AND LINEAR FORMS OF e IN TERMS OF INTEGRALS

AND LINEAR FORMS OF e IN TERMS OF INTEGRALS

TAKAO KOMATSU

Letpn/qn be thenth convergent ofα= [a0;a1, a2, . . .]. It is known thatpn/qn

approximatesαvery well. In other words,|pn/qn−α|is small and tends to 0 as ntends to infinity. The purpose of this paper is to expresspn−qnαin terms of integrals for a givenα. In special, some typical values of tanh, tan, and linear forms of e are considered. For example, if pn/qn is the nth convergent of the continued fraction expansion ofp

v/utanh(1/√

uv),u, v >0, then p2n−1−

rv

utanh 1

√uvq2n−1=− 1 e^2/√uv+ 1

4 uv

nZ1

0

x²ⁿ⁻¹(x−1)²ⁿ⁻¹ (2n−1)! e^2x/

√uv

dx forn≥1. Ifpn/qn is thenth convergent of the continued fraction expansion of (e + 1)/3, then

p18n+2−e + 1

3 q18n+2= 1 3

Z 1

0

(x+ 1)x⁶ⁿ⁺²(x−1)⁶ⁿ⁺²

(6n+ 2)! e^xdx, n≥0.

AMS 2000 Subject Classification: 11A55, 11J70.

Key words: continued fraction expansion, diophantine approximation.

1. INTRODUCTION

Let α = [a₀;a₁, a₂, . . .] denote the simple continued fraction expansion of a real α, where

α=a₀+ 1/α₁, a₀ =bαc, αn=an+ 1/αn+1, an=bα_nc, n≥1.

Thenth convergent of the continued fraction expansion is denoted byp_n/q_n= [a0;a1, . . . , an], andpnand qn satisfy the recurrence relations

pn=anpn−1+pn−2, n≥0, p−1 = 1, p−2= 0, q_n=a_nqn−1+qn−2, n≥0, q−1= 0, q−2 = 1.

REV. ROUMAINE MATH. PURES APPL.,54(2009),3, 223–242

pn/qn,n= 0,1,2, . . ., are good approximates of α in the sense that |p_n/qn− α| →0 as n→ ∞. More precisely, the equation

p_n−q_nα= (−1)ⁿ⁺¹ αn+1qn+qn−1

does hold (see, e.g., [1, Lemma 5.4]).

Osler [10] investigated the case where α = e^1/s, s≥ 2. When p_n/q_n is the nth convergent of the continued fraction of

e^1/s= [1; (2k−1)s−1,1,1 ]^∞_k=1, he showed that

p_3n−q_3ne^1/s=− 1 sⁿ⁺¹

Z 1 0

xⁿ(x−1)ⁿ

n! e^x/sdx, (1)

p3n+1−q3n+1e^1/s= 1 sⁿ⁺¹

Z 1 0

xⁿ⁺¹(x−1)ⁿ

n! e^x/sdx (2)

and

(3) p_3n+2−q_3n+2e^1/s= 1 sⁿ⁺¹

Z 1 0

xⁿ(x−1)ⁿ⁺¹

n! e^x/sdx

forn≥0. This was a direct extension of a result of Cohn [2] concerning e. In fact, a similar expression was given in [3].

In a similar manner, whenp_n/q_n is thenth convergent of the continued fraction of

e^2/s =

1;(6k−5)s−1

2 ,(12k−6)s,(6k−1)s−1 2 ,1,1

∞

k=1

, where s >1 is odd, for n≥0 we have

p_5n−q_5ne^2/s =− 2

s

3n+1Z 1 0

x³ⁿ(x−1)³ⁿ

(3n)! e^2x/sdx, (4)

p5n+1−q5n+1e^2/s=−2³ⁿ⁺¹ s³ⁿ⁺²

Z 1 0

x³ⁿ⁺¹(x−1)³ⁿ⁺¹

(3n+ 1)! e^2x/sdx, (5)

p5n+2−q5n+2e^2/s =− 2

s

3n+3Z 1 0

x³ⁿ⁺²(x−1)³ⁿ⁺²

(3n+ 2)! e^2x/sdx, (6)

p5n+3−q5n+3e^2/s = 2

s

3n+3Z 1 0

x³ⁿ⁺³(x−1)³ⁿ⁺²

(3n+ 2)! e^2x/sdx, (7)

and (see [7])

(8) p_5n+4−q_5n+4e^2/s = 2

s

3n+3Z ₁

0

x³ⁿ⁺²(x−1)³ⁿ⁺³

(3n+ 2)! e^2x/sdx.

