On inhomogeneous Diophantine approximation and the Nishioka–Shiokawa–Tamura algorithm

LXXXVI.4 (1998)

On inhomogeneous Diophantine approximation and the Nishioka–Shiokawa–Tamura algorithm

by

Takao Komatsu (Nagaoka)

In memory of Professor Takayoshi Mitsui

We obtain the values concerning M(θ, φ) = lim inf

|q|kqθ−φk using the algorithm by Nishioka, Shiokawa and Tamura. In application, we give the values M(θ, 1/2), M(θ, 1/a), M(θ, 1/ p

ab(ab + 4)) and so on when θ = ( p

ab(ab + 4) − ab)/(2a) = [0; a, b, a, b, . . .].

1. Introduction. Let θ be irrational and φ real. We suppose throughout that qθ − φ is never integral for any integer q. Define

M(θ, φ) = lim inf

|q|→∞

|q|kqθ − φk,

which is called the inhomogeneous approximation constant for the pair θ, φ.

It is convenient to introduce the functions M

(θ, φ) = lim inf

q→+∞

qkqθ − φk and

M

(θ, φ) = lim inf

q→+∞

qkqθ + φk = lim inf

q→−∞

|q|kqθ − φk.

Then M(θ, φ) = min(M

(θ, φ), M

(θ, φ)). These notations are introduced by Cusick, Rockett and Sz¨ usz [2]. M(θ, φ) or M

(θ, φ) has been treated by Cassels [1], Descombes [3], S´os [9], Cusick et al. [2] and the author [5] by using several algorithms for inhomogeneous Diophantine approximation in which φ is expressed by the continued fraction expansion of θ. However, it is not easy to evaluate M(θ, φ) if it exists for any given pair of θ and φ.

In this paper we establish the relationship between M(θ, φ) and the algorithm of Nishioka, Shiokawa and Tamura. Indeed, this was hinted at in [5] but has not been proved yet. If we use this result, we can find the

1991 Mathematics Subject Classification: 11J20, 11J70.

[305]

exact value of M(θ, φ) for any pair of θ and φ at least when θ is a positive real root of a quadratic equation and φ ∈ Q(θ). In fact, we give several applications when θ = ( p

ab(ab + 4) − ab)/(2a) = [0; a, b, a, b, . . .] = [0; a, b]

for fixed positive integers a and b.

2. Algorithm by Nishioka, Shiokawa and Tamura. We first in- troduce several notations and show how Nishioka, Shiokawa and Tamura [6] represent φ through the continued fraction expansion of θ. As usual θ = [a

; a

, a

, . . .] denotes the continued fraction expansion of θ, where

θ = a

+ θ

, 1/θ

= a

+ θ

,

a

= bθc,

a

= b1/θ

c (n = 1, 2, . . .).

The kth convergent p

/q

= [a

; a

, . . . , a

] of θ is then given by the recur- rence relations

p

= a

p

+ p

q

= a

q

+ q

(k = 0, 1, . . .), (k = 0, 1, . . .),

p

= 0, q

= 1,

p

= 1, q

= 0.

Denote by φ =

[b

; b

, b

, . . .] the expansion of φ in terms of the sequence {θ

, θ

, . . .}, where

φ = b

− φ

, φ

/θ

= b

− φ

,

b

= dφe,

b

= dφ

/θ

e (n = 1, 2, . . .).

Then φ is represented by

φ = b

− b

θ

+ b

θ

θ

− . . . + (−1)

b

θ

θ

. . . θ

− (−1)

θ

θ

. . . θ

φ

= b

− X

(−1)

b

θ

θ

. . . θ

= b

− X

b

D

,

where D

= q

θ − p

= (−1)

θ

θ

. . . θ

. As usual a

, a

, . . . , a

is the periodic sequence with period a

, a

, . . . , a

. We interpret b

, b

, . . . , b

similarly.

Now, we state our main theorem.

Theorem 1.

M

(θ, φ) = lim inf

n→+∞

min(B

kB

θ + φk, B

kB

θ + φk), where B

= P

k=1

b

q

and B

= B

− q

.

P r o o f. Let n be odd. Then kB

θ + φk = {B

θ + φ} = φ

D

. Take any integer k with 0 < k < q

. Since {(q

− jq

)θ} = 1 − |D

| − jD

is the jth largest value (j = 1, . . . , q

− 1) among {kθ} (see e.g. [4], [8]

for details),

{(B

+ q

− jq

)θ + φ} = {B

θ + φ} + {(q

− jq

)θ}

= 1 − (|D

| + (j − φ

)D

)

< 1 − φ

D

= 1 − kB

θ + φk

if j 6= 1. However, {(B

+ q

− q

)θ + φ} is less than 1, but less than 1 − kB

θ + φk if and only if θ

+ 1 > 2φ

. Thus, for any integer k with 0 < k < q

and k 6= q

− q

we have kB

θ + φk < k(B

+ k)θ + φk, yielding B

kB

θ + φk < (B

+ k)k(B

+ k)θ + φk.

