on the boundary of all Uk, and Vk,, we only need to calculate Hm,n for one point (m,n) on the boundary of Uk,1.

    Step 4 Reduction on k. Lemma 5.1 enable us to calculate Hm,n for (m,n) on the boundary of Uk+1, by using the values of Hm,n on the boundary of Uk, and Tk,. Lemma 5.2 enable us to calculate Hm,n for (m,n) on the boundary of Tk, by using the values of Hm,n on the boundary of Uk, and Uk1,. See Figure 4.

    Step 5 According to step 3 and step 4, to obtain the value of Hm,n for all (m,n), we only need to calculate Hm,n for (m,n) on the boundary of Uk,1, Vk,1 and Tk,2 for all k. These have been done by Theorem 5.3, Corollary 5.4 and Corollary 5.5.

    The paper is organized as follows. In Section 2, we introduce the f-representation of positive integers and give a criterion (Proposition 2.4) to determine sn according the f-representation of n. This criterion leads us to the key ingredient (Theorem 2.5) in calculating the Hankel determinants. Then we introduce the truncated f-representation which is essential in describing the parallelograms. In Section 3, we show that the parallelograms Uk,i, Vk,i and Tk,i tile all the integer points in the first quadrant. In Section 4, we first show that the Hankel determinants vanish when (m,n) is inside a parallelogram of those three types. Next we show the relations of Hankel determinants Hm,n on the boundary of a parallelogram Uk,i (or Vk,i, Tk,i) for a given k and i. Finally, for any k0, we describe the relation of values of Hankel determinants for parallelograms Uk,i for all i1. In Section 5, we give the expressions for Hankel determinants on the boundary of Uk,i (or Vk,i, Tk,i) for all k0. In the last section, we formulate and prove our main results.

    In this section, we first introduce the f-representation of positive integers according to the sequence (fn)n0. By understanding the occurrences of 0s in the sequence s, we prove a key result (Proposition 2.4) which can determine sn according to the f-representation of n. Then we give the essential result (Theorem 2.5). In subsection 2.3, we introduce the truncated f-representation which is useful in determining the parallelograms. In section 2.4, we investigate to some sub-sequences of (fn)n0 which are need in evaluating the coefficients of the Hankel determinants. Then we characterize two sub-sequences of s which helps us understand Hm,n.

    We introduce an auxiliary sequence (fn)n0 to determine the positions of 0's. For all n0, we define

    f2n=|τn(1)|andf2n+1=|τn(10)|.

    Then f0=1, f1=2, and for all n0,

    {f2n+2=f2n+f2n+1,f2n+3=f2n+f2n+2. (2.1)

    The first values are

    (fn)n0=(1,2,3,4,7,10,17,24,41,58,99,140,239,338,577,816,).

    Remark 2.1. The sequence (fn)n0 can be expressed in terms of Pell numbers (pn)n0 defined by the recurrence p0=1, p1=2 and pn+2=2pn+1+pn for all n0. See the sequence A000129 in [12]. Indeed, pn (resp. pn1) is the number of 1's (resp. 0's) in τn(1). It is easy to verify that f2n=pn+1pn and f2n+1=2pn.

    Since (fn)n0 is an increasing non-negative integer sequence, it is a numeration system in the following sense.

    Lemma 2.2 (Theorem 3.1.1 [2]). Let u0<u1<u2< be an increasing sequence of integers with u0=1. Every non-negative integer N has exactly one representation of the form 0iraiui where ar0, and for i0, the digits ai are non-negative integers satisfying the inequality

    a0u0+a1u1++aiui<ui+1.

    Proposition 2.3. Every integer n0 can be uniquely expressed as n=0iraifi with ai{0,1}, ar0, and

    {aiai+1=0for all0i<r,aiai+2=0for all even numbers0i<r1. (2.2)

    Proof. Suppose ai{0,1}. By Lemma 2.2, we only need to show that a0f0+a1f1++atft<ft+1 for all t if and only if the condition (2.2) holds.

    The 'only if' part. Suppose there is an index i such that aiai+1=1. Then

    a0f0+a1f1++aifi+ai+1fi+1fi+fi+1fi+2,

    which is a contradiction for t=i+1. Suppose there is an even index i such that aiai+2=1. Then

    a0f0+a1f1++aifi+ai+1fi+1+ai+2fi+2fi+fi+2=fi+3,

    which is a contradiction for t=i+2.

    The 'if' part. Suppose the condition (2.2) holds. When t is odd, the maximum possible value of a0f0+a1f1++atft occurs when atat1a0=101010, and this maximum value f1+f3++ft2+ft=ft+11. When t is even, the maximum possible value of a0f0+a1f1++atft occurs when atat1a0=100101010. In this case, the maximum value is f1+f3++ft5+ft3+ft=ft+11.

    Definition (f-representation). Let n0 be an integer. We call the representation n=0iraifi in Proposition 2.3 the f-representation of n. We also write n=+i=0aifi where ai=0 for all i>r. In the case that we need to emphasize that ai depends on n, we write ai=ai(n) as a function of n.

    Proposition 2.4. For any integer n0 with the f-representation ri=0ai(n)fi, we have sn=0 if and only if a0(n)=1.

    Proof. One can verify directly that the result holds for all n<f4=7. Assume that the result holds for n<f2k where k2. We only need to prove it for all f2kn<f2k+2.

    Suppose f2kn<f2k+1. One has a2k(n)=1 and hence a0(nf2k)=a0(n). Note that f2k=|τk(1)| and

    s0s1sf2k+21=τk+1(1)=τk(1)τk(0)τk(1). (2.3)

    We see that sn is the (n+1)-th letter of τk+1(1) and it is also the (n+1f2k)-th letter of τk(0)=τk1(1). Consequently, sn=snf2k. Since

    nf2k<f2k+1f2k=f2k2,

    by the inductive assumption, we have snf2k=0 if and only if a0(nf2k)=1. Therefore, sn=0 if and only if a0(n)=1.

    Suppose f2k+1n<f2k+2. In this case a2k+1(n)=1 and a0(nf2k+1)=a0(n). Since |τk(10)|=f2k+1, it follows from (2.3) that sn=snf2k+1. Note that

    nf2k+1<f2k+2f2k+1=f2k.

    By the inductive assumption, snf2k+1=0 if and only if a0(nf2k+1)=1 which implies the result also holds for all f2k+1n<f2k+2.

    We introduce the truncated f-representations (of positive integers) which are useful in telling two digits with a fixed gap in s are equal or not.

    Definition. (Truncated f-representation) Let n0 be an integer with the f-representation +i=0ai(n)fi. For all integers k0, the truncated f-representation of n is

    Φk(n):=2k+2i=0ai(n)fi.

    The next lemma gives a criterion that when two digits (with a fixed gap) in s are equal by using their positions.

    Theorem 2.5. Let n0 be an integer with the f-representation +i=0ai(n)fi. Then

    (i) for all k0, sn+f2ksn if and only if Φk(n){f2k+12,f2k+121};

    (ii) for all k1, sn+f2k+1sn if and only if Φk(n){f2k+32,f2k+321,f2k+32+f2k,f2k+32+f2k1}.

    Proof. (ⅰ) We prove by induction on k. When k=0, by Proposition 2.3, there are only four possible values for a0(n)a1(n)a2(n). By Eq (2.1), we have

    Φ0(n) 0 1 2 3
    a0(n)a1(n)a2(n) 000 100 010 001 .
    a0(n+f0) 1 0 0 0

     | Show Table
    DownLoad: CSV

    Then we see that a0(n)a0(n+f0) if and only if Φ0(n)=0 or 1. The result holds for k=0.

    When k=1, note that Φ1(n)4i=0fi<f5=10. We see

    Φ1(n) 0 1 2 3 4 5 6 7 8 9
    a0(n)a4(n) 00000 10000 01000 00100 00010 10010 01010 00001 10001 01001 .
    a0(n+f2) 0 0 1 0 0 1 0 0 1 0

     | Show Table
    DownLoad: CSV

    Then we have a0(n)a0(n+f2) if and only if Φ1(n)=1 or 2, that is f321 or f32. The result also holds for k=1.

    Now assume that the result holds for all 0k< with 2. We prove it for k=. Let w=a22(n)a21(n)a2+2(n) and v=a22(n+f2)a21(n+f2)a2+1(n+f2). According to Proposition 2.3, w can take only 10 different values.

    While w01000, one can determine v directly using Eq (2.1); thus in these cases, ai(n+f2)=ai(n) for all 0i23; see Table 1. For instance, when w=10010,

    n+f2=(23i=0ai(n)fi+f22+f2+1++i=2+3ai(n)fi)+f2=(23i=0ai(n)fi+f22)+(f2+2++i=2+3ai(n)fi).
    Table 1.  Values of v.
    w 00000 10000 01000 00100 00010 10010 01010 00001 10001 01001
    v 0010 0001 ? 0101 0000 1000 0100 0000 1000 0100

     | Show Table
    DownLoad: CSV

    Hence one can see that a0(n+f2)=a0(n).

    When w=01000, set n=22i=0ai(n)fi. Then a0(n)=a0(n) and

    n+f2=(n+f21++i=2+3ai(n)fi)+f2=(n+f24)+f2+1++i=2+3ai(n)fi.(by Eq. 2.1)

    Noticing that n+f24<f22, we have ai(n+f2)=ai(n+f24) for all 0i22. In particular, a0(n+f2)=a0(n+f24). By Proposition 2.4 and the inductive assumption,

    a0(n+f2)a0(n)a0(n+f24)a0(n)n{f232,f2321}Φ(n)=n+f21{f232+f21,f232+f211}.

