0. Introduction. We consider the function (1) f (θ, φ; x, y) =

LXXVI.2 (1996)

A certain power series associated with a Beatty sequence

by

Takao Komatsu (North Ryde, N.S.W.)

0. Introduction. We consider the function (1) f (θ, φ; x, y) =

X

X

1≤m≤kθ+φ

x

y

. Putting y = 1 entails that

(2) f (θ, φ; x, 1) =

X

[kθ + φ]x

.

The sequence {[kθ + φ]}

, which appears in this power series, is called a Beatty sequence. In that context it is natural to consider the sequence of differences

(3) {[(k + 1)θ + φ] − [kθ + φ]}

.

The function f (θ, 0; x, y) and the sequence {[(k + 1)θ] − [kθ]}

in the homogeneous case have been treated independently by many authors (see e.g. [1], [7], [8] and [2], [10] respectively). The inhomogeneous case of (3) has also been treated by several authors (see e.g. [3]–[5]).

In 1992 Nishioka, Shiokawa and Tamura [9] described the sequence (3) in the inhomogeneous case by using the characteristic properties of (1), but their result (Theorem 3 of [9]) is incorrect. The arguments only hold when φ is an integer or when b

= 1 for all positive integers n (for the definition of b

see the next section).

In this paper we base on the arguments corrected by the author [6] and describe the sequence (3) completely in the new form. Of course, Theorem 2 of [9] holds because φ = 0. Lemmas 2 and 3 of [9], which were used to prove Theorem 3 of [9], work and have meaning only in the original context. After

1991 Mathematics Subject Classification: Primary 11B83.

The author thanks Dr. Peter A. B. Pleasants for his useful advice.

[109]

correcting the arguments properly, both lemmas are no longer useful and we need different new arguments to obtain a correction to Theorem 3 of [9].

1. Preliminary remarks and notation. Throughout this paper θ > 0 is irrational and kθ + φ is never integral for any positive integer k. As usual, θ = [a

, a

, a

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

θ = a

+ θ

, a

= [θ],

1/θ

= a

+ θ

, a

= [1/θ

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

The nth convergent p

/q

= [a

, a

, . . . , a

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

p

= a

p

+ p

(n = 0, 1, . . .), p

= 0, p

= 1, q

= a

q

+ q

(n = 0, 1, . . .), q

= 1, q

= 0.

One now expands φ in terms of the sequence {θ

, θ

, . . .} by setting φ = b

− φ

, b

= dφe,

φ

/θ

= b

− φ

, b

= dφ

/θ

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

Furthermore, the quantities s

and t

are defined by s

=

X

b

p

(n = 0, 1, . . .), s

= 0 (n < 0),

t

= X

b

q

(n = 0, 1, . . .), t

= 0 (n < 0).

We can assume 0 < θ, φ < 1 without loss of generality. As shown in Sections 1 and 2 of [6],

f (θ, φ; x, y) = X

(−1)

x

y

(1 − x

y

)(1 − x

y

) , which yields

X

([(k + 1)θ + φ] − [kθ + φ])x

= 1 x lim

n→∞

P

(x), |x| < 1.

Here, P

(x) is defined recursively by

P

(x) = A

(x)P

(x) + x

P

(x) (n ≥ 1) with P

(x) = 1, P

(x) = 0, where

A

(x) = 1 − x

− x

(1 − x

)

1 − x

(n ≥ 1).

Let P

(x) = d

x + d

x

+ d

x

+ . . . be the power series expansion.

Put P

= d

d

d

. . . , which is the string of coefficients of the power series beginning from that of x

.

Define

Γ

= {a

− b

, a

− b

, . . . , a

− b

} (n ≥ 3)

and write π

= a

− b

if a

> b

, $

= a

− b

if a

≥ b

—to account for the case when the entry 0 is permitted.

We consider the following situations:

Γ

∈ O if Γ

= $

$

. . . $

,

Γ

∈ A

(or simply A) if Γ

ends in (−1)0

π

$ |

{z . . . $

}

l

, Γ

∈ B

(or simply B) if Γ

ends in (−1)0

,

Γ

∈ C

(or simply C) if Γ

ends in (−1)0

(k ≥ 2), Γ

∈ C

if Γ

ends in π

$

. . . $

| {z }

l

(−1), Γ

∈ D

(or simply D) if Γ

ends in (−1)0

(−1),

where k is a positive integer and l is a non-negative integer. (Note that Γ

∈ O if a

≥ b

and Γ

∈ C if a

= b

− 1.)