Next, consider a positive integer B

− k with 0 < k < q

. When b

≤ a

, then B

− (q

− q

) < B

+ q

. Hence, if B

− q

+ q

< B

, then

(B

− q

+ q

)k(B

− q

+ q

)θ + φk

≤ min(B

kB

θ+φk, (B

−q

+q

)k(B

−q

+q

)θ+φk).

Otherwise, it is sufficient to consider smaller n. When b

= a

+1, we obtain k(B

− q

+ q

)θ + φk = 1 + kB

θ + φk − {(q

− q

)θ}

= 1 + φ

|D

| − (1 − |D

| − |D

|)

= (1 + φ

+ θ

)|D

|

> |D

| + (1 − φ

)|D

|

= k(B

+ q

− q

)θ + φk, yielding

(B

− q

+ q

)k(B

− q

+ q

)θ + φk

> (B

+ q

− q

)k(B

+ q

− q

)θ + φk.

Therefore, it is sufficient to pay attention only to the small fractional parts {kθ}. Since {jq

θ} = jD

is the jth least (j = 1, . . . , q

− 1) (see [4], [8]), we have

k(B

− jq

)θ + φk = {jq

θ} − {B

θ + φ} = (j − φ

)D

> φ

D

= kB

θ + φk

if j 6= 1. However, k(B

− q

)θ + φk > kB

θ + φk if and only if φ

< 1/2.

Thus, we have kB

θ + φ

k < k(B

− k)θ + φk for any integer k with 0 < k <

q

and k 6= q

. When b

≤ 2, the assertion holds because B

− q

≤ B

+ q

. Thus, we can assume that b

≥ 3. Since k(B

− jq

)θ + φk is the jth least value among k(B

− k)θ + φk for all integers k with 0 < k < q

,

(B

− jq

)k(B

− jq

)θ + φk < (B

− k)k(B

− k)θ + φk

for (j − 1)q

< k < jq

(j = 1, . . . , b

− 1). Therefore, from B

=

B

+ b

q

> ((j + 1) − φ

)q

we have

(B

− jq

)k(B

− jq

)θ + φk = (B

− jq

)(j − φ

)D

> (B

− q

)(1 − φ

)D

= (B

− q

)k(B

− q

)θ + φk.

Finally, since

k(B

− q

)θ + φk = (1 − φ

)D

< |D

| + (1 − φ

)D

= k(B

+ q

− q

)θ + φk and B

− q

< B

+ q

− q

, we conclude that

(B

− q

)k(B

− q

)θ + φk < (B

+ q

− q

)k(B

+ q

− q

)θ + φk.

Let n be even. Then kB

θ + φk = φ

|D

| = 1 − {B

θ + φ}. The rest of the part is similar to the odd case.

Remark. (1) From the proof above, B

kB

θ + φk < B

kB

θ + φk if φ

> 1/2. But, B

kB

θ + φk > B

kB

θ + φk does not always hold even though φ

< 1/2.

(2) Together with M

(θ, φ) = M

(θ, 1 − φ), one can obtain the value M(θ, φ).

3. A special case. Without loss of generality, we assume hereafter that 0 < θ < 1 and 0 < φ ≤ 1/2. Set

θ =

√ D − ab

2a = [0; a, b, a, b, . . .] = [0; a, b]

where D = ab(ab + 4) is the discriminant of the quadratic equation aθ

+ abθ − b = 0. First of all, we consider an artificially made φ.

Theorem 2. Let φ = (1 − s) + sθ =

[1; sa, sb] with 0 < s < 1 satisfying sa, sb ∈ N. Then

M(θ, φ) = ksk

(ab − |a − b|)

√ D .

P r o o f. From the theory of continued fractions,

 0 1 1 0

   a 1 1 0

  b 1 1 0

 

=

 p

p

q

q



.

It follows that

 q

q

p

p



=

 ab + 1 a

b 1



= 1

α − β

 (α

− β

) − (α

− β

) a(α

− β

)

b(α

− β

) (α

− β

) − (α

− β

)



where

α = (ab + 2) + √ D

2 , β = (ab + 2) − √ D 2 satisfying α + β = ab + 2, αβ = 1 and α − β = √

D. Notice that φ

= s(1 − θ), φ

= s(1 − θ

) (n = 1, 2, . . .) where θ

= ( √

D − ab)/(2b) = [0; b, a].