    By Eq (2.1), we have

    f2+12=(f22+f2)/2=(f22+f22+f21)/2=f22+f212=f22+f24+f232=f21+f232. (2.4)

    Then we obtain that

    a0(n+f2)a0(n)Φ(n){f2+12,f2+121}.

    It follows from Proposition 2.4 that the result holds for k=.

    (ⅱ) For any k1, let u=a2k(n)a2k+1(n)a2k+2(n). It follows from Proposition 2.4 that u{100,010,001}. The proof is divided into the following three cases.

    ● When u=001, we also have a2k+3(n)=a2k+4(n)=0. So

    n+f2k+1=(2k1i=0ai(n)fi+f2k+2++i=2k+5ai(n)fi)+f2k+1=(2k1i=0ai(n)fi+f2k2)+(f2k+3++i=2k+5ai(n)fi).

    Let n=2k1i=0ai(n)fi. Then a0(n)=a0(n). Since n<f2k and n+f2k2<f2k+1, we have ai(n+f2k2)=ai(n+f2k+1) for all 0i2k. Thus

    a0(n+f2k+1)a0(n)ai(n+f2k2)a0(n)n=2ki=0ai(n)fi{f2k12,f2k121}

    where in the last step we use Theorem 2.5(ⅰ). By Eq (2.1) and Eq (2.4),

    f2k12+f2k+2=f2k12+f2k+1+f2k=f2k+32+f2k.

    So when u=001, a0(n+f2k+1)a0(n) if and only if

    Φk(n)=n+f2k+2{f2k+32+f2k,f2k+32+f2k1}.

    ● Suppose u=010. Applying Eq (2.1) twice, we obtain that

    2f2k+1=f2k2+f2k+f2k+1=f2k2+f2k+2.

    Then

    n+f2k+1=(2k1i=0ai(n)fi+f2k+1++i=2k+3ai(n)fi)+f2k+1=(2k1i=0ai(n)fi+f2k2)+(f2k+2++i=2k+3ai(n)fi).

    Let n=2k1i=0ai(n)fi. Using Theorem 2.5(ⅰ), the same argument as in the case u=001 leads us to the fact that

    a0(n+f2k+1)a0(n)n=2ki=0ai(n)fi{f2k12,f2k121}Φk(n)=n+f2k+1{f2k+32,f2k+321}.

    ● When u=100, we have a2k2(n)=a2k1(n)=0. Then

    n+f2k+1=(2k3i=0ai(n)fi+f2k++i=2k+3ai(n)fi)+f2k+1=(2k3i=0ai(n)fi)+(f2k+2++i=2k+3ai(n)fi)

    which implies that a0(n+f2k+1)=a0(n).

    To apply Theorem 2.5, we need to investigate the integers of the same truncated f-representations. The following two lemmas (Lemma 2.6 and Lemma 2.8) serve for this purpose.

    For all k0, denote

    Ek={xN:Φk(x)=f2k+32}andEk={xN:Φk(x)=f2k+32+f2k}.

    Let Ek=EkEk=(x(k)j)j1 where x(k)1<x(k)2<x(k)3<. The first values of Ek are

    (x(k)j)j1=(f2k+32,f2k+32+f2k,f2k+32+f2k+3,f2k+32+f2k+4,f2k+32+f2k+5,f2k+32+f2k+f2k+5,f2k+32+f2k+3+f2k+5,f2k+32+f2k+6,).

    Lemma 2.6. Let k0 and xEk with x=x(k)j for some j2. Then xf2k+2=x(k)j1Ek.

    Proof. Let xEk with x=x(k)j for some j2. Note that f2k+32+f2k=f2k+2+f2k12. By Proposition 2.3, have a2k+3(x)=a2k+4(x)=0. When 0<b<f2kf2k12, we see

    Φk(x+b)=(f2k+32+f2k)+b<f2k+3

    which implies that x+bEk. When f2kf2k12b<f2k+2,

    Φk(x+b)=(f2k+32+f2k)+bf2k+3<f2k+32.

    So x+bEk. Since Φk(x+f2k+2)=f2k+32, we have x+f2k+2=x(k)j+1Ek.

    Let xEk with x=x(k)j for some j1. According to Proposition 2.3, a2k+3(x)a2k+4(x)=00, 10 or 01, which can be divided into two sub-cases.

    a2k+3(x)a2k+4(x)=00. For 0<bf2k, we see Φk(x+b)=Φk(x)+b=f2k+32+b. Thus x+f2k=x(k)j+1Ek.

    a2k+3(x)a2k+4(x)=01 or 10. For 0<b<f2k, we see

    Φk(x+b)=Φk(x)+b=f2k+32+b<f2k+32+f2k.

    Thus x+bEk. For f2kb<f2k+2, we have

    Φk(x+b)=f2k+32+bf2k+2(f2k12,f2k+32)

    which yields that x+bEk. Noting that Φk(x+f2k+2)=Φk(x)=f2k+32, we obtain that x+f2k+2=x(k)j+1Ek.

    From the above argument, we see that if x=x(k)jEk for some j2, then either xf2k+2=x(k)j1Ek or xf2k+2=x(k)j1Ek with a2k+3(xf2k+2)a2k+4(xf2k+2)00. The result holds.

    Remark 2.7. From the proof of Lemma 2.6, we see the gaps between two adjacent elements in Ek are f2k and f2k+2. That is x(k)j+1x(k)j=f2k or f2k+2 for all j1. Moreover, the gaps between two adjacent elements in Ek are f2k+2 and f2k+3.

    For all k0, let

    Fk={yN:Φk(y)=f2k+12}=(y(k)j)j1andFk={yN:Φk+1(y)=f2k+12}

    where y(k)1<y(k)2<y(k)3<. Write Fk=FkFk. The first values of Fk are

    (y(k)j)j1=(f2k+12,f2k+12+f2k+3,f2k+12+f2k+4,f2k+12+f2k+5,f2k+12+f2k+5+f2k+3,f2k+12+f2k+6,).

    Lemma 2.8. For any yFk with y=y(k)j for some j1, we have

    y(k)j+1={y+f2k+3,ifyFk;y+f2k+2,ifyFk.

    Proof. We prove the result by giving the construction of Fk. It clear that y(k)1=f2k+12. Now suppose y=y(k)jFk where j1. According to Proposition 2.3, we see a2k+3(y)a2k+4(y)=00, 01 or 01.

    a2k+3(y)a2k+4(y)=00, i.e., yFk. Note that f2k+12=f2k1+f2k32. For 0<b<f2k+3f2k+12, we have Φk(y+b)=f2k+12+b, so y+bFk. For f2k+3f2k+12b<f2k+3,

    Φk(y+b)=f2k+12+bf2k+3<f2k+12,

    so y+bFk. Since Φk(y+f2k+3)=Φk(y), we obtain that y+f2k+3=y(k)j+1FkFk.

    a2k+3(y)a2k+4(y)=10 or 01. For 0<b<f2k+2f2k+12, we have Φk(y+b)=f2k+12+b, so y+bFk. For f2k+2f2k+12b<f2k+2, since

    Φk(y+b)=f2k+12+bf2k+2<f2k+12,

    we also have y+bFk. It follows from Φk(y+f2k+2)=Φk(y) that y+f2k+2=y(k)j+1Fk.

    The result follows from the above two sub-cases.

    Remark 2.9. From the proof of Lemma 2.8, we see the gaps between two adjacent elements in Fk are f2k+2 and f2k+3. That is y(k)j+1y(k)j=f2k+2 or f2k+3 for all j1. Moreover, the gaps between two adjacent elements in Fk are f2k+2 and f2k+4.

    The subsequences (sf2k+12)k0 and (sf2k+121)k0 can be determined according to the parity of k; see Lemma 2.11. We start with an auxiliary lemma which concerns the parity of f2k+12.

    Lemma 2.10. For all k0,

    (i) f2k{1,ifk0or3(mod4),3,ifk1or2(mod4),(mod4),

    (ii) f2k+1{2,ifkis even,0,ifkis odd,(mod4).

    Proof. (ⅰ) Note that f0=1 and f2=3. Since f2n is odd for all n0, using Eq. (2.1) twice, we have for all k2,

    f2k=f2k2+f2k1=2f2k2+f2k42+f2(k2)(mod4).

    The result follows by induction on k.

    (ⅱ) The initial value is f1=2. Using Eq. (2.1) and the previous result (ⅰ), we have for all k1,

    f2k+1=f2k+f2k2{2,k0,2(mod4),0,k1,3(mod4),(mod4)

    which is the desired result.

    In the calculation of Hm,n, we need to know sn explicitly for some n. The next lemma determines the values of two sub-sequences s.

    Lemma 2.11. For all k0,

    sf2k+12={1,ifkis odd,0,ifkis even,andsf2k+121={0,ifkis odd,1,ifkis even.

    Proof. By Eq (2.1), we obtain that for all k0,

    f2k+12=(f2k2+f2k)/2=(f2k2+f2k2+f2k1)/2=f2k2+f2k12==k1i=0f2i+f12. (2.5)

    When k is odd,

    f2k+12=(f2k2+f2k4)+(f2k6+f2k8)++(f4+f2)+f0+f12=f2k1+f2k5++f5+f1(by Eq. (2.1))=k12i=0f4i+1. (2.6)

    When k2 is even,

    f2k+12=(f2k2+f2k4)+(f2k6+f2k8)++(f2+f0)+f12=f2k1+f2k5++f3+f12(by Eq. (2.1))=k22i=0f4i+3+f12=k22i=0f4i+3+f0. (2.7)

    It follows from (2.6) and (2.7) that for all k0,

    a0(f2k+12)={0, if k is odd,1, if k is even,

    and

    a0(f2k+121)={1, if k is odd,0, if k is even.

