The Fibonacci word ε(a, b) on an alphabet {a, b} is the infinite sequence ε(a, b) = b ε

LXXXIX.2 (1999)

Hankel determinants for the Fibonacci word and Pad´ e approximation

by

Teturo Kamae (Osaka), Jun-ichi Tamura (Tokyo) and Zhi-Ying Wen (Beijing)

1. Introduction. The aim of this paper is to give a concrete and inter- esting example of the Pad´e approximation theory as well as to develop the general theory so as to find a quantitative relation between the Hankel de- terminant and the Pad´e pair. Our example is the formal power series related to the Fibonacci word.

The Fibonacci word ε(a, b) on an alphabet {a, b} is the infinite sequence ε(a, b) = b ε

ε b

. . . b ε

. . .

(1)

:= abaababaabaab . . . (b ε

∈ {a, b}), which is the fixed point of the substitution

(2) σ : a → ab, b → a.

The Hankel determinants for an infinite word (or sequence) ϕ = ϕ

ϕ

. . . (ϕ

∈ K) over a field K are

(3) H

(ϕ) := det(ϕ

)

(n = 0, 1, . . . ; m = 1, 2, . . .).

It is known [2] that the Hankel determinants play an important role in the theory of Pad´e approximation for the formal Laurent series

(4) ϕ(z) =

X

ϕ

z

.

Let K((z

)) be the set of formal Laurent series ϕ as above of z with coeffi- cients in K and h ∈ Z providing a nonarchimedean norm kϕk := exp(−k

+h) with k

= inf{k : ϕ

6= 0}. Let ϕ be as above with h = −1. We say that a

1991 Mathematics Subject Classification: 41A21, 11B39.

This research was partially supported by the Hatori Project of the Mathematical Society of Japan.

[123]

pair (P, Q) ∈ K[z]

of polynomials of z over K is a Pad´e pair of order m for ϕ if

(5) kQϕ − P k < exp(−m), Q 6= 0, deg Q ≤ m.

A Pad´e pair (P, Q) of order m for ϕ always exists and the rational func- tion P/Q ∈ K(z) is uniquely determined for each m = 0, 1, . . . The element P/Q ∈ K(z) with P, Q satisfying (5) is called the mth diagonal Pad´e ap- proximation for ϕ. A number m is called a normal index if (5) implies deg Q = m. Note that P/Q is irreducible if m is a normal index, although it can be reducible for a general m. A normal Pad´e pair (P, Q), i.e., deg Q is a normal index, is said to be normalized if the leading coefficient of Q is equal to 1. It is a classical result that m is a normal index for ϕ if and only if the Hankel determinant det(ϕ

)

is nonzero. Note that 0 is always a normal index and the determinant for the empty matrix is considered to be 1, so that the above statement remains valid for m = 0.

We succeed in obtaining a quantitative relation between the Hankel de- terminant and the normalized Pad´e pair. Namely,

(6) det(ϕ

)

= (−1)

Y

z; Q(z)=0

P (z)

for any normal index m with the normalized Pad´e pair (P, Q), where Q

z; Q(z)=0

indicates a product taken over all zeros z of Q with their multi- plicity (Theorem 6).

We are specially interested in the Pad´e approximation theory applied to the Fibonacci words ε := ε(1, 0) and ε := ε(0, 1), where 0, 1 are considered as elements in the field Q, since we have the following remark.

Remark 1. Let M be a matrix of size m×m with entries consisting of two independent variables a and b. Then det M = (a − b)

(pa + (−1)

qb), where p and q are integers defined by

p = det M |

, q = det M |

.

P r o o f. Subtracting the first column vector from all the other column vectors of M , we see that det M is divisible by (a − b)

as a polynomial in Z[a, b]. Hence, det M = (a − b)

(xa + yb) for integers x, y. Setting (a, b) = (1, 0), (0, 1), we get the assertion.

In Section 2, we study the structure of the Fibonacci word, in particular, its repetition property. The notion of singular words introduced in Z.-X. Wen and Z.-Y. Wen [5] plays an important role.

In Section 3, we give the value of the Hankel determinants H

(ε) and

H

(ε) for the Fibonacci words in some closed forms. It is a rare case where

the Hankel determinants are determined completely. Another such case is

for the Thue–Morse sequence ϕ consisting of 0 and 1, where the Hankel

determinants H

(ϕ) modulo 2 are obtained, and the function H

(ϕ) of (m, n) is proved to be 2-dimensionally automatic (see [1]).