Now, the relations B

=

X

(saq

+ sbq

)

= s

√ D X

(a(α

− β

) − a(α

− β

) + ba(α

− β

))

∼ sa

√ D · (1 + b)α − 1 α − 1 α

and kB

θ + φk = φ

|D

| = s(1 − θ)β

entail that B

kB

θ + φk → s

a

√ D · (1 + b)α − 1

α − 1 (1 − θ) = s

(ab + a − b)

√ D (n → ∞).

In a similar manner,

B

= B

− q

∼ a

√ D · s(1 + b)α − s − (α − 1)

α − 1 α

and kB

θ + φk = (1 − φ

)|D

| = (sθ − s + 1)β

entail that

n→∞

lim B

kB

θ + φk = sab + (2s − 1)a − s

(ab + a − b)

√ D .

Moreover, B

=

X

(saq

+ sbq

) − sbq

∼ sa

√ D · α − 1 + b α − 1 α

and kB

θ + φk = φ

|D

| = s(1 − θ

)β

entail that

n→∞

lim B

kB

θ + φk = s

(ab − a + b)

√ D .

Finally,

B

= B

− q

∼ a

√ D (s − (1 − s)θ) α

and kB

θ + φk = (1 − φ

)|D

| = (1 − s + sθ

)β

entail that

n→∞

lim B

kB

θ + φk = sab + (2s − 1)b − s

(ab − a + b)

√ D .

Since sb(1−2s)+(2s−1) ≥ 1·(1−2s)+(2s−1) = 0 and sa(1−2s)+(2s−1) ≥ 1 · (1 − 2s) + (2s − 1) = 0, one has

M

(θ, φ) = s

(ab − |a − b|)

√ D .

Next, one expands

1 − φ =

[1; (1 − s)a + 1, (1 − s)b, (1 − s)a]

and

φ

= (1 − s) + sθ,

φ

= (1 − s)(1 − θ

), φ

= (1 − s)(1 − θ) (n = 1, 2, . . .).

It follows that

n→∞

lim B

kB

θ − φk = (1 − s)

(ab − a + b)

√ D ,

n→∞

lim B

kB

θ − φk = (1 − s)ab + (1 − 2s)b − (1 − s)

(ab − a + b)

√ D ,

n→∞

lim B

kB

θ − φk = (1 − s)

(ab + a − b)

√ D ,

n→∞

lim B

kB

θ − φk = (1 − s)ab + (1 − 2s)a − (1 − s)

(ab + a − b)

√ D ,

yielding

M

(θ, φ) = M

(θ, 1 − φ) = (1 − s)

(ab − |a − b|)

√ D .

Example 1 ([7]). Let a = b be even, and s = 1/2. Then the pair of θ = [0; a] and φ = (1 + θ)/2 =

[1; a/2] gives

n→∞

lim B

kB

θ + φk = lim

n→∞

B

kB

θ + φk = a 4 √

a

+ 4 . Since 1 − φ = (1 − θ)/2 =

[1; a/2 + 1, a/2],

n→∞

lim B

kB

θ − φk = lim

n→∞

B

kB

θ − φk = a 4 √

a

+ 4 .

Therefore,

M(θ, φ) = M

(θ, φ) = a 4 √

a

+ 4 ,

which gives an answer to an open problem related to Khinchin’s results ([5]).

Example 2. Let a = b be a composite odd number, say a = a

a

, where a

and a

are also odd numbers with a

≤ a

and put s = 1/a

. Then

M(θ, φ) = M

(θ, φ) = a

a

√ a

+ 4 ,

which answers another open problem related to Khinchin’s results ([5]).

However, the case of odd prime a is not settled yet.

4. Some basic applications. In this section we compute M(θ, φ) for some basic φ’s as seen in [2] and [5]. Put Ξ

= B

kB

θ − φk √

D/φ

, Ξ

= B

kB

θ − φk √

D/φ

, Ψ

= B

kB

θ + φk √

D/φ

and Ψ

= B

kB

θ + φk √

D/φ

for simplicity.

Theorem 3.

M

 θ, 1

2



=

 

 

 



min(a, b) 4 √

D if both a and b are odd, a

4 √

D otherwise.

P r o o f. It is clear that Ξ

= Ψ

and Ξ

= Ψ

when φ = 1/2.