    Then by Proposition 2.4, the result follows.

    According to the values of the Hankel determinants of s, we tile the integer lattice using the following parallelograms. Given a k0, write the elements in Ek+1, Fk and Ek in ascending order as follows:

    Ek+1=(αi)i1,Fk=(βi)i1,Ek=(γi)i1.

    Moreover, let βi=βi+f2k for all i1. We define three different types of parallelograms: for i1,

    Uk,i={(m,n)N2:f2kn<f2k+3,αif2k+2<n+mαi},Vk,i={(m,n)N2:f2kn<f2k+2,βi<n+mβi+f2k+1},Tk,i={(m,n)N2:f2k+1n<f2k+2,γif2k<n+mγi}; (3.1)

    see Figure 1. Let Uk=i1Uk,i, Vk=i1Vk,i and Tk=i1Tk,i.

    Proposition 3.1. The parallelograms {Uk,i}, {Vk,i}, and {Tk,i} introduce a partition of pairs of positive integers. Namely, N×N1=k0(UkVkTk) where denotes the disjoint union.

    Proof. Let m0 and n1 be two integers. Since (fk)k0 and (γk)k1 are two increasing unbounded non-negative integer sequences, there exist k0 and 1 such that f2kn<f2k+2 and γ1<n+mγ where γ0:=0. The result clearly holds when =1. Now we assume that 2. From the proof of Lemma 2.6 we see that γγ1=f2k+2 or f2k+3 for all 2. When γf2k<n+mγ, we have

    (m,n){Uk1, if f2kn<f2k+1;Tk, if f2k+1n<f2k+2;

    see also Figure 5. When γ1<n+mγf2k, we have the following two cases:

    Figure 5.  Partition of the strip [0,+)×[f2k,f2k+2).

    Case 1: γγ1=f2k+2. In this case, we shall verify that (m,n)Vk. To do this, we only need to show that γ1f2kFk. Since γ1Ek, we have Φk(γ1)=f2k+32 and Φk(γ1f2k)=f2k+32f2k=f2k+12. So (γ1f2k)Fk. Suppose on the contrary that (γ1f2k)Fk. Then Φk+1(γ1f2k)=f2k+12 and Φk+1(γ1)=f2k+32. This implies Φk(γ1+f2k+2)=f2k+12 and (γ1+f2k+2)Ek. Note that in this case γ=γ1+f2k+2. We conclude that γEk which is a contradiction. Hence, (γ1f2k)Fk. The result follows.

    Case 2: γγ1=f2k+3. We assert that, in this case, γ1f2kFk. Since γ1Ek, we have Φk(γ1)=f2k+32. Consequently, Φk(γ1f2k)=f2k+12 and (γ1f2k)Fk. Suppose (γ1f2k)Fk. Then Φk(γ1)=f2k+32 and Φk+1(γ1)f2k+32. It follows that Φk(γ1+f2k+3)=f2k+32+f2k. Since γ1+f2k+3=γ, we obtain that γEk which is a contradiction. Now we have γ1f2kFk. This yields that Φk+1(γ1)=f2k+32. Observing that Φk+1(γf2k)=Φk+1(γ1+f2k+2)=f2k+52, we see γf2kEk+1. So (m,n)Uk.

    In this section, we use the Theorem 2.5 to show the determinant value inside Uk, Vk, Tk is 0. For some integer k0, we prove the relationship between the determinant value of the boundary of Uk, Vk, Tk. We assert that as long as we know one value of Uk(Vk or Tk), we can know all its values.

    The Hankel determinant Hm,n vanishes if (m,n) is not on the boundary of any parallelogram Uk,i, Vk,i or Tk,i.

    Lemma 4.1. Let m1 and n0 be two integer.

    (i) If (m,n) is inside Vk,i for some k0 and i1, i.e.,

    {f2k+1n<f2k+21,βi+1<n+mβi+f2k+11,

    then Hm,n=0.

    (ii) If (m,n) is inside Tk,i for some k0 and i1, i.e.,

    {f2k+1+1n<f2k+21,γif2k+1<n+mγi1,

    then Hm,n=0.

    (iii) If (m,n) is inside Uk,i for some k0 and i1, i.e.,

    {f2k+1n<f2k+31,αif2k+2+1<n+mαi1,

    then Hm,n=0.

    Proof. Let Am+i be the i-th row of Hm,n. Then

    Hm,n=det(smsm+1sm+n1sm+1sm+2.........sm+n1sm+2n2)=det(AmAm+1Am+n1).

    (ⅰ) When mβi+1, recall that βi=βif2kFk. Since nf2k+22, by Lemma 2.8, we have Φk(βk+j)f2k+12 or f2k+121 for all 1jn. Then it follows from Theorem 2.5(ⅰ) that

    Aβi+1=(sβi+1,sβi+2,,sβi+n)=(sβi+1,sβi+2,,sβi+n)=Aβi+1,

    which gives Hm,n=0. When m>βi+1, note that n+mβk+f2k+21. By Lemma 2.8, we have Φk(m+j)f2k+12 or f2k+121 for all 1jn. Then it follows from Theorem 2.5(ⅰ) that

    Am=(sm,sm+1,,sm+n1)=(sm+f2k,sm+f2k+1,,sm+f2k+n1)=Am+f2k.

    So Hm,n=0.

    (ⅱ) Recall that γiEk and by Lemma 2.6, γi and γif2k+2 are adjacent elements in Ek. Let

    r={γif2k+2+1m, if mγif2k+2+1,0, if m>γif2k+2+1.

    Combining Lemma 2.6 and Theorem 2.5(ⅱ), we have Am+r=Am+r+f2k+1 which means Hm,n=0.

    (ⅲ) Recall that αiEk+1 and by Lemma 2.6, αi and αif2k+4 are adjacent elements in Ek+1. When mαif2k+4+1, note that αif2k+4+n<αif2k+21. By Theorem 2.5(ⅱ), we have

    Aαif2k+4+1=(sαif2k+4+1,sαif2k+4+2,,sαif2k+4+n)=(sαif2k+2+1,sαif2k+2+2,,sαif2k+2+n)=Aαif2k+2+1.

    Thus Hm,n=0. When m>αif2k+4+1, since n+m1αi2, by Theorem 2.5(ⅱ), we obtain that

    Am=(sm,sm+1,,sm+n1)=(sm+f2k+3,sm+f2k+3+1,,sm+f2k+3+n1)=Am+f2k+3

    which also implies Hm,n=0.

    We first deal with the Hankel determinants Hm,n on the horizontal edges with n=f2k and f2k+1 where k0.

    Lemma 4.2. Let k0 and i1.

    (i) (Bottom edge of Vk,i) Hβi+r,f2k=Hβi+1,f2k for all 1rf2k+1.

    (ii) (Bottom edge of Uk,i) Hαif2k+3+r,f2k=Hαif2k,f2k for all 1rf2k+2.

    (iii) (Bottom edge of Tk,i) Hγif2k+2+r,f2k+1=(1)r+1Hγif2k+1,f2k+1 for all 1rf2k with γif2k+2+r0.

    Proof. (ⅰ) Let Aj=(sβi+j,sβi+j+1,,sβi+j+f2k1). Then for 1j<f2k+1,

    Hβi+j,f2k=det(AjAj+1Af2k+j1)andHβi+j+1,f2k=det(Aj+1Aj+2Af2k+j).

    Recall that βiFk. By Lemma 2.8, since j+f2k1f2k+22, we see Φk(βi+)f2k+12 or f2k+121 for all 1f2k+22. Applying Theorem 2.5(ⅰ), we have

    Aj=(sβi+j,sβi+j+1,,sβi+j+f2k1)=(sβi+j+f2k,sβi+j+1+f2k,,sβi+j+2f2k1)=Af2k+j.

    Therefore, for 1j<f2k+1,

    Hβi+j,f2k=|AjAj+1Af2k+j1|=|Af2k+jAj+1Af2k+j1|=(1)f2k1|Aj+1Aj+2Af2k+j|=Hβi+j+1,f2k

    where the last equality follows from Lemma 2.10(ⅰ).

    (ⅱ) Recall that αiEk+1 and Φk+1(αi)=f2k+52. Let y=αif2k+3. Then Φk+1(y)=f2k+12 and yFk. Let Bj=(sy+j,sy+j+1,,sy+j+f2k1). Then for 1j<f2k+2,

    Hy+j,f2k=det(BjBj+1Bf2k+j1)andHy+j+1,f2k=det(Bj+1Bj+2Bf2k+j).

    Since j+f2k1f2k+32, by Lemma 2.8 and Theorem 2.5(ⅰ),

    Bj=(sy+j,sy+j+1,,sy+j+f2k1)=(sy+j+f2k,sy+j+1+f2k,,sy+j+2f2k1)=Bf2k+j.

    Therefore, for 1j<f2k+2,

    Hy+j,f2k=(1)f2k1Hy+j+1,f2k=Hy+j+1,f2k

    where the last equality follows from Lemma 2.10(ⅰ).

    (ⅲ) Recall that γiEk. By Lemma 2.6, g:=γif2k+2Ek. Write

    Ag+j=(sg+j,sg+j+1,,sg+j+f2k+11).

    For 1r<f2k,

    Hg+r,f2k+1=|Ag+rAg+r+1Ag+r+f2k+11|andHg+r+1,f2k+1=|Ag+r+1Ag+r+2Ag+r+f2k+1|.