In Section 4, we consider the self-similar property of the values H

(ε) and H

(ε) for the Fibonacci words. The quarter plane {(n, m) : n ≥ 0, m ≥ 1} is tiled by 3 kinds of tiles with the values H

(ε) and H

(ε) on it with various scales.

In Section 5, we develop a general theory of Pad´e approximation. We also obtain the admissible continued fraction expansion of ϕ

and ϕ

, the formal Laurent series (4) with h = −1 for the sequences ε and ε, and determine all the convergents p

/q

of the continued fractions. It is known in general that the set of the convergents p

/q

for ϕ is the set of diagonal Pad´e approximations and the set of degrees of q

’s in z coincides with the set of normal indices for ϕ.

2. Structure of the Fibonacci word. In what follows, σ denotes the substitution defined by (2), and

b

ε = b ε

ε b

. . . b ε

. . . (b ε

∈ {a, b})

is the (infinite) Fibonacci word (1). A finite word over {a, b} is sometimes considered to be an element of the free group generated by a and b with inverses a

and b

. For n = 0, 1, . . . , we define the nth Fibonacci word F

and the nth singular word W

as follows:

(7) F

:= σ

(a) = σ

(b), W

:= β

F

α

, where we put

(8) α

= β

=

 a (n even, m odd), b (n odd, m even),

and we define W

to be the empty word and W

:= a for convenience.

Let (f

; n ∈ Z) be the Fibonacci sequence:

(9) f

= f

+ f

(n ∈ Z), f

= f

= 1.

Then |F

| = |W

| = f

(n ≥ 0), where |ξ| denotes the length of a finite word ξ.

For a finite word ξ = ξ

ξ

. . . ξ

and a finite or infinite word η = η

η

. . . over an alphabet, we denote

(10) ξ ≺

η

if ξ = η

η

. . . η

. We simply write

(11) ξ ≺ η

and say that ξ is a subword of η if ξ ≺

η for some k. For a finite word

ξ = ξ

ξ

. . . ξ

and i with 0 ≤ i < n, we denote the ith cyclic permutation

of ξ by C

(ξ) := ξ

ξ

. . . ξ

ξ

ξ

. . . ξ

. We also define C

(ξ) := C

(ξ) with i

:= i − n[i/n] for any i ∈ Z.

In this section, we study the structure of the Fibonacci word b ε and discuss the repetition property. The following two lemmas were obtained by Z.-X.

Wen and Z.-Y. Wen [5] and we omit the proofs.

Lemma 1. We have the following statements:

(1) b ε = F

F

F

F

F

. . . (n ≥ 1),

(2) F

= F

F

= F

F

β

α

β

α

(n ≥ 2), (3) F

F

≺ b ε (n ≥ 3),

(4) b ε = W

W

W

W

W

. . . , (5) W

= W

W

W

(n ≥ 1),

(6) W

is a palindrome, that is, W

stays invariant under reading the letters from the end (n ≥ −2),

(7) C

(F

) ≺ b ε (n ≥ 0, 0 ≤ i < f

),

(8) C

(F

) 6= C

(F

) for any i 6= j, moreover , they are different already before their last places (n ≥ 1, 0 ≤ i < f

),

(9) W

6= C

(F

) (n ≥ 0, 0 ≤ i < f

),

(10) ξ ≺ b ε and |ξ| = f

imply that either ξ = C

(F

) for some i with 0 ≤ i < f

or ξ = W

(n ≥ 0).

Lemma 2. For any k ≥ −1, we have the decomposition of b ε as follows:

b

ε = (W

W

. . . W

)W

γ

W

γ

. . . W

γ

. . . ,

where all the occurrences of W

in b ε are picked up and γ

is either W

or W

corresponding to b ε

is a or b, respectively. That is, any two different occurrences of W

do not overlap and are separated by W

or W

.

We introduce another method to discuss the repetition property of b ε.

Let N be the set of nonnegative integers. For n ∈ N, let

(12) n =

X

τ

(n)f

,

τ

(n) ∈ {0, 1} and τ

(n)τ

(n) = 0 (i ∈ N)

be the regular expression of n in the Fibonacci base due to Zeckendorf. For m, n ∈ N and a positive integer k, we define

(13) m ≡

n

if τ

(m) = τ

(n) for all i < k.