If a = 1 and b is even with b ≥ 4, then φ = 1/2 is expanded as 1

2 =

θ

 1; 1, b

2 , 1, b + 1, 1, b 2 − 1



and for n = 1, 2, . . . ,

φ

= b + 2 − √ D

4 ,

φ

= b + 3 − √ D

2 ,

φ

= 5b − √ D 4b , φ

= 5b − 3 √

D 4b .

By the inhomogeneous continued fraction expansion above, for n = 1, 2, . . . , B

= 1 + b

2 q

+

n−1

X

i=1



q

+ (b + 1)q

+ q

+

 b 2 − 1

 q

 , B

= B

+ q

, B

= B

+ q

+ (b + 1)q

, B

= B

+ q

+ (b + 1)q

+ q

.

Now,

kB

θ + φk = φ

|D

|

= φ

θ

θ

. . . θ

=

β(θ

θ

)

=

β

,

and

√ DB

= √ D

 1 + b

2

 +

n−1

X

i=1



(α

− β

) − (α

− β

) + (b + 1)(α

− β

) + (α

− β

) − (α

− β

) +

 b 2 − 1



(α

− β

)



= √ D

 1 + b

2

 +

 b

2 α

+ (b + 1)α − 1



α · α

− 1 α

− 1

−

 b

2 β

+ (b + 1)β − 1



β · β

− 1 β

− 1

= 1

2 (bα

+ 2(b + 1)α − 2) α

α

− 1 + (others).

Here “others” tend to 0 in Ψ

as n tends to infinity. Therefore, we obtain

n→∞

lim Ψ

= bα + 2(b + 1) − 2β α

− 1 = 1.

Similarly, one finds lim

Ψ

= 2b − 1. Since B

kB

θ + φk <

B

kB

θ + φk or Ψ

< Ψ

as φ

> 1/2, there is no necessity to evaluate the right-hand side (in fact, one can find lim

Ψ

= 6b−1).

And B

= B

entails Ψ

= Ψ

→ 1 (n → ∞).

Since B

kB

θ + φk < B

kB

θ + φk or Ψ

< Ψ

as φ

> 1/2, it is sufficient to evaluate the left-hand side (the right-hand side tends to 2b + 9).

The relations

kB

θ + φk = (1 − φ

)|D

| =

− β

θ

θ

=

− β
β

and √

D B

= √

D(B

+ q

+ bq

)

= 1

2 (bα

+ 2(b + 1)α − 2) α

α

− 1

+ (α

− β

) − (α

− β

) + b(α

− β

) + (others) entail that

n→∞

lim Ψ

= 2

 1 2 − β

 bα + 2(b + 1) − 2β

α

− 1 + (2 − 2β) + 2b



= (1 − 2β)(3 − 2β + 2b) = 2b − 1.

Since B

= B

, one obtains Ψ

= Ψ

→ 2b + 9 (n → ∞). On the other hand, the relations

kB

θ + φk = φ

|D

| = 5b − 3 √ D

4b θ

θ

=

 1 2 − 2β



β

and

√ DB

= 1

2 (bα

+ 2(b + 1)α − 2) α

α

− 1 + (α

− α

) + (b + 1)α

+ (α

− α

) + (others) entail that

n→∞

lim Ψ

= 4

 1 2 − 2β



×

 bα + 2(b + 1) − 2β

2(α

− 1) + (1 − β) + (b + 1) + (α − 1)



= (−(2b + 3) + 2 √

D)(2b + 3 + 2 √

D) = 4b − 9.

If a = 1 and b is odd with b ≥ 3, then φ = 1/2 is expanded as 1

2 =

θ



1; 1, b + 1

2 , 1, b − 1

2 , 1, b + 1, 1, b − 1 2



and for n = 1, 2, . . . , φ

= b + 4 − √

D

4 ,

φ

= 5b − √ D 4b ,

φ

= 2b − √ D 2b , φ

= b + 3 − √

D

2 ,

φ

= b + 2 − √ D

4 ,

φ

= 5b − 3 √ D 4b . Hence, for n = 1, 2, . . . one can similarly find

Ψ

→ b, Ψ

→ 3b − 4, Ψ

= Ψ

→ b + 4, Ψ

= Ψ

→ 1, Ψ

→ 2b − 1, Ψ

→ 6b − 1,

Ψ

= Ψ

→ 2b + 9, Ψ

→ 2b − 1, Ψ

→ 4b − 9 (n → ∞).