    By Theorem 2.5, Ag+r=Ag+r+f2k+1. Then using Lemma 2.10, for all 1r<f2k,

    Hg+r,f2k+1=(1)f2k+11Hg+r+1,f2k+1=Hg+r+1,f2k+1

    and Hg+r,f2k+1=(1)f2krHg+r+f2k,f2k+1=(1)1+rHg+r+f2k,f2k+1.

    In fact, for all i1, the Hankel determinants on the bottom of Uk,i and Vk,i take the same value which depends only on k. The following lemma helps us to connect the determinants on the bottom of Uk, and Vk,.

    Lemma 4.3. Let k0 and i1. If γi+1γi=f2k+3, then Hγi+f2k+1,f2k=Hγi+1f2k+1,f2k. If γi+1γi=f2k+2, then Hγi+1,f2k=Hγi+1f2k+1,f2k.

    Proof. Suppose γi+1γi=f2k+3. Then Φk(γi+f2k)=f2k+32+f2k. Since 3f2k=f2k+2+f2k1<f2k+3, by Theorem 2.5(ⅱ), we have

    (sγi+f2k+1sγi+3f2k1)=(sγi+f2k+1+f2k+1sγi+3f2k1+f2k+1)=(sγi+1f2k+1sγi+1+f2k1).

    Therefore

    Hγi+f2k+1,f2k=|sγi+f2k+1sγi+2f2ksγi+2f2ksγi+3f2k1|=|sγi+1f2k+1sγi+1sγi+1sγi+1+f2k1|=Hγi+1f2k+1,f2k.

    When γi+1γi=f2k+2, we have Φk(γi)=f2k+32. By Theorem 2.5(ⅱ),

    (sγi+1sγi+2f2k1)=(sγi+1+f2k+1sγi+2f2k1+f2k+1)=(sγi+1f2k+1sγi+1+f2k1).

    So Hγi+1,f2k=Hγi+1f2k+1,f2k.

    Next we give the connection between Tk,i and Tk,i+1.

    Lemma 4.4. For all i1, Hγif2k+1,f2k+1=Hγi+1f2k+1,f2k+1.

    Proof. If γi+1γi=f2k+3, then Φk+1(γi)=f2k+32 and Φk+1(γi+f2k+1)<f2k+52. By Theorem 2.5(ⅱ), we have

    (sγif2k+1sγi+f2k+12)=(sγif2k+1+f2k+3sγi+f2k+12+f2k+3)=(sγi+1f2k+1sγi+1+f2k+12).

    Consequently, Hγif2k+1,f2k+1=Hγi+1f2k+1,f2k+1.

    If γi+1γi=f2k+2, then Φk+1(γi)=f2k+32+f2k+3 or f2k+32+f2k+4. By Theorem 2.5(ⅰ), we have

    (sγif2k+1sγi+f2k+12)=(sγif2k+1+f2k+2sγi+f2k+12+f2k+2)=(sγi+1f2k+1sγi+1+f2k+12).

    Consequently, Hγif2k+1,f2k+1=Hγi+1f2k+1,f2k+1.

    According to Lemma 4.3 and Lemma 4.4, the values of the determinants on the bottom edges of Uk,i and Vk,i only depends on k. We improve Lemma 4.2 to the following proposition.

    Proposition 4.5. Let k0. For all i1,

    (i) (Bottom edges of Uk,i and Vk,i) for all 1rf2k+1 and 1rf2k+2,

    Hαif2k+3+r,f2k=Hβi+r,f2k=Hα1f2k,f2k;

    (ii) (Bottom edge of Tk,i) Hγif2k+2+r,f2k+1=(1)r+1Hγ1f2k+1,f2k+1 for all 1rf2k with γif2k+2+r0.

    Proof. Since αif2k+2Ek and βi+f2kEk, Lemma 4.3 shows that the values of two determinants on the bottom edge of two adjacent parallelograms in {Uk,j}j1{Vk,j}j1 are the same. Then Lemma 4.2 implies the result (ⅰ). The result (ⅱ) follows from Lemma 4.2(ⅲ) and Lemma 4.4.

    Lemma 4.6. Let k0 and i1. For all 0rf2k+21 with αif2k+4+2+r0,

    (i) (Right edge of Uk,i) Hαif2k+3+1+r,f2k+31r=(1)rk(1)r(r1)2Hαif2k+3+1,f2k+31,

    (ii) (Left edge of Uk,i) Hαif2k+4+2+r,f2k+31r=(1)rk(1)r(r1)2Hαif2k+3+1,f2k+31,

    (iii) (Upper edge of Uk,i) Hαif2k+4+2+r,f2k+31=(1)rHαif2k+3+1,f2k+31.

    Proof. Write y=αif2k+3. Recall that αiEk+1. So Φk+1(y)=f2k+12 and yFk.

    (ⅰ) For 0r<f2k+2, let Ay+r+j be the j-th column of My+1+r,f2k+31r. Applying Lemma 2.8 and Theorem 2.5(ⅰ), we see sy+r+=sy+r++f2k for 1f2k+3r2 and sy+f2k+31sy+f2k+31+f2k. Then Proposition 2.4 and Lemma 2.11 yields sy+f2k+31sy+f2k+31+f2k=(1)k. Therefore,

    Ay+r+1Ay+r+f2k=(sy+r+1sy+r+2sy+f2k+32sy+f2k+31)(sy+r+1+f2ksy+r+2+f2ksy+f2k+32+f2ksy+f2k+31+f2k)=(000(1)k)

    and

    Hy+1+r,f2k+31r=|Ay+r+1Ay+r+2Ay+f2k+31|=|(Ay+r+1Ay+r+f2k)Ay+r+2Ay+f2k+31|=|0f2k+32r,1My+r+2,f2k+3r2(1)k|=(1)k(1)1+f2k+31rHy+r+2,f2k+3r2=(1)k+rHy+r+2,f2k+3r2 (4.1)

    where in the last equality we apply Lemma 2.10 and 0i,j denotes the i×j zero matrix. It follows from Eq (4.1) that

    Hy+1,f2k+31=(1)kHy+2,f2k+32=(1)k(1)k+1Hy+3,f2k+33=(1)k(1)k+1(1)k+r1Hy+1+r,f2k+31r=(1)rk(1)r(r1)2Hy+1+r,f2k+31r.

    (ⅱ) Let Byf2k+2+1+r+j be the j-th row of Myf2k+2+2+r,f2k+31r. Combining Lemma 2.8, Theorem 2.5(ⅰ), Proposition 2.4 and Lemma 2.11, a similar argument as above yields

    Hyf2k+2+2+r,f2k+31r=|Byf2k+2+1+r+1ByBy+f2k|=|Byf2k+2+1+r+1ByBy+f2kBy|=|Myf2k+2+2+(r+1),f2k+31(r+1)(1)k01,f2k+32r|=(1)k+rHyf2k+2+2+(r+1),f2k+31(r+1). (4.2)

    Applying Eq (4.2), we have

    Hyf2k+2+2+r,f2k+31r=(1)k+r(1)k+r+1(1)k+f2k+22Hy+1,f2k=(1)(2k+r+f2k+22)(f2k+2r1)2Hy+1,f2k=(1)(2k+r+f2k+22)(f2k+2r1)2Hy+f2k+2,f2k(by Lemma 4.2(ii))=(1)rk(1)r(r1)2Hy+1,f2k+31.(by Lemma 4.6(i))

    (ⅲ) Let Cj be the j-th column of Myf2k+2+2+r,f2k+31. Then

    Hyf2k+2+2+r,f2k+31=det(C1,C2,,Cf2k+31)=det(C1,C2,,Cf2k+31r,C1,C2,,Cr)

    where Cp=Cf2k+31r+pCf2k+31r+pf2k for 1pr. According to Lemma 2.8, Theorem 2.5(ⅰ), we have sy+=sy++f2k for all 1f2k+32 and f2k+3+1f2k+3+r2. By Proposition 2.4 and Lemma 2.11, we obtain that sy+f2k+31+f2ksy+f2k+31=(1)k+1 and sy+f2k+3+f2ksy+f2k+3=(1)k. Thus

    (C1,C2,,Cr)=(0f2k+31r,rX)

    where X is the r×r matrix

    (00(1)k+1......(1)k......(1)k+1(1)k0).

    Expanding by the last r columns, we have

    Hyf2k+2+2+r,f2k+31=det(Myf2k+2+2+r,f2k+31r0f2k+31r,rX)=(1)(k+1)r(1)(r1)r2Hyf2k+2+2+r,f2k+31r=(1)(k+1)r(1)(r1)r2(1)rk(1)r(r1)2Hy+1,f2k+31(by Lemma 4.6(ii))=(1)rHy+1,f2k+31.

    Remark 4.7. From Proposition 4.5(ⅰ) and Lemma 4.6, Hankel determinants on the boundary of Uk,i can be determined by Hα1f2k+3+1,f2k+31=Hf2k+12+1,f2k+31 (the upper right corner of Uk,1).

    Lemma 4.8. Let k0 and i1. For all 0rf2k+11,

    (i) (Left edge of Vk,i) Hβif2k+1+2+r,f2k+21r=(1)rk(1)r(r+1)2Hβif2k+1+2,f2k+21,

    (ii) (Right edge of Vk,i) Hβi+1+r,f2k+21r=(1)rk(1)r(r+1)2Hβi+1,f2k+21,

    (iii) (Upper edge of Vk,i) Hβif2k+1+2+r,f2k+21=Hβi+1,f2k+21.