These results were extended to the cases of e^1/s/3,se^1/(ls), and e^1/(ls)/s in [9]. In this paper, we establish a method to express p_n−q_nα in terms of integrals ifα is a function of e and its continued fraction expansion has some kind of regularity like quasi-periodicity. In particular, we shall consider the following values of α

rv

utanh 1

√uv, rv

utan 1

√uv, e^1/s+ 1 3 for u, v >0 and s≥1.

2. MAIN LEMMA

Ifpn/qn denotes the nth convergent of the continued fraction of e^1/s =

1; (2k−1)s−1,1,1∞

k=0, s≥1, then it can be proved by induction that

p_3n=

n

X

k=0

(n+k)!

k!(n−k)!s^k, q_3n=

n

X

k=0

(−1)^n−k (n+k)!

k!(n−k)!s^k, p3n+1 =

n

X

k=0

(n+k+ 1)!

k!(n−k)! s^k+1, q3n+1= (n+1)

n+1

X

k=0

(−1)^n−k+1 (n+k)!

k!(n−k+1)!s^k, p3n+2 = (n+ 1)

n+1

X

k=0

(n+k)!

k!(n−k+ 1)!s^k, q3n+2=

n

X

k=0

(−1)^n−k(n+k+ 1)!

k!(n−k)! s^k+1 for n≥0 (see [6, Example 1]). By substituting these expressions into Osler’s identities (1), (2) and (3), we get the following lemma, which shall be used again and again to obtain the main results in this paper. An independent proof can be found in [9, Theorem 2].

Lemma 1. Forn≥0 we have

n

X

k=0

(n+k)!

k!(n−k)!s^k−e^1/s

n

X

k=0

(−1)^n−k (n+k)!

k!(n−k)!s^k (9)

=− 1 sⁿ⁺¹

Z 1 0

xⁿ(x−1)ⁿ

n! e^x/sdx,

n

X

k=0

(n+k+ 1)!

k!(n−k)! s^k+1−e^1/s(n+ 1)

n+1

X

k=0

(−1)^n−k+1 (n+k)!

k!(n−k+ 1)!s^k (10)

= 1

sⁿ⁺¹ Z 1

0

xⁿ⁺¹(x−1)ⁿ

n! e^x/sdx,

(n+ 1)

n+1

X

k=0

(n+k)!

k!(n−k+ 1)!s^k−e^1/s

n

X

k=0

(−1)^n−k(n+k+ 1)!

k!(n−k)! s^k+1 (11)

= 1

sⁿ⁺¹ Z 1

0

xⁿ(x−1)ⁿ⁺¹

n! e^x/sdx .

3. DIOPHANTINE APPROXIMATION OF tanh Letp_n/q_n be the nth convergent of the continued fraction

rv

utanh 1

√uv =

0; (4k−3)u,(4k−1)v ∞

k=1, u, v >0.

Our main result can be stated as follows.

Theorem 1. For n= 1,2, . . . we have p2n−1−

rv

utanh 1

√uvq2n−1

(12)

=− 1

e^2/√uv+ 1 4

uv nZ 1

0

x²ⁿ⁻¹(x−1)²ⁿ⁻¹ (2n−1)! e^2x/

√uvdx,

p2n− rv

utanh 1

√uvq2n

(13)

=− 2

(e^2/√uv+ 1)u 4

uv nZ 1

0

x²ⁿ(x−1)²ⁿ (2n)! e^2x/

√uvdx .

The proof is based upon a combinatorial expression of the leaping con- vergents together with Lemma 1.

Lemma 2[8, Corollary 1]. Forn= 1,2, . . . we have p2n−1 =

n−1

X

i=0

(2n+ 2i−1)!

(2i)!(2n−2i−1)!

u 2

iv 2

i

,

p2n=

n−1

X

i=0

(2n+ 2i+ 1)!

(2i+ 1)!(2n−2i−1)!

u 2

iv 2

i+1

,

q2n−1 =

n−1

X

i=0

(2n+ 2i)!

(2i+ 1)!(2n−2i−2)!

u 2

i+1v 2

i

, q_2n=

n

X

i=0

(2n+ 2i)!

(2i)!(2n−2i)!

u 2

iv 2

i

.

Theproof is by induction and we omit it.

Proof of Theorem1. First, notice that

2n−1

X

k=0

(2n+k−1)!

k!(2n−k−1)!

√ uv 2

k

=

n−1

X

i=0

(2n+ 2i−1)!

(2i)!(2n−2i−1)!

uv 4

i

+

n−1

X

i=0

(2n+ 2i)!