Let β

= t

− q

− b

+ 1 = (b

− 1)q

+ b

q

+ . . . + b

q

+ 1.

We define the words u, v and ∆

as u = 0 . . . 0 | {z }

a1−1

1, v = 0 . . . 0 | {z }

b₁−1

1 and ∆

= 0 . . . 0 | {z }

β_n−2

(−1)

(−1)

.

2. Main results. Our main result, which replaces the alleged Theorem 3 of [9], is

Theorem. Let θ be irrational with 0 < θ, φ < 1. Then either {[(k + 1)θ + φ] − [kθ + φ]}

= lim

n→∞

P

or

{[(k + 1)θ + φ] − [kθ + φ]}

= lim

n→∞

0 . . . 01 | {z }

b1−1

w

.

Here (w

) is the sequence of words of respective lengths q

, with letters 0 or 1, given inductively by

w

= u, w

= w

0w

, w

= w

w

w

, where

c

=

 



 

b

+ 1 if Γ

∈ B and a

> b

,

0 if Γ

∈ C,

1 if Γ

∈ D,

min(a

, b

) otherwise.

R e m a r k. By Lemma 1 below, a

≤ b

if Γ

∈ C, D. Other possible cases are limited to Γ

∈ B and a

= b

, and Γ

∈ O, A.

The Theorem is a direct consequence of the following Proposition, which describes P

. From now on the underline means to add (−1) to the last one part in that word. For example, if W = 00101, then W = 00100. If W = 00100, then W = 0010(−1), W

= 00100 0010(−1) and (W )

= 0010(−1)0010(−1).

Proposition. For every n = 1, 2, . . . , we have P

= vw

w

. Here,

|w

| = q

for every n, and w

= u, w

= u

0u

; w

and w

are empty; and w

and w

(n ≥ 3) are determined as follows:

(1) If n = 3 and Γ

∈ O or A (n ≥ 4), then

 w

= w

w

w

and w

= empty if a

≥ b

, w

= w

w

and w

= ∆

if a

= b

− 1.

(2) If Γ

∈ B (n ≥ 5), then

 w

= w

w

w

and w

= empty if a

> b

, w

= w

w

and w

= ∆

if a

= b

, (k = 1 if Γ

∈ D).

(3) If Γ

∈ C (n ≥ 4), then w

= w

w

and w

=

 empty if a

= b

,

∆

if a

= b

− 1.

(4) If Γ

∈ D (n ≥ 5), then

w

= w

w

w

and w

=

 empty if a

= b

,

∆

if a

= b

− 1.

We detail the initial cases n = 1, 2, 3 here. We notice that A

(x) = 1 + x

+ . . . + x

+

 



 

x

(1 + x

+ . . . + x

) if a

> b

,

0 if a

= b

,

−x

if a

= b

− 1.

Since P

(x) = x

, we have P

= v = vu. Since P

(x) = x

A

(x), we have

P

=

 



vu

0u

= vu

0u

u if a

> b

,

vu

= vu

0u if a

= b

,

vu

(−1) if a

= b

− 1.

Thus, w

= u

0u

. Since P

(x) = x

(A

(x)A

(x) + 1), we have

P

=

 

 

 



vw

uw

u

0u

if a

> b

, vw

= vw

u if a

= b

, vw

00 . . . 00 | {z }

a₁+β₂−2

(−1)1 if a

= b

− 1.

3. Lemmas. We need the following lemmas to complete the proof of the Proposition.

Lemma 1. (1) If Γ

∈ C or D, then a

≤ b

. (2) If Γ

∈ B, then a

≥ b

.

Fig. 1

P r o o f. We prove (1) and (2) together. Notice that as long as a

≥ b

for i = 3, 4, . . . , always Γ

∈ O. Suppose that a

≥ b

, . . . , a

≥ b

and a

= b

− 1 for some fixed n ≥ 4, which means Γ

∈ C

. From the definition we have

θ

+ φ

=

 1 θ

− a

 +



b

− φ

θ



= 1 − φ

θ

+ 1 > 1 or

0 < 1

θ

− φ

θ

< 1.