Lemma 3. We have b ε

= a if and only if τ

(n) = 0.

P r o o f. We use induction on n. The lemma holds for n = 0, 1, 2. Assume

that it holds for any n ∈ N with n < f

for some k ≥ 2. Take any n ∈ N

with f

≤ n < f

. Then, since 0 ≤ n − f

< f

, we have

n =

k−1

X

i=0

τ

(n − f

)f

+ f

,

which gives the regular expression if τ

(n − f

) = 0. If τ

(n − f

) = 1, then we have the regular expression n = P

i=0

τ

(n − f

)f

+ f

. In any case, we have τ

(n) = τ

(n − f

). On the other hand, since b ε starts with F

F

by Lemma 1, we have b ε

= b ε

. Hence, b ε

= a if and only if τ

(n) = 0 by the induction hypothesis. Thus, we have the assertion for any n < f

, and by induction, we complete the proof.

Lemma 4. Let n = P

i=0

n

f

with n

∈ {0, 1} (i ∈ N). Assume that n

n

= 0 for 0 ≤ i < k. Then n

= τ

(n) for 0 ≤ i < k.

P r o o f. If there exists i ∈ N such that n

n

= 1, let i

be the maximum such i. Take the maximum j such that n

= n

= n

= . . . = n

= 1. Then, replacing f

+ f

+ f

+ f

+ . . . + f

by f

, we have a new expression of n:

n = X

n

f

:=

i

X

n

f

+ f

+ X

n

f

.

This new expression is unchanged at the indices less than k, and is either regular or has a smaller maximum index i with n

n

= 1. By continuing this procedure, we finally get the regular expression of n, which does not differ from the original expression at the indices less than k. Thus, n

= τ

(n) for any 0 ≤ i < k.

Lemma 5. For any n ∈ N and k ≥ 0, τ

(n + f

) 6= τ

(n) if and only if either n ≡

f

− 2 or n ≡

f

− 1. Moreover ,

b

ε

− b ε

=

 (−1)

(a − b) (n ≡

f

− 2), (−1)

(a − b) (n ≡

f

− 1), where a and b are considered as independent variables.

P r o o f. If k = 0, we can verify the statement by a direct calculation.

Assume that k ≥ 1 and τ

(n) = 0. Then n + f

=

k−1

X

i=0

τ

(n)f

+ f

+ X

τ

(n)f

.

By Lemma 4, we have τ

(n + f

) = τ

(n) if k ≥ 2 or if k = 1 and τ

(n) = 0.

In the case where k = 1, τ

(n) = 1 and τ

(n) = 0, since n + f

= 1 + 2 +

X

τ

(n)f

= f

+ X

τ

(n)f

,

we have τ

(n + f

) = 0 by Lemma 4. On the other hand, in the case where

k = 1, τ

(n) = 1 and τ

(n) = 1, since

n + f

= 1 + 2 + 3 + X

τ

(n)f

= f

+ f

+ X

τ

(n)f

, we have τ

(n + f

) = 1 by Lemma 4.

Thus, in the case where k ≥ 1 and τ

(n) = 0, τ

(n + f

) 6= τ

(n) if and only if k = 1, τ

(n) = 1 and τ

(n) = 0, or equivalently, if and only if n ≡

f

− 2 with k = 1. Note that n ≡

f

− 1 with k = 1 contradicts τ

(n) = 0.

Now assume that k ≥ 1 and τ

(n) = 1. Take the minimum j ≥ 0 such that τ

(n) = τ

(n) = τ

(n) = . . . = τ

(n) = 1. Then since 2f

= f

+ f

for any i ∈ N, we have

n + f

=

j−3

X

i=0

τ

(n)f

+ f

(14)

+ f

+ f

+ f

+ . . . + f

+ X

τ

(n)f

,

where the first term on the right-hand side vanishes if j = 0, 1, 2. Hence by Lemma 4, τ

(n + f

) = τ

(n) if j ≥ 4.

In the case where j = 3, τ

(n + f

) = τ

(n) holds if τ

(n) = 0 by (14) and Lemma 4. If τ

(n) = 1, then by (14) and Lemma 4, τ

(n + f

) = 0.

Thus, for j = 3, τ

(n + f

) 6= τ

(n) if and only if τ

(n) = 1.