    Proof. (ⅰ) Denote by Aβif2k+1+1+r+j the j-th row of Mβif2k+1+2+r,f2k+21r. Then

    Hβif2k+1+2+r,f2k+21r=det(Aβif2k+1+2+rAβif2k+1+2+r+1Aβi+f2k)=det(Aβif2k+1+2+rAβif2k+1+2+r+1Aβi+f2kAβi).

    From Lemma 2.8, Theorem 2.5 and Lemma 2.11, we have

    Aβi+f2kAβi=((1)k,0,,0).

    For 0rf2k+11,

    Hβif2k+1+2+r,f2k+21r=(Mβif2k+1+2+(r+1),f2k+21(r+1)(1)k01,f2k+22r)=(1)k(1)1+f2k+21rHβif2k+1+2+(r+1),f2k+21(r+1)=(1)k+1+rHβif2k+1+2+(r+1),f2k+21(r+1).(by Lemma 2.10(i))

    Thus

    Hβif2k+1+2+r,f2k+21r=(1)k+1+r1(1)k+1+r2(1)k+1Hβi+2f2k+1,f2k+21=(1)r(k+1)(1)r(r1)2Hβi+2f2k+1,f2k+21.

    (ⅱ) Let Bβi+r+j be the j-th column of Mβi+1+r,f2k+21r.

    Hβi+1+r,f2k+21r=det(sβi+1+rsβi+2+rsβi+f2k+21...sβi+f2k+21sβi+f2k+2sβi+2f2k+23r)=det(Bβi+1+rBβi+2+rBβi+f2k+21)=det(Bβi+1+rBβi+1+r+f2kBβi+2+rBβi+f2k+21).

    Recall that βiFk. By Lemma 2.8, βi and βi+f2k+2 are adjacent elements in Fk. It follows from Theorem 2.5(ⅰ) and Lemma 2.11 that

    Hβi+1+r,f2k+21r=det(0f2k+22r,1Mβi+1+(r+1),f2k+21(r+1)(1)k)=(1)k(1)1+f2k+21rHβi+1+(r+1),f2k+21(r+1)=(1)k+1+rHβi+1+(r+1),f2k+21(r+1).(by Lemma 2.10(i))

    Hence Hβi+1+r,f2k+21r=(1)r(k+1)(1)r(r1)2Hβi+1,f2k+21.

    (ⅲ) Let Cβif2k+1+1+r+j be the j-th column of Mβif2k+1+2+r,f2k+21. Then

    Hβif2k+1+2+r,f2k+21=det(Cβif2k+1+2+rCβif2k+1+2+r+1Cβi+f2k+r)=det(Cβif2k+1+2+rCβi+f2kC1Cr)

    where Cp=Cβi+f2k+pCβi+p for 1pr. By Lemma 2.8 and Theorem 2.5(ⅰ), we have sβi+=sβi++f2k for 1f2k+22 and f2k+2+1r+f2k+21. Moreover, by Proposition 2.4 and Lemma 2.11, we have sβi+f2k+f2k+21sβi+f2k+21=(1)k+1 and sβi+f2k+f2k+2sβi+f2k+2=(1)k. Thus

    (C1Cr)=(0f2k+21r,rX)

    where X is the r×r matrix

    (00(1)k+1......(1)k......(1)k+1(1)k0).

    Now expanding Hβif2k+1+2+r,f2k+21 by its last r columns, we obtain that for 0rf2k+11,

    Hβif2k+1+2+r,f2k+21=det(Mβif2k+1+2+r,f2k+21r0f2k+21r,rX)=(1)(k+1)r(1)(r1)r2Hβif2k+1+2+r,f2k+21r=Hβif2k+1+2,f2k+21.(by Lemma 4.8(i))

    Remark 4.9. From Proposition 4.5(ⅰ) and Lemma 4.8, Hankel determinants on the boundary of V_{k, i} can be determined by H_{\frac{f_{2k+1}}{2}+f_{2k+3}+1, f_{2k}} (the lower left corner of V_{k, 1} ).

    Lemma 4.10. Let k\ge 0 and i\ge 1 .

    (i) \textrm{(Left edge of $ T_{k, i} $)} For all 0\le r\le f_{2k}-1 with \gamma_i-f_{2k+3}+2+r\ge 0 ,

    H_{\gamma_i-f_{2k+3}+2+r, \, f_{2k+2}-1-r} = (-1)^{rk}(-1)^{\frac{r(r-1)}{2}}H_{\gamma_i-f_{2k+3}+2, \, f_{2k+2}-1}.

    (ii) \textrm{(Right edge of $ T_{k, i} $)} For all 0\le r\le f_{2k}-1 with \gamma_i-f_{2k+2}+1+r\ge 0 ,

    H_{\gamma_i-f_{2k+2}+1+r, \, f_{2k+2}-1-r} = (-1)^{rk}(-1)^{\frac{r(r-1)}{2}}H_{\gamma_i-f_{2k+2}+1, \, f_{2k+2}-1}.

    (iii) \textrm{(Upper edge of $ T_{k, i} $)} For all 0\le r\le f_{2k}-1 with \gamma_i-f_{2k+3}+2+r\ge 0 ,

    H_{\gamma_i-f_{2k+3}+2+r, \, f_{2k+2}-1} = H_{\gamma_i-f_{2k+3}+2, \, f_{2k+2}-1}.

    Proof. To shorten the notation, write x = \gamma_i-f_{2k+3}+2 and x' = \gamma_i-f_{2k+2}+1 .

    (ⅰ) Let \max\{0, -x\}\leq \ell \leq f_{2k}-1 and let A_{j} be the j th row of H_{x+\ell, f_{2k+2}-1-\ell} . By Theorem 2.5(ⅱ) and Lemma 2.11, we see

    A_{f_{2k+2}-1-\ell}-A_{f_{2k}-1-\ell} = \begin{pmatrix} (-1)^{k+1} & 0 & \dots & 0 \end{pmatrix}.

    Then for \max\{0, -x\}\leq \ell\leq f_{2k}-1 ,

    \begin{align*} H_{x+\ell, f_{2k+2}-1-\ell} & = \det\begin{pmatrix} A_1\\ A_2\\ \vdots\\ A_{f_{2k+2}-2-\ell}\\ A_{f_{2k+2}-1-\ell} \end{pmatrix} = \det\begin{pmatrix} A_1\\ A_2\\ \vdots\\ A_{f_{2k+2}-2-\ell}\\ A_{f_{2k+2}-1-\ell}-A_{f_{2k}-1-\ell} \end{pmatrix}\\ & = \begin{vmatrix} * & M_{x+(\ell+1), f_{2k+2}-1-(\ell+1)}\\ (-1)^{k+1}& {\bf{0}}_{1, f_{2k+2}-2-\ell} \end{vmatrix}\\ & = (-1)^{k+1}(-1)^{1+f_{2k+2}-1-\ell}H_{x+(\ell+1), f_{2k+2}-1-(\ell+1)}\\ & = (-1)^{k+\ell}H_{x+(\ell+1), f_{2k+2}-1-(\ell+1)}. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 2.10(i))} \end{align*}

    Applying the above equality r times, one has

    \begin{align*} H_{x+r, f_{2k+2}-1-r}& = (-1)^{k+r-1}(-1)^{k+r-2}\cdots(-1)^{k}H_{x, f_{2k+2}-1}\\ & = (-1)^{rk}(-1)^{\frac{r(r-1)}{2}}H_{x, f_{2k+2}-1}. \end{align*}

    (ⅱ) Let \max\{0, -x'\}\leq \ell \leq f_{2k}-1 and let B_j be the j th column of H_{x'+\ell, f_{2k+2}-1-\ell} . By Theorem 2.5(ⅱ) and Lemma 2.11, we see

    B_1-B_{1+f_{2k+1}} = \begin{pmatrix} 0\\ \vdots\\ 0 \\ (-1)^{k+1} \end{pmatrix}.

    Therefore, for \max\{0, -x'\}\leq \ell \leq f_{2k}-1 ,

    \begin{align*} H_{x'+\ell, f_{2k+2}-1-\ell} & = \det\begin{pmatrix} B_1 & B_2 & \cdots & B_{f_{2k+2}-1-\ell} \end{pmatrix}\\ & = \det\begin{pmatrix} B_1-B_{1+f_{2k+1}} & B_2 & \cdots & B_{f_{2k+2}-1-\ell} \end{pmatrix}\\ & = \det \begin{pmatrix} {\bf{0}}_{ f_{2k+2}-2-\ell, 1} & M_{x'+(\ell+1), f_{2k+2}-1-(\ell+1)}\\ (-1)^{k+1}& *\\ \end{pmatrix}\\ & = (-1)^{k+1}(-1)^{1+f_{2k+2}-1-\ell}H_{x'+(\ell+1), f_{2k+2}-1-(\ell+1)}\\ & = (-1)^{k+\ell}H_{x'+(\ell+1), f_{2k+2}-1-(\ell+1)}. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\text{(by Lemma 2.10(i))} \end{align*}

    Applying the above equality r times, one has

    \begin{align*} H_{x'+r, f_{2k+2}-1-r}& = (-1)^{k+r-1}(-1)^{k+r-2}\cdots(-1)^{k}H_{x', f_{2k+2}-1}\\ & = (-1)^{rk}(-1)^{\frac{r(r-1)}{2}}H_{x', f_{2k+2}-1}. \end{align*}

    (ⅲ) Let \max\{0, -x\}\leq r \leq f_{2k}-1 and let C_j be the j th column of M_{x+r, f_{2k+2}-1} . Then

    \begin{align*} H_{x+r, f_{2k+2}-1} & = \det\begin{pmatrix} C_1 & C_2 & \cdots & C_{f_{2k+2}-1} \end{pmatrix}\\ & = \det\begin{pmatrix} C_1 & C_2 & \cdots & C_{f_{2k+2}-r-1} & C'_1 & \cdots & C'_r \end{pmatrix} \end{align*}

    where C'_p = C_{f_{2k+2}-r-1+p}-C_{f_{2k}-r-1+p} for 1\le p \le r . Note that

    \begin{pmatrix} C'_1 & \cdots & C'_r \end{pmatrix} = \begin{pmatrix} s_{\gamma_i-f_{2k}+1} & \cdots & s_{\gamma_i-f_{2k}+r}\\ \vdots & \ddots & \vdots\\ s_{\gamma_i+f_{2k+1}-1} & \cdots & s_{\gamma_i+f_{2k+1}+r-2} \end{pmatrix}- \begin{pmatrix} s_{\gamma_i-f_{2k+2}+1} & \cdots & s_{\gamma_i-f_{2k+2}+r}\\ \vdots & \ddots & \vdots\\ s_{\gamma_i-1} & \cdots & s_{\gamma_i+r-2} \end{pmatrix}.