(2i+ 1)!(2n−2i−2)!

√ uv 4

i√ uv

2 =p2n−1+ rv

uq2n−1

and

2n−1

X

k=0

(−1)^k+1 (2n+k−1)!

k!(2n−k−1)!

√ uv 2

k

=−p_2n−1+ rv

uq2n−1. Replacing nby 2n−1 and sby√

uv/2 in (9), we have

p2n−1+ rv

uq2n−1

−e^2/

√uv

−p_2n−1+ rv

uq2n−1

=− 4

uv nZ 1

0

x²ⁿ⁻¹(x−1)²ⁿ⁻¹ (2n−1)! e^2x/

√uvdx .

Dividing both sides by e^2/

√uv+ 1, we get (12).

Next, notice that

2n

X

k=0

(2n+k)!

k!(2n−k)!

√ uv 2

k

=

n

X

i=0

(2n+ 2i)!

(2i)!(2n−2i)!

uv 4

i

+

n−1

X

i=0

(2n+ 2i+ 1)!

(2i+ 1)!(2n−2i−1)!

√ uv 4

i √ uv

2 =q2n+ ru

v p2n

and

2n

X

k=0

(−1)^k (2n+k)!

k!(2n−k)!

√ uv 2

k

=q2n− ru

v p2n. Replacing nby 2n ands by√

uv/2 in (9), we have

q2n+ ru

v p2n

−e^2/

√uv

q2n−

ru vp2n

=− 2

√uv 4

uv nZ 1

0

x²ⁿ(x−1)²ⁿ (2n)! e^2x/

√uvdx .

Dividing both sides by (e^2/

√uv+ 1)p

u/v, we get (13).

4. DIOPHANTINE APPROXIMATION OF tan Letpn/qn be the nth convergent of the continued fraction rv

utan 1

√uv =

0;u−1,1,(4k−1)v−2,1,(4k+ 1)u−2 ∞

k=1, u, v, >0.

Now, our main result can be stated as follows.

Theorem 2. For n= 1,2, . . . we have p4n−3−

rv

utan 1

√uvq4n−3 = 1 2(e²

√−1/√ uv+ 1)

4 uv

n

(14) ×

× Z 1

0

(2x²−2x+ (2n−1)v)x²ⁿ⁻²(x−1)²ⁿ⁻²

(2n−1)! e²

√−1x/√ uvdx, p4n−2−

rv

utan 1

√uvq4n−2= 1 e²

√−1/√ uv+ 1

4 uv

n

(15) ×

× Z 1

0

x²ⁿ⁻¹(x−1)²ⁿ⁻¹ (2n−1)! e²

√−1x/√ uvdx, p4n−1−

rv

utan 1

√uvq4n−1 =− 2 (e²

√−1/√

uv+ 1)u 4

uv n

(16) ×

× Z 1

0

(x²−x+nu)x²ⁿ⁻¹(x−1)²ⁿ⁻¹

(2n)! e²

√−1x/√ uvdx, p4n−

rv

utan 1

√uvq4n

(17)

=− 2

(e²

√−1/√

uv+ 1)u 4

uv nZ 1

0

x²ⁿ(x−1)²ⁿ (2n)! e²

√−1x/√ uvdx .

The proof is based upon a combinatorial expression of the leaping con- vergents together with Lemma 1 again.

Lemma 3[8, Corollary 2]. Forn= 1,2, . . . we have p4n−3 =

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i−1)!

(2i)!(2n−2i−1)!

u 2

iv 2

i

−

n−2

X

i=0

(−1)ⁿ⁻ⁱ (2n+ 2i−1)!

(2i+ 1)!(2n−2i−3)!

u 2

iv 2

i+1

,

p4n−2 =

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i−1)!

(2i)!(2n−2i−1)!

u 2

iv 2

i

,

p4n−1 =

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i+ 1)!

(2i+ 1)!(2n−2i−1)!

u 2

iv 2

i+1

−

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i−1)!

(2i)!(2n−2i−1)!

u 2

iv 2

i

,

p_4n=

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i+ 1)!

(2i+ 1)!(2n−2i−1)!

u 2

iv 2

i+1

,

q4n−3 =

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i)!

(2i+ 1)!(2n−2i−2)!

u 2

i+1v 2

i

−

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i−2)!

(2i)!(2n−2i−2)!

u 2

iv 2

i

,

q4n−2 =

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i)!

(2i+ 1)!(2n−2i−2)!

u 2

i+1v 2

i

, q4n−1 =

n

X

i=0

(−1)ⁿ⁻ⁱ (2n+ 2i)!