Therefore,

a

=

 1 θ



≤ b

=

 φ

θ

 . The case Γ

∈ C

is proved.

If a

= b

, that is, Γ

∈ B

, we get θ

+ φ

=

 1

θ

− a

 +



b

− φ

θ



= 1

θ

− φ

θ

< 1.

Therefore,

a

=

 1 θ



≥ b

=

 φ

θ

 .

The case Γ

∈ B

is proved. If a

> b

, Γ

∈ A

. If a

= b

, Γ

∈ C

.

If a

< b

, that is, Γ

∈ D

, similarly to the case Γ

∈ C

, we get θ

+ φ

> 1 and a

≤ b

. The case Γ

∈ D

is proved. If a

= b

, Γ

∈ B

. If a

< b

, Γ

∈ D

again.

Now, we consider each case for an arbitrary positive integer k (≥ 2). Let Γ

∈ C

for some integer i (≥ 6). Since Γ

∈ B

,

1

θ

− φ

θ

> 1.

Hence,

θ

+ φ

=

 1

θ

− a

 +



b

− φ

θ



= 1 − φ

θ

> 1 or

0 < 1

θ

− φ

θ

< 1.

Therefore, a

≤ b

. If a

= b

, Γ

∈ B

. If a

< b

, Γ

∈ D

. The general case Γ

∈ B

or Γ

∈ D

is treated similarly.

The situation in Lemma 1 is illustrated in Figure 1.

Lemma 2. (1) If Γ

∈ O or A, then β

≤ q

. (2) If Γ

∈ C

and Γ

∈ B

, then β

≤ q

. (3) If Γ

∈ D

and Γ

∈ B

, then β

≤ q

+ q

. (4) If Γ

∈ C

, then β

≤ q

.

(5) If Γ

∈ D

, then β

≤ q

+ q

. P r o o f. If a

≥ b

for any i = 3, 4, . . . , n, then

β

= (b

− 1)q

+ b

q

+ . . . + b

q

+ b

q

+ 1

≤ (a

− 1)q

+ a

q

+ . . . + a

q

+ (a

+ 1)q

+ 1 = q

.

The other cases will be proved inductively in the proof of the Proposition.

Lemma 3. (1) If Γ

∈ O or A and a

≥ b

, then w

w

− w

w

= 0 . . . 0 | {z }

qn−1

∆

.

(2) If Γ

∈ O or A and a

< b

, then

−w

w

+ w

w

= 0 . . . 0 | {z }

qn

∆

.

(3) If Γ

∈ B

and a

> b

, then

−w

w

+ w

w

= 00 . . . 00 | {z }

(b_n+1)q_n−1+q_n−2

∆

.

(4) If Γ

∈ B

and a

= b

, then

−w

w

+ w

w

= 0 . . . 0 | {z }

q_n

∆

.

(5) If Γ

∈ C

and a

≤ b

, then

w

w

− w

w

= 0 . . . 0 | {z }

qn−1

∆

.

(6) If Γ

∈ D

and a

≤ b

, then

w

w

− w

w

= 0 . . . 0 | {z }

qn−1

∆

.

P r o o f. Here, we shall prove only the case when Γ

∈ O and a

≥ b

. The others will be proved inductively in the proof of the Proposition. Both w

and w

are empty by induction. Set X = x

for brevity. If a

> b

, then

P

(x) = (1 + X + . . . + X

+ X

x

(1 + X + . . . + X

))

× P

(x) + X

P

(x), which yields

P

= vw

w

w

.

If a

= b

, we have P

(x) = (1+X +. . .+X

)P

(x)+X

P

(x), yielding P

= vw

w

. Hence, we have w

= w

w

w

. There- fore, if n is odd, then

w

w

− w

w

= w

w

w

w

− w

w

w

w

= 0 . . . 0 | {z }

bnqn−1

(w

w

− w

w

)

= 0 . . . 0 | {z }

bnqn−1

(w

w

w

w

− w

w

w

w

)

= 00 . . . 00 | {z }

bnqn−1+bn−1qn−2

(w

w

− w

w

) = . . .