    By Lemma 2.6 and Theorem 2.5(ⅱ), for 1\leq q\leq f_{2k+2}-2 and f_{2k+2}+1\leq q \leq f_{2k+2}+r-2 ,

    s_{\gamma_i-f_{2k}+q} = s_{\gamma_i-f_{2k+2}+q}.

    Moreover, by Lemma 2.11, s_{\gamma_i+f_{2k+1}-1}-s_{\gamma_i-1} = (-1)^{k} and s_{\gamma_i+f_{2k+1}}-s_{\gamma_i} = (-1)^{k+1} . Then

    \begin{pmatrix} C'_1 & \cdots & C'_r \end{pmatrix} = \begin{pmatrix} {\bf{0}}_{f_{2k+2}-1-r, \, r}\\ X \end{pmatrix}

    where X is the r\times r matrix

    \begin{pmatrix} 0 & \cdots & 0 & (-1)^{k}\\ \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & (-1)^{k+1}\\ \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ (-1)^{k} & (-1)^{k+1} & \cdots & 0 \end{pmatrix}.

    Therefore,

    \begin{align*} H_{x+r, f_{2k+2}-1} & = \det \begin{pmatrix} M_{x+r, f_{2k+2}-1-r}& {\bf{0}}_{f_{2k+2}-1-r, \, r}\\ * & X \end{pmatrix}\\ & = (-1)^{kr}(-1)^{\frac{(r-1)r}{2}}H_{x+r, f_{2k+2}-1-r}\\ & = H_{x, f_{2k+2}-1}. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 4.10(i))} \end{align*}

    Remark 4.11. From Proposition 4.5(ⅱ) and Lemma 4.10, Hankel determinants on the boundary of T_{k, i} can be determined by H_{\frac{f_{2k+3}}{2}+f_{2k}+1, f_{2k+1}} (the lower left corner of T_{k, 2} ).

    In section 4, we show that for any k\ge 0 , to know all the determinants on the boundary of U_{k, i} (resp. V_{k, i} , T_{k, i} ) for all i , it is enough to know the value of one determinant on the boundary U_{k, i} (resp. V_{k, i} or T_{k, i} ) for some i . In this section, for certain i , we shall give the expression of a determinant on the boundary U_{k, i} (resp. V_{k, i} or T_{k, i} ) for all k .

    The next result allows us to determine the determinant on the lower left corner of U_{k, i} by using the determinants on the boundary of U_{k-1, *} and T_{k-1, *} .

    Lemma 5.1. \textrm{(Lower left corner of $ U_{k, i} $)} For all k\ge 1 and i\ge 1 ,

    H_{\alpha_i-f_{2k+3}+1, \, f_{2k}} = (-1)^{k}(H_{\alpha_i-f_{2k+3}+2, \, f_{2k}-1}-H_{\alpha_i-f_{2k+3}+1, \, f_{2k}-1}).

    Proof. Let y = \alpha_i-f_{2k+3} and let A_j be the j th column of H_{y+1, f_{2k}} . Then

    \begin{align*} H_{y+1, f_{2k}} & = \begin{vmatrix} s_{y+1} & s_{y+2} & \cdots & s_{y+f_{2k}}\\ \vdots & \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ s_{y+f_{2k}} & s_{y+f_{2k}+1} & \cdots & s_{y+2f_{2k}-1} \end{vmatrix} = \det \begin{pmatrix} A_1 & A_2 & \cdots & A_{f_{2k}-1} \end{pmatrix}. \end{align*}

    Recall that \alpha_i\in E'_{k+1} . Then \Phi_{k+1}(y) = \frac{f_{2k+1}}{2} and \Phi_{k-1}(y+f_{2k}) = \frac{f_{2k-1}}{2} . This implies y\in F'_k and y+f_{2k}\in F_{k-1} . By Lemma 2.8 and Theorem 2.5(ⅰ), the fact y+f_{2k}\in F_{k-1} yields that s_{y+\ell} = s_{y+f_{2k-2}+\ell} for 1\leq \ell \leq f_{2k}-2 . By Lemma 2.11, s_{y+f_{2k}-1}-s_{y+f_{2k}+f_{2k-2}-1} = (-1)^{k+1} and s_{y+f_{2k}}-s_{y+f_{2k}+f_{2k-2}} = (-1)^{k} . So

    \begin{align} H_{y+1, f_{2k}} & = \det \begin{pmatrix} A_1-A_{1+f_{2k-2}} & A_2 & \cdots & A_{f_{2k}-1} \end{pmatrix}\\ & = \begin{vmatrix} 0 & s_{y+2} & \cdots & s_{y+f_{2k}}\\ \vdots & \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ 0 & s_{y+f_{2k}-1} & \cdots & s_{y+2f_{2k}-3}\\ (-1)^{k+1} & s_{y+f_{2k}} & \cdots & s_{y+2f_{2k}-2}\\ (-1)^{k} & s_{y+f_{2k}+1} & \cdots & s_{y+2f_{2k}-1} \end{vmatrix}\\ & = (-1)^{k}(-1)^{1+f_{2k}} H_{y+2, f_{2k}-1} + (-1)^{k+1}(-1)^{f_{2k}}X \end{align} (5.1)

    where

    X = \begin{vmatrix} s_{y+2} & \cdots & s_{y+f_{2k}}\\ \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ s_{y+f_{2k}-1} & \cdots & s_{y+2f_{2k}-3}\\ s_{y+f_{2k}+1} & \cdots & s_{y+2f_{2k}-1} \end{vmatrix} = (-1)^{f_{2k}-2}\begin{vmatrix} s_{y+f_{2k}+1} & \cdots & s_{y+2f_{2k}-1}\\ s_{y+2} & \cdots & s_{y+f_{2k}}\\ \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ s_{y+f_{2k}-1} & \cdots & s_{y+2f_{2k}-3} \end{vmatrix}.

    Since y+f_{2k}\in F'_{k} , by Lemma 2.8 and Theorem 2.5(ⅰ), we see

    \begin{pmatrix} s_{y+f_{2k}+1} & \cdots & s_{y+2f_{2k}-1} \end{pmatrix} = \begin{pmatrix} s_{y+1} & \cdots & s_{y+f_{2k}-1} \end{pmatrix}

    and X = (-1)^{f_{2k}-2}H_{y+1, f_{2k}-1} . Then the result follows from Eq (5.1) and Lemma 2.10.

    Now we show how to obtain the determinant on the lower left corner of T_{k, i} by using determinants on the boundary of U_{k, i-1} and U_{k-1, i+1} .

    Lemma 5.2. \textrm{(Lower left corner of $ T_{k, i}) $} For all k\ge 1 and i\ge 2 ,

    H_{\gamma_i-f_{2k+2}+1, \, f_{2k+1}} = (-1)^{k}(H_{\gamma_i-f_{2k+2}+2, \, f_{2k+1}-1}+H_{\gamma_i-f_{2k+2}+1, \, f_{2k+1}-1}).

    Proof. Let y = \gamma_i-f_{2k+2} and let A_j be the j th column of H_{y+1, f_{2k+1}} . Then

    \begin{align*} H_{y+1, f_{2k+1}} & = \begin{vmatrix} s_{y+1} & s_{y+2} & \cdots & s_{y+f_{2k+1}}\\ \vdots & \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ s_{y+f_{2k+1}} & s_{y+f_{2k+1}+1} & \cdots & s_{y+2f_{2k+1}-1} \end{vmatrix} = \det \begin{pmatrix} A_1 & A_2 & \cdots & A_{f_{2k+1}-1} \end{pmatrix}. \end{align*}