If a = 1 and b = 2, then φ = 1/2 is expanded as 1

2 =

[1; 1, 1, 1, 3, 2, 3, 1, 2]

and for n = 1, 2, . . . , φ

= 5 − 2 √

3

2 , φ

= 9 − 3 √ 3

4 , φ

= 6 − 3 √ 3

2 , φ

= 7 − 3 √ 3

4 .

In a similar manner, one finds

Ψ

→ 13, Ψ

→ 3, Ψ

→ 27, Ψ

→ 1,

Ψ

= Ψ

→ 9, Ψ

→ 11, Ψ

→ 11 (n → ∞).

If a = b = 1, then φ = 1/2 is expanded as 1

2 =

[1; 1, 1, 2, 2, 1, 1]

and for n = 1, 2, . . . , φ

= 9 − 3 √

5

4 , φ

= 7 − 3 √ 5

4 , φ

= 4 − √ 5 2 . Similarly one finds that for n = 1, 2, . . . ,

Ψ

= Ψ

→ 9, Ψ

→ 5, Ψ

→ 11, Ψ

= Ψ

→ 1 as n tends to infinity. Therefore, we obtain

M(θ, φ) = M

(θ, φ) = 1 4 √

5 .

If a is odd with a ≥ 3 and b is even, then φ = 1/2 is represented as 1

2 =

θ



1; a + 1 2 , b

2 , a + 1

2 , b, a − 1 2

 . Then φ

= 1/2, and for n = 1, 2, . . . ,

φ

= a + 2

4 −

√ D 4b , φ

= a + 4

4 −

√ D 4b ,

φ

= b + 2

4 −

√ D 4a , φ

= b + 1

2 −

√ D 2a . It follows that

Ψ

→ (2b − 1)a + 4b, Ψ

→ (2b + 1)a − 4b, Ψ

→ (b − 1)a + b, Ψ

→ (b + 1)a − b,

Ψ

→ a,

Ψ

→ (2b − 1)a + 4b, Ψ

→ (b + 1)a − b,

Ψ

→ (b − 1)a + b (n → ∞).

If a is odd with a ≥ 3 and b is odd with b ≥ 3, then φ = 1/2 is represented as

1 2 =

θ



1; a + 1 2 , b + 1

2 , a, b − 1 2 , a + 1

2 , b, a − 1 2

 . Then φ

= 1/2, and for n = 1, 2, . . . ,

φ

= a + 2

4 −

√ D 4b , φ

= b + 2

4 −

√ D 4a ,

φ

= b + 4

4 −

√ D 4a , φ

= a + 4

4 −

√ D 4b ,

φ

= a + 1

2 −

√ D 2b , φ

= b + 1

2 −

√ D 2a . It follows that

Ψ

→ (2b + 4)a − b, Ψ

→ b,

Ψ

→ (2b − 4)a + b, Ψ

→ (2b + 4)a − b,

Ψ

→ (b + 1)a − b, Ψ

→ (b − 1)a + b,

Ψ

→ (2b − 1)a + 4b, Ψ

→ a,

Ψ

→ (2b + 1)a − 4b, Ψ

→ (2b − 1)a + 4b,

Ψ

→ (b − 1)a + b, Ψ

→ (b + 1)a − b (n → ∞).

If a is even, then φ = 1/2 is expanded as 1

2 =

θ

 1; a

2 + 1, b, a 2

 . Then φ

= 1/2, and for n = 1, 2, . . . ,

φ

= a + 4

4 −

√ D

4b and φ

= b + 1

2 −

√ D 2a . It follows that

Ψ

→ (2b + 1)a − 4b, Ψ

→ (2b − 1)a + 4b,

Ψ

→ (2b − 1)a + 4b, Ψ

→ a (n → ∞).

Theorem 4.

M

 θ, 1

√ D



= a

D √ D . P r o o f. If a ≥ 2, then φ = 1/ √

D is represented as

√ 1

D =

[1; a, 1, a + 1, b, a − 1].

Then φ

= 1 − 1/ √

D, and for n = 1, 2, . . . , φ

= ab + 1

2b − a(ab + 3) 2 √

D , φ

= ab + 2b + 1

2b − a(ab + 5) 2 √

D ,

φ

= 3

2 − ab + 2 2 √

D , φ

= b + 2

2 − ab

+ 4b + 2 2 √

D .

It follows that

Ψ

→ a(b

(b + 1)a

− b(b + 1)(b − 4)a − (2b + 1)

), Ψ

→ a(b

(b + 1)a + (2b + 1)

),

Ψ

→ a,

Ψ

→ a(b

a

− b

(b − 5)a − (2b − 1)

), Ψ

→ a(2b

a

+ 8ba − 1),

Ψ

→ a,

Ψ

→ a(b

(b − 1)a

+ b(b

+ 5b − 4)a + (2b + 1)

),

Ψ

→ a(b

a

+ 4ba − 1) (n → ∞).