If j = 2, then by the assumption on j, we have τ

(n) = 0. On the other hand, since f

= 1, by (14) and Lemma 4, we have τ

(n + f

) = 1. Thus, τ

(n + f

) 6= τ

(n).

If j = 1, then τ

(n) = 0 since τ

(n) = 1 by the assumption on j. On the other hand, since f

= 1, we have τ

(n + f

) = 1 by (14) and Lemma 4.

Thus, τ

(n + f

) 6= τ

(n).

If j = 0, then by the assumption on j, τ

(n) = 1. On the other hand, since f

= 0, we have τ

(n + f

) = 0 by (14) and Lemma 4. Thus, τ

(n + f

) 6=

τ

(n).

By combining all the results as above, we get the first part.

The second part follows from Lemma 3 and the fact that for any k ≥ 0, f

− 1 = f

+ f

+ . . . + f

with i = 0 if k is even and i = 1 if k is odd. Hence, τ

(f

− 1) = τ

(f

− 2) =

 a (k odd, h even), b (k even, h odd).

Lemma 6. For any k ≥ 0, W

≺

ε if and only if n ≡ b

f

− 1.

P r o o f. By Lemma 2, the smallest n ∈ N such that W

≺

b ε is

f

+ f

+ f

+ . . . + f

= f

− 1,

which is the smallest n ∈ N such that n ≡

f

− 1. Let n

:= f

− 1.

Then the regular expression of n

is

n

= f

+ f

+ f

+ . . . + f

,

where d = (1 − (−1)

)/2. The next n with n ≡

n

is clearly n = f

+ f

+ f

+ . . . + f

,

which is, by Lemma 2, the next n such that W

≺

ε since f b

+f

= f

. For i = 1, 2, . . . , let

n

= n

+ X

τ

(i)f

.

Then it is easy to see that n

is the ith n after n

such that n ≡

f

− 1.

We prove by induction on i that n

is the ith n after n

such that W

≺

ε. b Assume that it is so for i. Then by Lemma 4, W

γ

W

≺

ε. Hence, the b next n after n

such that W

≺

ε is n b

+ f

+ |γ

|. Thus, we have

n

+ f

+ |γ

| = n

+ f

+ f

1

+ f

1

= n

+ f

1

+ f

1

= n

, which completes the proof.

Lemma 7. Let k ≥ 0 and n, i ∈ N satisfy n ≡

i.

(1) If 0 ≤ i < f

, then τ

(n + j) = τ

(i + j) for any j = 0, 1, . . . , f

− i − 3.

(2) If f

≤ i < f

, then τ

(n + j) = τ

(i + j) for any j = 0, 1, . . . , f

− i − 3.

P r o o f. (1) We prove the lemma by induction on k. The assertion holds for k = 0. Let k ≥ 1 and assume that the assertion is valid for k − 1. For j = 0, 1, . . . , f

− i, we have n + j ≡

i + j and hence, τ

(n + j) = τ

(i + j).

Let j

= f

− i. Then, since n + j

≡

i + j

≡

0, we have τ

(n + j

+ j) = τ

(i + j

+ j) = τ

(j) for any j = 0, 1, . . . , f

− 3 by the induction hypothesis. Thus, τ

(n + j) = τ

(i + j) for any j = 0, 1, . . . , f

− i − 3.

This proves (1).

(2) In this case, τ

(n) = 0. Hence, n ≡

i. Therefore, we can apply (1) with k + 1 for k. Thus, we get (2).

Let n, m, i ∈ N with m ≥ 2 and 0 < i < m. We call n an (m, i)-shift invariant place in b ε if

b

ε

b ε

. . . b ε

= b ε

ε b

. . . b ε

.

We call n an m-repetitive place in b ε if there exist i, j ∈ N with i > 0 and

i + j < m such that n + j is an (m, i)-shift invariant place in b ε. Let R

be

the set of m-repetitive places in b ε.

Lemma 8. (1) Let n ≡

0 for some k ≥ 1. Then n is an (f

− 2, f

)- shift invariant place in b ε.

(2) Let n ≡

f

for some k ≥ 2. Then n is an (f

− 2, f

)-shift invariant place in b ε.

P r o o f. (1) Since the least i ≥ n such that either i ≡

f

− 1 or i ≡

f

− 2 is not less than n + f

− 2, by Lemma 5, we have

b

ε

b ε

. . . b ε

= b ε

ε b

. . . b ε

.