(2i)!(2n−2i)!

u 2

iv 2

i

−

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i)!

(2i+ 1)!(2n−2i−2)!

u 2

i+1v 2

i

, q4n=

n

X

i=0

(−1)ⁿ⁻ⁱ (2n+ 2i)!

(2i)!(2n−2i)!

u 2

iv 2

i

.

Proof of Theorem2. Notice that

(−1)ⁿ⁻¹

2n−1

X

k=0

(2n+k−1)!

k!(2n−k−1)!

√ uv 2√

−1 k

=

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i−1)!

(2i)!(2n−2i−1)!

u 2

iv 2

i

+

n−1

X

i=0

(−1)ⁿ⁻ⁱ⁻¹ (2n+ 2i)!

(2i+ 1)!(2n−2i−2)!

u 2

iv 2

i √ uv 2√

−1

=p4n−2+ rv

u

√1

−1q4n−2

and (−1)ⁿ⁻¹

2n−1

X

k=0

(−1)^k+1 (2n+k−1)!

k!(2n−k−1)!

√ uv 2√

−1 k

=−p4n−2+ rv

u

√1

−1q4n−2. Replacing nby 2n−1 and sby√

uv/(2√

−1) in (9), we obtain

p4n−2+ rv

u

√1

−1q4n−2

−e²

√−1/√ uv

−p_4n−2+ rv

u

√1

−1q4n−2

=−(−1)ⁿ⁻¹(−1)ⁿ 4

uv nZ 1

0

x²ⁿ⁻¹(x−1)²ⁿ⁻¹ (2n−1)! e²

√−1/√ uvdx . Dividing both sides by e²

√−1/√

uv+ 1, we get (15).

In order to get (17), notice that (−1)ⁿ

2n

X

k=0

(2n+k)!

k!(2n−k)!

√ uv 2√

−1 k

=

n

X

i=0

(−1)ⁿ⁻ⁱ (2n+ 2i)!

(2i)!(2n−2i)!

u 2

iv 2

i

+

n−1

X

i=0

(−1)ⁿ⁻ⁱ (2n+ 2i+ 1)!

(2i+1)!(2n−2i−1)!

u 2

iv 2

i √ uv 2√

−1 =q4n+ ru

v

√−1p4n

and

(−1)ⁿ

2n

X

k=0

(−1)^k (2n+k)!

k!(2n−k)!

√ uv 2√

−1 k

=q4n− ru

v

√−1p4n.

Replacing nby 2n ands by√

uv/(2√

−1) in (9), we obtain

q_4n+

ru v

√−1p_4n

−e²

√−1/√ uv

q_4n−

ru v

√−1p_4n

=−(−1)ⁿ(−1)ⁿ 4

uv n

2√

√−1 uv

Z 1 0

x²ⁿ(x−1)²ⁿ (2n)! e²

√−1/√ uvdx . Dividing both sides by (e²

√−1/√

uv+ 1)p u/v√

−1, we get (17).

The identity (16) follows from (15) and (17) because p4n−1−

rv

utan 1

√uvq4n−1

=

p_4n− rv

utan 1

√uvq_4n

−

p4n−2− rv

utan 1

√uvq4n−2

.

The identity (14) also follows from (15) and (2) as p4n−3−

rv

utan 1

√uvq4n−3

=

p4n−2− rv

utan 1

√uvq4n−2

−

p4n−4− rv

utan 1

√uv q4n−4

.

5. DIOPHANTINE APPROXIMATION OF (e^1/s+ 1)/3 In this section we shall consider a linear form ofe. The method developed here can be applied to any linear form of e^1/s, s ≥ 1, e^2/2s+1, s ≥ 0, and any value of a function of e if the corresponding continued fraction can be expressed explicitly as a so-called Hurwitz continued fraction. For example, the continued fraction expansion

e + 1 3 =h

1; 4,5,4k−3,1,1,36k−16,1,1,4k−2,1,1,36k−4, 1,1,4k−1,1,5,4k,1

i∞ k=1

does hold ([4, p. 294, (19)], [12]). In fact, this is a special case of e^1/(3s+1)+ 1