    Recall that \gamma_i\in E'_{k} . By Lemma 2.6, y\in E_k and \Phi_{k}(y+f_{2k+1}) = \frac{f_{2k+1}}{2} . This implies y+f_{2k+1}\in F_{k} . By Lemma 2.8 and Theorem 2.5(ⅰ), the fact y+f_{2k+1}\in F_{k} yields that s_{y+\ell} = s_{y+f_{2k}+\ell} for 1\leq \ell \leq f_{2k+1}-2 . By Lemma 2.11, s_{y+f_{2k+1}-1}-s_{y+f_{2k+1}+f_{2k}-1} = (-1)^{k} and s_{y+f_{2k+1}}-s_{y+f_{2k+1}+f_{2k}} = (-1)^{k+1} . So

    \begin{align} H_{y+1, f_{2k+1}} & = \det \begin{pmatrix} A_1-A_{1+f_{2k}} & A_2 & \cdots & A_{f_{2k+1}-1} \end{pmatrix}\\ & = \begin{vmatrix} 0 & s_{y+2} & \cdots & s_{y+f_{2k+1}}\\ \vdots & \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ 0 & s_{y+f_{2k+1}-1} & \cdots & s_{y+2f_{2k+1}-3}\\ (-1)^{k} & s_{y+f_{2k+1}} & \cdots & s_{y+2f_{2k+1}-2}\\ (-1)^{k+1} & s_{y+f_{2k+1}+1} & \cdots & s_{y+2f_{2k+1}-1} \end{vmatrix}\\ & = (-1)^{k+1}(-1)^{1+f_{2k+1}} H_{y+2, f_{2k+1}-1} + (-1)^{k}(-1)^{f_{2k+1}}X \end{align} (5.2)

    where

    X = \begin{vmatrix} s_{y+2} & \cdots & s_{y+f_{2k+1}}\\ \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ s_{y+f_{2k+1}-1} & \cdots & s_{y+2f_{2k+1}-3}\\ s_{y+f_{2k+1}+1} & \cdots & s_{y+2f_{2k+1}-1} \end{vmatrix} = (-1)^{f_{2k+1}-2}\begin{vmatrix} s_{y+f_{2k+1}+1} & \cdots & s_{y+2f_{2k+1}-1}\\ s_{y+2} & \cdots & s_{y+f_{2k+1}}\\ \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots\\ s_{y+f_{2k+1}-1} & \cdots & s_{y+2f_{2k+1}-3} \end{vmatrix}.

    Since y\in E_{k} , by Lemma 2.6 and Theorem 2.5(ⅱ), we see

    \begin{pmatrix} s_{y+f_{2k+1}+1} & \cdots & s_{y+2f_{2k+1}-1} \end{pmatrix} = \begin{pmatrix} s_{y+1} & \cdots & s_{y+f_{2k+1}-1} \end{pmatrix}

    and X = (-1)^{f_{2k+1}-2}H_{y+1, f_{2k+1}-1} . Then the result follows from Eq (5.2) and Lemma 2.10.

    Now We are able to give the exact value of the Hanker determinant on the upper right corner of U_{k, i} , and hence we know all the determinants on the boundary of U_{k, i} .

    Theorem 5.3. \textrm{(Upper right corner of $ U_{k, 1} $)} Let k\ge 1 . Then

    H_{\frac{f_{2k+1}}{2}+1, \, f_{2k+3}-1} = (-1)^{k+1}\frac{f_{2k+1}}{2}.

    Proof. We can check directly that the result holds for k = 1, 2 . Now suppose k\ge 3 . Let h_{k} = H_{\frac{f_{2k+1}}{2}+1, \, f_{2k+3}-1} . Then

    \begin{align*} h_{k} & = (-1)^{(f_{2k+2}-1)k}(-1)^{\frac{(f_{2k+2}-1)(f_{2k+2}-2)}{2}} H_{\frac{f_{2k+5}}{2}-f_{2k}, f_{2k}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 4.6(i))}\\ & = (-1)^{\frac{f_{2k+2}-1}{2}} H_{\frac{f_{2k+1}}{2}+1, f_{2k}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 2.10 and Lemma 4.2(ii))}\\ & = (-1)^{\frac{f_{2k+2}-1}{2}}(-1)^{k}(H_{\frac{f_{2k+1}}{2}+2, \, f_{2k}-1}-H_{\frac{f_{2k+1}}{2}+1, \, f_{2k}-1})\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 5.1)}. \end{align*}

    Applying Lemma 4.6(ⅰ) for k-1 * and r = f_{2k-2} ,

    *We need to mention that \alpha_i 's depend also on k . For example, \alpha_1^{(k-1)} = \frac{f_{2k+3}}{2} and \alpha_1^{(k)} = \frac{f_{2k+5}}{2} .

    \begin{align*} H_{\frac{f_{2k+1}}{2}+1, \, f_{2k}-1} & = (-1)^{f_{2k-2}(k-1)}(-1)^{\frac{f_{2k-2}(f_{2k-2}-1)}{2}}H_{\frac{f_{2k-1}}{2}+1, \, f_{2k+1}-1}\\ & = (-1)^{k-1}(-1)^\frac{f_{2k-2}-1}{2}h_{k-1}.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 2.10)} \end{align*}

    Applying Lemma 4.10(ⅰ) for k-1 and r = f_{2k-2} ,

    \begin{align*} H_{\frac{f_{2k+1}}{2}+2, \, f_{2k}-1} & = (-1)^{(f_{2k-2}-1)(k-1)}(-1)^{\frac{(f_{2k-2}-1)(f_{2k-2}-2)}{2}}H_{\frac{f_{2k+1}}{2}+f_{2k-2}+1, \, f_{2k-1}}\\ & = (-1)^{\frac{f_{2k-2}-1}{2}}H_{\frac{f_{2k+1}}{2}+f_{2k-2}+1, \, f_{2k-1}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\text{(by Lemma 2.10)}\\ & = (-1)^{\frac{f_{2k-2}-1}{2}}(-1)^{k-1}\left(H_{\frac{f_{2k+1}}{2}+f_{2k-2}+2, \, f_{2k-1}-1}+H_{\frac{f_{2k+1}}{2}+f_{2k-2}+1, \, f_{2k-1}-1}\right)\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 5.2)}\\ & = (-1)^{\frac{f_{2k-2}-1}{2}}(-1)^{k-1}\left(h_{k-2}+H_{\frac{f_{2k+1}}{2}+f_{2k-2}+1, \, f_{2k-1}-1}\right) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 4.6(iii))}\\ & = (-1)^{\frac{f_{2k-2}-1}{2}}(-1)^{k-1}\left(h_{k-2}-h_{k-1}\right). \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 4.6(i) and Lemma 2.10)} \end{align*}

    Combing previous equations, we have

    \begin{align*} h_{k} & = (-1)^{\frac{f_{2k+2}-1}{2}}(-1)^{k}(H_{\alpha_1-f_{2k+3}+2, \, f_{2k}-1}-H_{\alpha_1-f_{2k+3}+1, \, f_{2k}-1})\\ & = (-1)^{\frac{f_{2k+2}-1}{2}}(-1)^{k}(-1)^{\frac{f_{2k-2}-1}{2}}(-1)^{k-1}(h_{k-2}-2h_{k-1})\\ & = h_{k-2}-2h_{k-1}. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 2.10)} \end{align*}

    The initial values are h_1 = 2 , h_2 = -5 . The result follows from the recurrence relation of h_k and its initial values.

    Corollary 5.4. \textrm{(Lower left corner of $ V_{k, 1} $)} For all k\ge 1 ,

    H_{\frac{f_{2k+1}}{2}+f_{2k+3}+1, \, f_{2k}} = (-1)^{k+\frac{f_{2k+2}+1}{2}}\frac{f_{2k+1}}{2}.

    Proof. By Proposition 4.5,

    \begin{align*} H_{\frac{f_{2k+1}}{2}+f_{2k+3}+1, \, f_{2k}} & = H_{\frac{f_{2k+5}}{2}-f_{2k}, \, f_{2k}}\\ & = (-1)^{(f_{2k+2}-1)k}(-1)^{\frac{(f_{2k+2}-1)(f_{2k+2}-2)}{2}}H_{\frac{f_{2k+1}}{2}+1, f_{2k+3}-1}\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 4.6(i))}\\ & = (-1)^{\frac{f_{2k+2}-1}{2}}H_{\frac{f_{2k+1}}{2}+1, f_{2k+3}-1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{(by Lemma 2.10)}\\ & = (-1)^{k+\frac{f_{2k+2}+1}{2}}\frac{f_{2k+1}}{2}. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Theorem 5.3)} \end{align*}

    Corollary 5.5. \textrm{(Lower left corner of $ T_{k, 2} $)} For all k\ge 1 , H_{\frac{f_{2k+3}}{2}+f_{2k}+1, f_{2k+1}} = f_{2k} .

    Proof. From Lemma 5.2, we have

    \begin{align} H_{\frac{f_{2k+3}}{2}+f_{2k}+1, f_{2k+1}}& = (-1)^k(H_{\frac{f_{2k+3}}{2}+f_{2k}+2, f_{2k+1}-1}+H_{\frac{f_{2k+3}}{2}+f_{2k}+1, f_{2k+1}-1}). \end{align} (5.3)

    Note that H_{\frac{f_{2k+3}}{2}+f_{2k}+2, f_{2k+1}-1} is on the upper left corner of U_{k-1, 2} . By Lemma 4.6(ⅲ),

    H_{\frac{f_{2k+3}}{2}+f_{2k}+2, f_{2k+1}-1} = H_{\frac{f_{2k-1}}{2}+f_{2k+3}+1, f_{2k+1}-1}.