(2) Since the minimum i ≥ n such that either i ≡

f

− 1 or i ≡

f

− 2 is n + f

− 2, by Lemma 5, we have b

ε

b ε

. . . b ε

= b ε

ε b

. . . b ε

.

Theorem 1. The pair (n, m) of nonnegative integers satisfies n ∈ R

if one of the following two conditions holds:

(1) f

+ 1 ≤ m ≤ f

− 2, n − i ≡

0 and i ≤ n for some k ≥ 1 and i ∈ Z with f

+ 1 ≤ m + i ≤ f

− 2.

(2) f

+ 1 ≤ m ≤ f

− 2, i ≤ n and n − i ≡

f

for some k ≥ 2 and i ∈ Z with f

+ 1 ≤ m + i ≤ f

− 2.

Remark 2. The “if and only if” statement actually holds in Theorem 1 in place of “if” since we will prove later that H

6= 0 if none of the conditions (1) and (2) hold.

P r o o f (of Theorem 1). Assume (1) and i ≥ 0. By Lemma 8(1), n − i is an (f

− 2, f

)-shift invariant place. Then n is an (m, f

)-shift invariant place since i + m ≤ f

− 2. Thus, n ∈ R

as f

< m.

Assume (1) and i < 0. Then, since n−i is an (f

−2, f

)-shift invariant place and m ≤ f

−2, it is an (m, f

)-shift invariant place. Moreover, since f

− i < m, n is an m-repetitive place.

Assume (2) and i ≥ 0. Then, n − i is an (f

− 2, f

)-shift invariant place by Lemma 8(2). Then, n is an (m, f

)-shift invariant place since i + m ≤ f

− 2. Thus, n is an m-repetitive place as f

< m.

Assume (2) and i < 0. Then, since n − i is an (f

− 2, f

)-shift invariant place and m ≤ f

− 2, it is an (m, f

)-shift invariant place.

Then n is an m-repetitive place, since f

− i < m. Thus, n ∈ R

. Corollary 1. The place 0 is m-repetitive for an m ≥ 2 if m 6∈

S

k=1

{f

− 1, f

}.

Remark 3. The “if and only if” statement actually holds in Corollary 1 in place of “if” since we prove later that H

6= 0 if m ∈ S

k=1

{f

− 1, f

}.

P r o o f (of Corollary 1). Let i = 0 in (1) of Theorem 1. Then 0 is

m-repetitive if f

+ 1 ≤ m ≤ f

− 2 for some k ≥ 1.

Corollary 2. Let k ≥ 2. The place n is f

-repetitive if W

≺ b ε

ε b

. . . b ε

.

P r o o f. By (2) of Theorem 1, for any k ≥ 2, n is an f

-repetitive place if n − i ≡

f

for some i with i ≤ n and −f

+ 1 ≤ i ≤ f

− 2. Since the condition n − i ≡

f

is equivalent to n − i ≡

f

and there is no carry in addition of −i to both sides of n ≡

f

+ i, the condition n − i ≡

f

is equivalent to n ≡

f

+ i. Hence, the place n is f

-repetitive if n ≡

j for some j with f

+ 1 ≤ j ≤ f

− 2. By Lemma 6, this condition is equivalent to W

starting at one of the places in {n + 1, n + 2, . . . , f

− 2}, which completes the proof.

3. Hankel determinants. The aim of this section is to find the value of the Hankel determinants

H

:= H

(ε) = det(ε

)

, H

:= H

(ε) = det(ε

)

(n = 0, 1, . . . ; m = 1, 2, . . .) for the Fibonacci word ε(a, b) at (a, b) = (1, 0) and (a, b) = (0, 1):

ε := ε(1, 0) = 10110101101101 . . . , ε := ε(0, 1) = 01001010010010 . . .

It is clear that H

(ε(a, b)) = 0 if n is the m-repetitive place in ε(a, b), where a, b are considered to be two independent variables, and that, in gen- eral, H

(ε(a, b)) becomes a polynomial in a and b as stated in Remark 1.

In the following lemmas, theorems and corollary, we give parallel state- ments for ε and ε, while we give the proofs only for ε since those for ε are similar. The only difference is the starting point, Lemma 5, where a − b on the right-hand side is 1 for ε and −1 for ε.

We use the following notation: for every subset S of {0, 1, 2, 3, 4, 5}, χ(k : S) is the function on k ∈ Z such that

χ(k : S) =

n −1 if k ≡ s (mod 6) for some s ∈ S, 1 otherwise.