3 =h

0; 1,2,(12k−11)s+ (4k−5),1,5,(12k−9)s+ (4k−4),1,5, (12k−7)s+ (4k−3),1,1,9(12k−5)s+ 4(9k−4),1,1, (12k−3)s+ (4k−2),1,1,9(12k−1)s+ 4(9k−1),1,1

i∞ k=1, where s≥2. For s= 0, by the rule [. . . , a,−b, γ] = [. . . , a−1,1, b−1,−γ], we get the above expression because

e + 1 3 =h

0; 1,2,−1,1,5,0,1,5,4k−3,1,1,36k−16, 1,1,4k−2,1,1,36k−4,1,1,4k−1,1,5,4k,1i∞

k=1. Furthermore, for s= 1 we have

e^1/4+ 1

3 =

h

0; 1,2,0,1,5,16k−13,1,5,16k−10,1,1,

144k−61,1,1,16k−5,1,1,144k−13,1,1,16k,1i∞ k=1

=h

0; 1,3,5,16k−13,1,5,16k−10,1,1,

144k−61,1,1,16k−5,1,1,144k−13,1,1,16k,1i∞ k=1.

There are two more different patterns having the same form. Lets≥2.

Then

e^1/(3s)+ 1

3 =

h

0; 1,2,(4k−3)s−1,1,1,9(4k−1)s−1,1,1 i∞

k=1. For s= 1 by the rule [. . . , a,0, b, . . .] = [. . . , a+b, . . .], we have

e^1/3+ 1

3 =h

0; 1,2,0,1,1,9(4k−1)−1,1,1,4k,1i∞ k=1

=h

0; 1,3,1,36k−10,1,1,4k,1i∞ k=1. Lets≥1. Then

e^1/(3s+2)+ 1

3 =

h

0; 1,2,(12k−11)s+ (8k−8),5,1,(12k−9)s+ (8k−7),5,1, (12k−7)s+ (8k−6),1,1,9(12k−5)s+ (72k−31),1,1, (12k−3)s+ (8k−3),1,1,9(12k−1)s+ (72k−7),1,1i∞

k=1. For s= 0 we have

e^1/2+ 1

3 =h

0; 1,2,0,5,1,8k−7,5,1,8k−6,1,1,

72k−31,1,1,8k−3,1,1,72k−7,1,1,8k,5i∞ k=1

= h

0; 1,7,1,8k−7,5,1,8k−6,1,1,

72k−31,1,1,8k−3,1,1,72k−7,1,1,8k,5i∞ k=1. Now, our main results can be stated in the following three theorems.

Theorem 3. Let pn/qn be the nth convergent of the continued fraction expansion of (e^1/(3s)+ 1)/3, s≥2. Then, forn≥0,

p_6n+2−e^1/(3s)+ 1

3 q_6n+2 =− 1 (3s)²ⁿ⁺¹

Z 1 0

x²ⁿ(x−1)²ⁿ

(2n)! e^x/(3s)dx, (18)

p6n+3− e^1/(3s)+ 1

3 q6n+3= 1 3(3s)²ⁿ⁺¹

Z 1 0

(x+1)x²ⁿ(x−1)²ⁿ

(2n)! e^x/(3s)dx, (19)

p6n+4− e^1/(3s)+ 1

3 q6n+4= 1 3(3s)²ⁿ⁺¹

Z 1 0

(x−2)x²ⁿ(x−1)²ⁿ

(2n)! e^x/(3s)dx, (20)

p_6n+5− e^1/(3s)+ 1

3 q_6n+5=− 1 3(3s)²ⁿ⁺²

Z 1 0

x²ⁿ⁺¹(x−1)²ⁿ⁺¹

(2n+ 1)! e^x/(3s)dx, (21)

p_6n+6−e^1/(3s)+ 1

3 q_6n+6 = 1 3(3s)²ⁿ⁺²

Z ₁

0

(3x−1)x²ⁿ⁺¹(x−1)²ⁿ⁺¹

(2n+ 1)! e^x/(3s)dx, (22)

p_6n+7−e^1/(3s)+ 1

3 q_6n+7 = 1 3(3s)²ⁿ⁺²

Z ₁

0

(3x−2)x²ⁿ⁺¹(x−1)²ⁿ⁺¹

(2n+ 1)! e^x/(3s)dx . (23)

Remark 1. For s = 1, denote by p^∗_n/q^∗_n, the nth convergent of the continued fraction expansion of (e^1/3+ 1)/3. Then every equation in Theo- rem 3 holds if p_n+2 and q_n+2 are replaced by p^∗_n and q^∗_n, respectively, for n= 0,1,2, . . ..