If a = 1, then φ = 1/ √

D is expanded as

√ 1

D =

[1; 1, 1, 2, b + 1, 2, b, 1]

and for n = 1, 2, . . . , φ

= (b + 4) √

D − (b

+ 4b + 2) 2 √

D ,

φ

= (b + 3) √

D − (b

+ 5b + 2) 2 √

D ,

φ

= (4b + 1) √

D − (2b

+ 7b) 2b √

D ,

φ

= (3b + 1) √

D − (b

+ 5b) 2b √

D .

Thus, one can find

Ψ

→ b

+ 7b

+ 12b − 1, Ψ

→ 1, Ψ

→ 3b

+ 11b

− 4b + 1, Ψ

→ 2b

+ 8b − 1, Ψ

= Ψ

→ b

+ 4b − 1,

Ψ

→ b

+ 5b

+ 4b + 1, Ψ

→ 2b

+ 8b − 1 (n → ∞).

If b ≥ 2, then 1 − φ = 1 − 1/ √

D is represented as 1 − 1

√ D =

[1; 1, b, a, b, a, b − 1].

Then φ

= 1/ √

D, φ

= (2b − 1)/(2b) − a/(2 √

D), and for n = 1, 2, . . . , φ

= b + 1

2b − ab

− ab + 4b − 2 2 √

D ,

φ

= b + 2

2 − ab

+ 4b − 2 2 √

D ,

φ

= ab + 2b − 1

2b − a(ab + 3) 2 √

D , φ

= ab + 2b − 1

2b − a(ab + 5) 2 √

D . In a similar manner, one finds that

Ξ

→ a(b

(b + 1)a

− b(b

− 5b − 4)a − (2b − 1)

), Ξ

→ a(b

(b − 1)a + (2b − 1)

),

Ξ

→ a(b

(b − 2)a

+ b(b

+ 3b − 8)a + (2b − 1)

), Ξ

→ a(2b

a

+ 8ba − 1),

Ξ

→ a(b

a

− b

(b − 5)a − (2b − 1)

), Ξ

→ a(b

(b − 1)a + (2b − 1)

),

Ξ

→ a(b

(b − 1)a

+ b(b + 4)(b − 1)a + (2b − 1)

), Ξ

→ a(b

a

+ 4ba − 1) (n → ∞).

If a ≥ 2 and b = 1, then 1 − φ = 1 − 1/ √

D is expanded as 1 − 1

√ D =

[1; 1, 1, a, 1, a + 1, 1, a − 1]

and

φ

= (a + 1) √

D − (a

+ 3a) 2 √

D ,

φ

= (a + 3) √

D − (a

+ 5a) 2 √

D ,

φ

= 3 √

D − (a + 2) 2 √

D ,

φ

= 3 √

D − (a + 6) 2 √

D .

Hence, one can obtain

Ξ

= Ξ

→ a, Ξ

→ a(a

+ 4a − 1),

Ξ

→ a(2a

+ 8a − 1), Ξ

= Ξ

→ a(2a + 9),

Ξ

→ a(a

+ 4a − 1), Ξ

→ a(2a

+ 6a − 9) (n → ∞).

If a = b = 1, then 1 − φ = 1 − 1/ √

5 is expanded as 1 − 1

√ 5 =

[1; 1, 1, 1, 1, 2, 2, 2, 1, 1]

and

φ

= 5 √ 5 − 7 2 √

5 , φ

= 2 √

5 − 4

√ 5 ,

φ

= 5 √ 5 − 9 2 √

5 , φ

= 2 √

5 − 3

√ 5 . Hence, one can find

Ξ

→ 19, Ξ

→ 1, Ξ

= Ξ

→ 11,

Ξ

→ 9, Ξ

→ 11, Ξ

= Ξ

→ 4 (n → ∞).

Next, we find the value M(θ, 1/a). In fact, we can do more. Concerning an arbitrary divisor of a, say d (≥ 2), we have the following theorem:

Theorem 5.

M

 θ, 1

d



= a

d

√ D . P r o o f. φ = 1/d is represented as

1 d =

θ

 1;

 1 − 1

d



a + 1, b,

 1 − 1

d

 a



and

φ

= 1 − 1/d, φ

= 1 −

 1 − 1

d



θ

, φ

=

 1 − 1

d



− θ (n = 1, 2, . . .).