= 000 . . . 000 | {z }

bnqn−1+bn−1qn−2+...+b3q2

(w

w

− w

w

)

= 000 . . . 000 | {z }

bnqn−1+bn−1qn−2+...+b3q2

(uu

0u

− u

0u

u)

= 0000 . . . 0000 | {z }

bnqn−1+bn−1qn−2+...+b3q2+(b2−1)q1

(0 . . . 0 | {z }

q1−1

10 − 0 0 . . . 0 | {z }

q1−1

1)

= 00 . . . 00 | {z }

q_n−1+β_n−2

1(−1).

If n is even, then w

and w

above are interchanged, so we obtain 00 . . . 00

| {z }

qn−1+βn−2

(−1)1.

4. Proof of Proposition. We prove the Proposition together with Lemmas 2 and 3. We write [B

C

D

] for brevity when Γ

∈ B

, Γ

∈ C

and Γ

∈ D

. From Lemma 1 all cases are classified into one of O, A, B, C, D and the number of patterns like [B

C

D

] is limited.

We denote by S the sequence of the patterns of [Γ

, Γ

, Γ

].

4.1. Case Γ

∈ O. The only possible pattern is [OOO]. Then, both w

and w

are empty. As we have already seen in the proof of Lemma 3,

w

= w

w

w

and w

= empty if a

≥ b

.

If a

= b

− 1, by using Lemma 3(1) with Γ

∈ O and β

= (b

− 1)q

+ q

+ β

we have P

(x) = (1 + X + . . . + X

− x

)P

(x) + X

P

(x), which yields

P

= vw

w

− 0 . . . 0 | {z }

q_n

vw

= v w

| {z }

firstq_n

w

− 0 . . . 0 | {z }

b1+qn

w

= vw

w

00 . . . 00 | {z }

(b_n−1−1)q_n−2

(w

w

− w

w

)

= vw

w

∆

.

Therefore, w

= w

w

and w

= ∆

.

Using the results here and Lemma 3(1) with Γ

∈ O, we obtain Lemma 3(2), that is,

−w

w

+ w

w

= − w

w

w

+ w

w

w

= 0 . . . 0 | {z }

anqn−1

0 . . . 0

| {z }

qn−2

∆

= 0 . . . 0 | {z }

qn

∆

.

As long as a

≥ b

for i = 3, 4, . . . , there is no other pattern. But once a

< b

for some n, the pattern [OOC

] follows [OOO] in the sequence S and the loop starts. The situation after this can be seen in Figure 2. “//”

stands for a

≥ b

, “/” for a

> b

, “−” for a

= b

, “\” for a

< b

. Once we encounter C

(or B

, D

) again, the situation after that is the same as the situation after the first C

(or B

, D

).

We shall indicate the loop in all patterns according to the class of Γ

. Some initial cases are omitted, but it is easy to see that they are special cases of the general ones and they are included in them.

4.2. Case Γ

∈ C. From Lemma 1 the possible patterns are [OOC

], [B

A

C

], [A

A

C

], [C

B

C

], [D

B

C

].

• [OOC

]. This follows [OOO] in the sequence S.

Since Γ

= $

. . . $

(−1) (Γ

= (−1) when n = 4), w

is empty and w

= w

w

and w

= ∆

.

If a

= b

, we have P

(x) = (1+X +. . .+X

)P

(x)+X

P

(x).

Since the string of coefficients of

(1 + X + . . . + X

) × x

X((−1)

x

+ (−1)

x

) is

0 . . . 0

| {z }

b₁+β_n−2

( 0 . . . 0 | {z }

qn−1−2

(−1)

(−1)

)

, we obtain

P

= vw

w

+ 0 . . . 0 | {z }

b₁+β_n−2

( 0 . . . 0 | {z }

qn−1−2

(−1)

(−1)

)

.

From Lemma 2(1) with Γ

∈ O we get 0 < β

≤ q

. Therefore, w

is empty and the conclusion of Lemma 2(4) is satisfied.

If a

= b

− 1, we have P

(x) = (1 + X + . . . + X

− x

)P

(x) + X

P

(x). Since the string of coefficients of

(1 + X + . . . + X

− x

) × x

X((−1)

x

+ (−1)

x

) is

0 . . . 0

| {z }

b1+βn−2

( 0 . . . 0 | {z }

qn−1−2

(−1)

(−1)

)

0 . . . 0 | {z }

qn−2−2

(−1)

(−1)

,

Fig. 2