Theorem 4. Let pn/qn be the nth convergent of the continued fraction expansion of (e^1/(3s+1)+1)/3, s≥2. Then, for n≥0,

p_18n+2−e^1/(3s+1)+1

3 q_18n+2=− 1

(3s+1)⁶ⁿ⁺¹ Z 1

0

x⁶ⁿ(x−1)⁶ⁿ

(6n)! e^x/(3s+1)dx, (24)

p_18n+3−e^1/(3s+1)+1

3 q_18n+3= 1

3(3s+1)⁶ⁿ⁺¹ Z 1

0

(x+2)x⁶ⁿ(x−1)⁶ⁿ

(6n)! e^x/(3s+1)dx, (25)

p_18n+4−e^1/(3s+1)+1

3 q_18n+4= 1

3(3s+1)⁶ⁿ⁺¹ Z 1

0

x⁶ⁿ(x−1)⁶ⁿ⁺¹

(6n)! e^x/(3s+1)dx, (26)

p_18n+5−e^1/(3s+1)+1

3 q_18n+5 =− 1

(3s+1)⁶ⁿ⁺² Z 1

0

x⁶ⁿ⁺¹(x−1)⁶ⁿ⁺¹

(6n+1)! e^x/(3s+1)dx, (27)

p_18n+6−e^1/(3s+1)+1

3 q_18n+6= (28)

= 1

3(3s+1)⁶ⁿ⁺² Z 1

0

(x+2)x⁶ⁿ⁺¹(x−1)⁶ⁿ⁺¹

(6n+1)! e^x/(3s+1)dx,

p18n+7−e^1/(3s+1)+1

3 q18n+7= 1

3(3s+1)⁶ⁿ⁺² Z 1

0

x⁶ⁿ⁺¹(x−1)⁶ⁿ⁺²

(6n+1)! e^x/(3s+1)dx, (29)

p_18n+8−e^1/(3s+1)+ 1

3 q_18n+8= (30)

=− 1

(3s+ 1)⁶ⁿ⁺³ Z 1

0

x⁶ⁿ⁺²(x−1)⁶ⁿ⁺²

(6n+ 2)! e^x/(3s+1)dx,

p18n+9−e^1/(3s+1)+ 1

3 q18n+9= (31)

= 1

3(3s+ 1)⁶ⁿ⁺³ Z 1

0

(x+ 1)x⁶ⁿ⁺²(x−1)⁶ⁿ⁺²

(6n+ 2)! e^x/(3s+1)dx, p_18n+10−e^1/(3s+1)+ 1

3 q_18n+10= (32)

= 1

3(3s+ 1)⁶ⁿ⁺³ Z 1

0

(x−2)x⁶ⁿ⁺²(x−1)⁶ⁿ⁺²

(6n+ 2)! e^x/(3s+1)dx, p18n+11−e^1/(3s+1)+ 1

3 q18n+11= (33)

=− 1

3(3s+ 1)⁶ⁿ⁺⁴ Z 1

0

x⁶ⁿ⁺³(x−1)⁶ⁿ⁺³

(6n+ 3)! e^x/(3s+1)dx, p_18n+12−e^1/(3s+1)+ 1

3 q_18n+12= (34)

= 1

3(3s+ 1)⁶ⁿ⁺⁴ Z 1

0

(3x−1)x⁶ⁿ⁺³(x−1)⁶ⁿ⁺³

(6n+ 3)! e^x/(3s+1)dx, p18n+13−e^1/(3s+1)+ 1

3 q18n+13= (35)

= 1

3(3s+ 1)⁶ⁿ⁺⁴ Z 1

0

(3x−2)x⁶ⁿ⁺³(x−1)⁶ⁿ⁺³

(6n+ 3)! e^x/(3s+1)dx, p_18n+14−e^1/(3s+1)+ 1

3 q_18n+14= (36)

=− 1

(3s+ 1)⁶ⁿ⁺⁵ Z 1

0

x⁶ⁿ⁺⁴(x−1)⁶ⁿ⁺⁴

(6n+ 4)! e^x/(3s+1)dx, p_18n+15−e^1/(3s+1)+ 1

3 q_18n+15= (37)

= 1

3(3s+ 1)⁶ⁿ⁺⁵ Z 1

0

(x+ 1)x⁶ⁿ⁺⁴(x−1)⁶ⁿ⁺⁴

(6n+ 4)! e^x/(3s+1)dx, p18n+16−e^1/(3s+1)+ 1

3 q18n+16= (38)

= 1

3(3s+ 1)⁶ⁿ⁺⁵ Z 1

0

(x−2)x⁶ⁿ⁺⁴(x−1)⁶ⁿ⁺⁴

(6n+ 4)! e^x/(3s+1)dx, p_18n+17−e^1/(3s+1)+ 1

3 q_18n+17= (39)