It follows that

Ψ

→ a(bd + d − 1)(d − 1) − bd

, Ψ

→ a(bd − d + 1)(d − 1) + bd

,

Ψ

→ (bd − 1)a + bd

,

Ψ

→ a(d − 1)

(n → ∞).

1 − φ = 1 − 1/d is represented as 1 − 1

d =

θ

 1; a

d + 1, b, a d



and

φ

= 1/d, φ

= 1 − 1

d θ

, φ

= 1

d − θ (n = 1, 2, . . .).

In a similar manner, one finds that Ξ

→ (bd + 1)a − bd

, Ξ

→ (bd − 1)a + bd

,

Ξ

→ a(bd − d + 1)(d − 1) + bd

, Ξ

→ a (n → ∞).

Putting a = d yields the desired result.

Corollary 1. For a ≥ 2, M

 θ, 1

a



= 1

a √ D .

Compared with this result, it is not easy to find M(θ, 1/b), M(θ, 1/(2a)) or M(θ, 1/k) with k ∈ N because these inhomogeneous continued fraction expansions are not so simple.

Theorem 6. If a = 1 and b > 1, then M

 θ, 1

b



= M

 θ, 1

b



= 1

b

√ D . P r o o f. Let b ≥ 4. Then φ = 1/b is expanded as

1

b =

[1; 1, 1, 1, 2, 1, 3, 1, 4, . . . , 1, b − 1, 1, b + 1, 2, b + 1, 1, 1

| {z }

2b

] and

φ

=

 

 

 



(i + 2)b − 2 − i √ D

2b (i = 0, 1, 2, . . . , b − 1), (i + 4)b − 2 − i √

D

2b (i = b, b + 1), φ

= b

+ (2i + 1)b − (b − 1) √

D

2b

(i = 0, 1, 2, . . . , b − 1), φ

= 4b

+ b − (2b − 1) √

D

2b

, φ

= 2b

+ 3b − (2b − 1) √ D

2b

,

φ

= φ

(k = 4, 5, . . . , 2b + 3; n = 1, 2, . . .).

Hence, for n = 0, 1, . . . and i = 2, 3, . . . , b − 1 one gets Ψ

= Ψ

→ (i + 1)b

− (i

+ i + 2)b + 1,

Ψ

→ (i

+ i)b − 1, Ψ

→ ib

+ (i

+ i + 2)b − 1

and for n = 0, 1, . . . ,

Ψ

→ b

+ 3b

− 4b + 1, Ψ

→ 2b − 1,

Ψ

→ 3b

− b

+ 4b − 1, Ψ

→ 2b

− 4b + 1,

Ψ

= Ψ

→ b

+ 3b

− 6b + 1, Ψ

→ b

+ 4b − 1, Ψ

= Ψ

→ 6b − 1, Ψ

→ 3b

− 8b + 1 (n → ∞).

Therefore, M

(θ, 1/b) = (2b − 1)/(b

√

b

+ 4b).

1 − φ = 1 − 1/b is expanded as 1 − 1

b =

[1; 1, b, 1, b − 2, 1, b − 3, . . . , 1, 4, 1, 3, 1, 2, 2, b + 1, 1, b, 1, b − 1

| {z }

2b

]

and

φ

=

 

 

 



b

− (i − 2)b + 2 − (b − i) √ D

2b (i = 1, 2, . . . , b − 2), 2b

− (i − 2)b + 2 − (2b − i) √

D

2b (i = b − 1, b),

φ

=

 

 

 



3b

− (2i + 1)b − (b + 1) √ D

2b

(i = 1, 2, . . . , b − 3), 5b

− (2i + 1)b − (b + 1) √

D

2b

(i = b − 2, b − 1, b), φ

= φ

(k = 2, 3, . . . , 2b + 1; n = 1, 2, . . .).

Hence, for n = 0, 1, . . . and for i = 1, 2, . . . , b − 3 one gets Ξ

= Ξ

→ (i + 2)b

− (i

+ i − 2)b + 1, Ξ

→ b

− (2i + 1)b

+ (i

+ i)b − 1,

Ξ

→ 2b

− 3(i + 1)b

+ (i

+ i − 2)b − 1 (n → ∞), and for n = 0, 1, . . . ,

Ξ

→ 3b

+ 1, Ξ

→ 2b − 1, Ξ

→ 2b

+ 3b

− 1, Ξ

→ 1, Ξ

= Ξ

→ b

+ 2b + 1,

Ξ

→ b

+ b

− 1, Ξ

→ 2b

− 2b − 1,

Ξ

= Ξ

→ 2b

+ 2b + 1, Ξ

→ b

− b

− 1,

Ξ

→ 2b

− 3b

− 2b − 1 (n → ∞).