    According to Proposition 4.5 and Lemma 4.6, the determinants on the upper left corner of U_{k-1, 1} and U_{k-1, 1} are equal. Namely, H_{\frac{f_{2k-1}}{2}+f_{2k+3}+1, f_{2k+1}-1} = H_{\frac{f_{2k-1}}{2}+1, f_{2k+1}-1} . Therefore,

    \begin{equation} H_{\frac{f_{2k+3}}{2}+f_{2k}+2, f_{2k+1}-1} = H_{\frac{f_{2k-1}}{2}+1, f_{2k+1}-1}. \end{equation} (5.4)

    It follows from Lemma 4.6(ⅰ) and Lemma 2.10 that

    \begin{align} H_{\frac{f_{2k+3}}{2}+f_{2k}+1, , \, f_{2k+1}-1} & = (-1)^{2f_{2k}k}(-1)^{\frac{2f_{2k}(2f_{2k}-1)}{2}}H_{\frac{f_{2k+1}}{2}+1, \, f_{2k+3}-1}\\ & = - H_{\frac{f_{2k+1}}{2}+1, \, f_{2k+3}-1}. \end{align} (5.5)

    Using Eq (5.3), Eq (5.4) and Eq (5.5), we have

    \begin{align*} H_{\frac{f_{2k+3}}{2}+f_{2k}+1, f_{2k+1}} & = (-1)^{k}\left(H_{\frac{f_{2k-1}}{2}+1, f_{2k+1}-1}-H_{\frac{f_{2k+1}}{2}+1, \, f_{2k+3}-1}\right)\\ & = (-1)^k\left((-1)^k\frac{f_{2k-1}}{2}-(-1)^{k+1}\frac{f_{2k+1}}{2}\right)\;\;\;\;\;\;\;\;\;\;\text{(by Theorem 5.3)}\\ & = f_{2k}. \end{align*}

    Proof of Theorem 1.1. Suppose (m, n)\in U_{k, i} for some i . Then \alpha_{i}-f_{2k+2} < n+m\leq \alpha_i and m = \alpha_i-f_{2k+2}+1-n+r where 0\leq r < f_{2k+2} .

    Case 1: n = f_{2k+3}-1 . Applying Lemma 4.6(ⅰ), Proposition 4.5 and then Lemma 4.6(ⅰ) again, we have

    \begin{equation} H_{\alpha_i-f_{2k+3}+1, f_{2k+3}-1} = H_{\alpha_1-f_{2k+3}+1, f_{2k+3}-1} = (-1)^{k+1}\frac{f_{2k+1}}{2}. \end{equation} (6.1)

    where the last equality follows from Theorem 5.3. Since r = \alpha_i-m-n = \frac{f_{2k+5}}{2}-\Phi_{k+1}(m+n) , by Lemma 4.6(ⅲ),

    \begin{align*} H_{m, n}& = H_{\alpha_i-f_{2k+4}+2+r, f_{2k+3}-1}\\ & = (-1)^r H_{\alpha_i-f_{2k+3}+1, f_{2k+3}-1}\\ & = (-1)^{\frac{f_{2k+5}}{2}-\Phi_{k+1}(m+n)}H_{\alpha_i-f_{2k+3}+1, f_{2k+3}-1}\\ & = (-1)^{k+1}(-1)^{\frac{f_{2k+5}}{2}-\Phi_{k+1}(m+n)}\frac{f_{2k+1}}{2} \end{align*}

    where the last equality follows from Eq (6.1).

    Case 2: n = f_{2k} . By Proposition 4.5, we have

    \begin{align*} H_{m, n}& = H_{\alpha_1-f_{2k}, \, f_{2k}} \\ & = (-1)^{\ell k}(-1)^{\frac{\ell(\ell-1)}{2}}H_{\alpha_1-f_{2k+3}+1, f_{2k+3}-1} \;\;\;\;\;\;\;\;\;\;\;\;\text{(by Lemma 4.6(iii))}\\ & = (-1)^{k+1}(-1)^{\frac{f_{2k+2}-1}{2}}\cdot \frac{f_{2k+1}}{2} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{(by Theorem 5.3)} \end{align*}

    where \ell = f_{2k+3}-1-f_{2k} is even by Lemma 2.10.

    Case 3: f_{2k} < n < f_{2k+3}-1 . If m+n = \alpha_i (or \alpha_i-f_{2k+2}+1 ), then applying Lemma 4.6(ⅰ) (or Lemma 4.6(ⅱ)) and then Eq (6.1), we have

    \begin{align*} H_{m, n} & = (-1)^{\ell k}(-1)^{\frac{\ell(\ell-1)}{2}}H_{\alpha_i-f_{2k+3}+1, f_{2k+3}-1}\\ & = (-1)^{\ell k}(-1)^{\frac{\ell(\ell-1)}{2}}(-1)^{k+1}\frac{f_{2k+1}}{2} \end{align*}

    where \ell = f_{2k+3}-1-n . If \alpha_i-f_{2k+2}+1 < m+n < \alpha_i , then Lemma 4.1 yields H_{m, n} = 0 .

    Proof of Theorem 1.2. Suppose (m, n)\in V_{k, i} for some i . Then \beta_i < n+m\leq \beta_i+f_{2k+1} .

    Case 1: n = f_{2k+2}-1 . By Lemma 4.8(ⅲ) & (ⅰ), we have

    \begin{align} H_{m, n} & = H_{\beta'_i-f_{2k+1}+2, \, f_{2k+2}-1} \\ & = (-1)^{(f_{2k+1}-1)k}(-1)^{\frac{(f_{2k+1}-1)f_{2k+1}}{2}}H_{\beta'_i+1, \, f_{2k}}. \end{align} (6.2)

    According to Proposition 4.5(ⅰ) and Corollary 5.4,

    \begin{equation} H_{\beta'_i+1, \, f_{2k}} = H_{\beta'_1+1, \, f_{2k}} = (-1)^{k+\frac{f_{2k+2}+1}{2}}\frac{f_{2k+1}}{2}. \end{equation} (6.3)

    The result follows from Eq (6.2) and Eq (6.3).

    Case 2: n = f_{2k} . By Proposition 4.5, H_{m, n} = H_{\alpha_1-f_{2k}, \, f_{2k}} . Then the result follows from Theorem 1.1(ⅱ).

    Case 3: f_{2k} < n < f_{2k+2}-1 . If m+n = \beta_i+1 (or \beta_i+f_{2k+1} ), then by Lemma 4.8(ⅰ) (or Lemma 4.8(ⅱ)), we have

    H_{m, n} = -(-1)^{(f_{2k+2}-n)(k+1)}(-1)^{\frac{(f_{2k+2}-1-n)(f_{2k+2}-2-n)}{2}} \frac{f_{2k+1}}{2}.

    If \beta_i+1 < m+n < \beta_i+f_{2k+1} , then Lemma 4.1 shows H_{m, n} = 0 .

    Proof of Theorem 1.3. Suppose (m, n)\in T_{k, i} for some i . Then \gamma_i-f_{2k} < n+m\leq \gamma_i .

    Case 1: n = f_{2k+2}-1 . By Lemma 4.10(ⅲ) & (ⅰ),

    \begin{align} H_{m, n}& = H_{\gamma_i-f_{2k+3}+2, \, f_{2k+2}-1} \\ & = (-1)^{(f_{2k}-1)k}(-1)^{\frac{(f_{2k}-1)(f_{2k}-2)}{2}}H_{\gamma_i-f_{2k+2}+1, \, f_{2k+1}}\\ & = (-1)^{\frac{f_{2k}-1}{2}}H_{\gamma_i-f_{2k+2}+1, \, f_{2k+1}}. \end{align} (6.4)

    where the last equality follows from Lemma 2.10(ⅰ). Using Proposition 4.5(ⅱ) and Corollary 5.5,

    \begin{equation} H_{\gamma_i-f_{2k+2}+1, \, f_{2k+1}} = H_{\gamma_2-f_{2k+2}+1, \, f_{2k+1}} = f_{2k}. \end{equation} (6.5)

    The result follows from Eq (6.4) and Eq (6.5).

    Case 2: n = f_{2k+1} . Write m = \gamma_i-f_{2k+2}+r with 1\leq r\leq f_{2k} . By Proposition 4.5(ⅱ) and Corollary 5.5,

    \begin{align*} H_{m, n} & = (-1)^{r+1}H_{\gamma_{2}-f_{2k+2}+1, \, f_{2k+1}} = (-1)^{r+1}f_{2k}. \end{align*}

    Note that \gamma_i = m+n+f_{2k}-r and 1\leq r\leq f_{2k} . Consequently,

    \frac{f_{2k+3}}{2} = \Phi_k(\gamma_i) = \Phi_{k}(m+n+f_{2k}-r) = \Phi_{k}(m+n)+f_{2k}-r

    which gives r = \Phi_{k}(m+n)-\frac{f_{2k+1}}{2} . Then the result follows.

    Case 3: f_{2k} < n < f_{2k+2}-1 . If m+n = \gamma_i-f_{2k}+1 (or \gamma_i ), then by Lemma 4.10(ⅰ) (or Lemma 4.10(ⅱ)),

    H_{m, n} = (-1)^{(f_{2k+2}-1-n)k}(-1)^{\frac{(f_{2k+2}-1-n)(f_{2k+2}-2-n)}{2}}(-1)^{\frac{f_{2k}-1}{2}}\cdot f_{2k}.

    If \gamma_i-f_{2k}+1 < m+n < \gamma_i , then Lemma 4.1 yields H_{m, n} = 0 .

    In this paper, we study the Hankel determinants H_{m, n} of the Sturmian sequence {\bf{s}} = \tau^{\infty}(1) . In Theorem 1.1, 1.2 and 1.3, we give the closed form of the Hankel determinants H_{m, n} for all m\ge 0 and n\ge 1 . To extend the results to other Sturmian sequences, the difficulty is to locate the parallelograms that are composed by (m, n) 's such that H_{m, n} = 0 . This will need further effort.

    We thank the referees for their valuable comments and suggestions. This work was supported by Guangzhou Science and Technology program (202102020294), Guangdong Basic and Applied Basic Research Foundation (2021A1515010056) and the Fundamental Research Funds for the Central Universities from SCUT (2020ZYGXZR041).

    The authors declare that they have no conflict of interest.



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