=− 1

3(3s+ 1)⁶ⁿ⁺⁶ Z 1

0

x⁶ⁿ⁺⁵(x−1)⁶ⁿ⁺⁵

(6n+ 5)! e^x/(3s+1)dx,

p18n+18−e^1/(3s+1)+ 1

3 q18n+18= (40)

= 1

3(3s+ 1)⁶ⁿ⁺⁶ Z 1

0

(3x−1)x⁶ⁿ⁺⁵(x−1)⁶ⁿ⁺⁵

(6n+ 5)! e^x/(3s+1)dx, p_18n+19−e^1/(3s+1)+ 1

3 q_18n+19= (41)

= 1

3(3s+ 1)⁶ⁿ⁺⁶ Z 1

0

(3x−2)x⁶ⁿ⁺⁵(x−1)⁶ⁿ⁺⁵

(6n+ 5)! e^x/(3s+1)dx .

Remark 2. Fors= 1, denote byp^∗_n/q_n^∗, thenth convergent of the contin- ued fraction expansion of (e^1/4+1)/3. Then every equation in Theorem 4 holds ifpn+2 and qn+2 are replaced by p^∗_n and q_n^∗, respectively, for n= 0,1,2, . . ..

Fors= 0, denote byp^∗∗_n/q^∗∗_n , thenth convergent of the continued fraction expansion of (e^1/3+ 1)/3. Then every equation in Theorem 4 holds ifp_n+6 and qn+6 are replaced byp^∗_nand q_n^∗, respectively, for n= 0,1,2, . . .. For example, for n≥0 we have

p^∗∗_18n+3−e+ 1

3 q^∗∗_18n+3 =p_18n+9−e^1/(3s+1)+ 1 3 q_18n+9

= 1 3

Z 1 0

(x+ 1)x⁶ⁿ⁺²(x−1)⁶ⁿ⁺² (6n+ 2)! e^xdx .

Theorem 5. Let pn/qn be the nth convergent of the continued fraction expansion of (e^1/(3s+2)+ 1)/3, s≥1. Then, forn≥0,

p_18n+2−e^1/(3s+2)+ 1

3 q_18n+2=− 1

(3s+ 1)⁶ⁿ⁺¹ Z 1

0

x⁶ⁿ(x−1)⁶ⁿ

(6n)! e^x/(3s+2)dx, (42)

p_18n+3−e^1/(3s+2)+ 1

3 q_18n+3 = 1

3(3s+ 2)⁶ⁿ⁺¹ Z 1

0

x⁶ⁿ⁺¹(x−1)⁶ⁿ

(6n)! e^x/(3s+2)dx, (43)

p_18n+4−e^1/(3s+2)+ 1

3 q_18n+4= (44)

= 1

3(3s+ 2)⁶ⁿ⁺¹ Z 1

0

(5x−3)x⁶ⁿ(x−1)⁶ⁿ

(6n)! e^x/(3s+2)dx, p_18n+5−e^1/(3s+2)+ 1

3 q_18n+5= (45)

=− 1

(3s+ 2)⁶ⁿ⁺² Z 1

0

x⁶ⁿ⁺¹(x−1)⁶ⁿ⁺¹

(6n+ 1)! e^x/(3s+2)dx,

p18n+6−e^1/(3s+2)+1

3 q18n+6 = (46)

= 1

3(3s+2)⁶ⁿ⁺² Z 1

0

x⁶ⁿ⁺²(x−1)⁶ⁿ⁺¹

(6n+ 1)! e^x/(3s+2)dx, p_18n+7−e^1/(3s+2)+ 1

3 q_18n+7= (47)

= 1

3(3s+ 2)⁶ⁿ⁺² Z 1

0

(5x−3)x⁶ⁿ⁺¹(x−1)⁶ⁿ⁺¹

(6n+ 1)! e^x/(3s+2)dx, p18n+8−e^1/(3s+2)+ 1

3 q18n+8= (48)

=− 1

(3s+ 2)⁶ⁿ⁺³ Z 1

0

x⁶ⁿ⁺²(x−1)⁶ⁿ⁺²

(6n+ 2)! e^x/(3s+2)dx, p_18n+9−e^1/(3s+2)+ 1

3 q_18n+9= (49)

= 1

3(3s+ 2)⁶ⁿ⁺³ Z 1

0

(x+ 1)x⁶ⁿ⁺²(x−1)⁶ⁿ⁺²

(6n+ 2)! e^x/(3s+2)dx, p18n+10−e^1/(3s+2)+ 1

3 q18n+10= (50)