If b = 3, then φ = 1/3 is expanded as 1

3 =

[1; 1, 1, 1, 2, 1, 4, 2, 4, 1, 1]

and for n = 1, 2, . . . , φ

= 5 − √

21

3 ,

φ

= 39 − 5 √ 21

18 ,

φ

= 12 − √ 21

9 ,

φ

= 11 − 2 √ 21

3 ,

φ

= 19 − 3 √ 21

6 ,

φ

= 27 − 5 √ 21

18 .

It follows that

Ψ

= Ψ

→ 4, Ψ

→ 17, Ψ

→ 41, Ψ

→ 43, Ψ

→ 5, Ψ

→ 83, Ψ

→ 7, Ψ

= Ψ

→ 37, Ψ

→ 20, Ψ

→ 17 as n tends to infinity.

1 − φ = 2/3 is expanded as 2

3 =

[1; 1, 1, 1, 2, 1, 4, 2, 4, 1, 1]

and for n = 1, 2, . . . , φ

= 7 − √

21

3 ,

φ

= 15 − 2 √ 21

9 ,

φ

= 18 − 2 √ 21

9 ,

φ

= 17 − 3 √ 21

6 ,

φ

= 10 − 2 √ 21

3 ,

φ

= 12 − 2 √ 21

9 .

It follows that

Ξ

→ 28, Ξ

→ 5, Ξ

→ 80, Ξ

→ 1, Ξ

= Ξ

→ 16, Ξ

→ 35, Ξ

→ 47, Ξ

= Ξ

→ 25, Ξ

→ 17, Ξ

→ 20 as n tends to infinity.

The case b = 2 is included in Theorem 3.

Theorem 7. If b = 1, then M

 θ, 1

2a



= M

 θ, 1

2a



= 1

4a √

a

+ 4a .

P r o o f. If a is even with a ≥ 4, then φ = 1/(2a) is expanded as 1

2a =

θ



1; a + 1, 1, a

2 − 1, 1, a

2 − 1, 1, a, 1, a



and

φ

= 2a

+ 5a − (2a − 1) √ D

4a ,

φ

= a

+ 3a − (a − 1) √ D

4a ,

φ

= a

+ a − (a − 1) √ D

4a ,

φ

= 2a

+ 3a − (2a − 1) √ D

4a ,

φ

= 5a − 2 − 3 √ D

4a ,

φ

= 2a − 1 − √ D

2a ,

φ

= 5a − 2 − √ D

4a ,

φ

= 3a − 1 − √ D

2a .

It follows that

Ψ

= Ψ

→ a(2a

+ 10a − 1), Ψ

→ a(2a

− 4a + 1), Ψ

→ a(4a

− 14a + 1), Ψ

= Ψ

→ a(a

+ 4a − 1),

Ψ

→ a(a − 1)

, Ψ

→ a(3a

− 8a + 1), Ψ

= Ψ

→ a(2a − 1), Ψ

→ a(2a

− 4a + 1), Ψ

→ a(6a

− 6a + 1),

Ψ

= Ψ

→ a(6a − 1), Ψ

→ a(2a − 1)

, Ψ

→ a(8a

− 10a + 1) as n tends to infinity.

If a is odd with a ≥ 5, then φ = 1/(2a) is expanded as 1

2a =

θ



1; a + 1, 1, a − 3

2 , 1, a, 1, a



and for n = 1, 2, . . . ,

φ

= 2a

+ 5a − (2a − 1) √ D

4a ,

φ

= a

+ a − (a − 1) √ D

4a ,

φ

= 2a

+ 3a − (2a − 1) √ D

4a ,

φ

= 5a − 2 − 3 √ D

4a ,

φ

= 5a − 2 − √ D

4a ,

φ

= 3a − 1 − √ D

2a .

It follows that

Ψ

= Ψ

→ a(2a

+ 10a − 1), Ψ

→ a(2a

− 4a + 1), Ψ

→ a(4a

− 14a + 1), Ψ

= Ψ

→ a(2a − 1), Ψ

→ a(2a

− 4a + 1), Ψ

→ a(6a

− 6a + 1),

Ψ

= Ψ

→ a(6a − 1), Ψ

→ a(2a − 1)

, Ψ

→ a(8a

− 10a + 1) as n tends to infinity.

If a = 3, then φ = 1/6 is expanded as 1

6 =

[1; 4, 2, 4, 1, 2, 1, 3]