The following corollary follows from Theorem 1.

Corollary 3. H

= 0 if one of the conditions (1), (2) in Theorem 1 is satisfied. The same statement holds for H

.

Lemma 9. For any k ≥ 2, we have

H

= χ(k : 2, 3)(H

− (−1)

H

),

H

= χ(k : 1, 3, 4, 5)(H

− (−1)

H

).

P r o o f. The matrix (ε

)

is decomposed into three parts:

(ε

)

=



 A A

B



 ,

where

A = (ε

)

, A

= (ε

)

,

B = (ε

)

. By Lemma 5, the following two subwords of ε:

ε

ε

. . . ε

and ε

ε

. . . ε

differ only at two places, namely, ε

6= ε

and ε

6=

ε

. Thus, we get

(15) B − A =



 



(−1)

(−1)

(−1)

(−1)

0 . . .

(−1)

(−1)

0



 

 .

Let A

, A

, . . . , A

be the columns of the matrix

in order from the left. Since

(A

A

. . . A

) = (ε

)

, (A

A

. . . A

) = (ε

)

and

ε

ε

. . . ε

= ε

ε

. . . ε

by Lemma 5, we get

(16) (A

A

. . . A

) = (A

A

. . . A

).

Thus, from (15) and (16) we obtain

H

= det



 

 



A

. . . A

A

. . . A

A

(−1)

(−1)

(−1)

(−1)

0 ^{. . . . . .}

(−1)

(−1)

0



 

  (17) 

= det



 

 



A

. . . A

0 . . . 0 A

(−1)

(−1)

(−1)

(−1)

0 ^{. . . . . .}

(−1)

(−1)

0



 

 



= (−1)

(−1)

det(A

A

. . . A

)

+ (−1)

(−1)

det(A

A

A

. . . A

).

Since

ε

ε

. . . ε

= ε

ε

. . . ε

by Lemma 5, we get

det(A

A

A

. . . A

) = det(ε

)

= H

. Thus we get

H

= (−1)

(−1)

H

+ (−1)

(−1)

H

= χ(k : 2, 3)(H

− (−1)

H

), where we have used the fact that

(−1)

(−1)

= χ(k : 2, 3).

Lemma 10. For k ≥ 2, we have

H

= χ(k : 1, 3, 4, 5)H

, H

= χ(k : 2, 3)H

.

P r o o f. Just as in the proof of Lemma 9, we decompose the matrix (ε

)

into three parts:

(ε

)

=



 A A

B



 ,

where

A = (ε

)

, A

= (ε

)

,

B = (ε

)

. By Lemma 5, the following two subwords of ε:

ε

ε

. . . ε

and

ε

ε

. . . ε

differ only at two places. Namely, ε

6= ε

and ε

6= ε

. Therefore, we get

B − A =



 

 



(−1)

(−1)

(−1)

0 ^{. . .}

. . .

(−1)

(−1)

0



 

 

 .

Thus, we have

(18) det(ε

)

= det



 

 



A

A

. . . A

A

. . . A

A

(−1)

(−1)

(−1)

0 ^{. . . . . .}

(−1)

(−1)

0



 

 



= (−1)

(−1)

det(A

A

. . . A

)

= χ(k : 1, 3, 4, 5)H

.

Lemma 11. For any k ≥ 2, we have

H

= χ(k : 2, 5)H

, H

= χ(k : 2, 5)H

. P r o o f. Since, by Lemma 5,

ε

ε

. . . ε

= ε

ε

. . . ε

, we get

(ε

)

=



 

 

0 0 1

1 · 0

· ·

· ·

0 ^{1 0}



 

  (ε

)

.

Also, by Lemma 5,

(ε

)

= (ε

)

. Thus we obtain

H

= det(ε

)

= (−1)

det(ε

)

= χ(k : 2, 5)H

. Lemma 12. For any k ≥ 3, we have

H

= χ(k : 2, 3)H

+ χ(k : 2, 4)H

,

H

= χ(k : 1, 3, 4, 5)H

+ χ(k : 0, 1, 2, 3)H

. P r o o f. Clear from Lemmas 9–11.

Lemma 13. For any k ≥ 0, we have

H

= χ(k : 2)f

,

H

= χ(k : 1, 2, 4)f

.