= 1

3(3s+ 2)⁶ⁿ⁺³ Z 1

0

(x−2)x⁶ⁿ⁺²(x−1)⁶ⁿ⁺²

(6n+ 2)! e^x/(3s+2)dx, p_18n+11−e^1/(3s+2)+ 1

3 q_18n+11= (51)

=− 1

3(3s+ 2)⁶ⁿ⁺⁴ Z 1

0

x⁶ⁿ⁺³(x−1)⁶ⁿ⁺³

(6n+ 3)! e^x/(3s+2)dx, p_18n+12−e^1/(3s+2)+ 1

3 q_18n+12= (52)

= 1

3(3s+ 2)⁶ⁿ⁺⁴ Z 1

0

(3x−1)x⁶ⁿ⁺³(x−1)⁶ⁿ⁺³

(6n+ 3)! e^x/(3s+2)dx, p18n+13−e^1/(3s+2)+ 1

3 q18n+13= (53)

= 1

3(3s+ 2)⁶ⁿ⁺⁴ Z 1

0

(3x−2)x⁶ⁿ⁺³(x−1)⁶ⁿ⁺³

(6n+ 3)! e^x/(3s+2)dx, p_18n+14−e^1/(3s+2)+ 1

3 q_18n+14= (54)

=− 1

(3s+ 2)⁶ⁿ⁺⁵ Z 1

0

x⁶ⁿ⁺⁴(x−1)⁶ⁿ⁺⁴

(6n+ 4)! e^x/(3s+2)dx,

p18n+15−e^1/(3s+2)+ 1

3 q18n+15= (55)

= 1

3(3s+ 2)⁶ⁿ⁺⁵ Z 1

0

(x+ 1)x⁶ⁿ⁺⁴(x−1)⁶ⁿ⁺⁴

(6n+ 4)! e^x/(3s+2)dx, p_18n+16−e^1/(3s+2)+ 1

3 q_18n+16= (56)

= 1

3(3s+ 2)⁶ⁿ⁺⁵ Z 1

0

(x−2)x⁶ⁿ⁺⁴(x−1)⁶ⁿ⁺⁴

(6n+ 4)! e^x/(3s+2)dx, p18n+17−e^1/(3s+2)+ 1

3 q18n+17= (57)

=− 1

3(3s+ 2)⁶ⁿ⁺⁶ Z 1

0

x⁶ⁿ⁺⁵(x−1)⁶ⁿ⁺⁵

(6n+ 5)! e^x/(3s+2)dx, p18n+18−e^1/(3s+2)+ 1

3 q18n+18= (58)

= 1

3(3s+ 2)⁶ⁿ⁺⁶ Z 1

0

(3x−1)x⁶ⁿ⁺⁵(x−1)⁶ⁿ⁺⁵

(6n+ 5)! e^x/(3s+2)dx, p_18n+19−e^1/(3s+2)+ 1

3 q_18n+19= (59)

= 1

3(3s+ 2)⁶ⁿ⁺⁶ Z 1

0

(3x−2)x⁶ⁿ⁺⁵(x−1)⁶ⁿ⁺⁵

(6n+ 5)! e^x/(3s+2)dx .

Remark. Fors= 0, denote byp^∗_n/q_n^∗ thenth convergent of the continued fraction expansion of (e^1/2+ 1)/3. Then every equation in Theorem 5 holds if p_n+2 and q_n+2 are replaced by p^∗_n and q_n^∗, respectively, for n= 0,1,2, . . ..

Note that equations (43), (44), (46) and (47) in Theorem 5 are a little different from equations (25), (26), (28) and (29) in Theorem 4, respectively.

Other equations have the same forms which differ only by having 3s+2 instead of 3s+ 1.

6. PROOF OF THEOREMS 3–5

We shall only prove Theorem 3. The proofs of Theorems 4 and 5 are similar. Lemma 1 is used again and again combined with a combinatorial expression of leaping convergents of e^1/(3s) as shown in the following lemma.

The proof of this lemma is by induction on nand we omit it.

Lemma 4. Let pn/qn be the nth convergent of the continued fraction expansion of (e^1/(3s)+ 1)/3. Then, forn≥0,

p6n+2 = 2

n

X

i=0

(2n+ 2i)!

(2i)!(2n−2i)!(3s)²ⁱ, p6n+3 =

n

X

i=0

(2n+ 2i+ 2)!

(2i+ 1)!(2n−2i)!

(3s)²ⁱ⁺¹

3 −

n

X

i=0

(2n+ 2i)!

(2i)!(2n−2i)!(3s)²ⁱ, p6n+4 =

n

X

i=0

(2n+ 2i+ 2)!

(2i+ 1)!(2n−2i)!

(3s)²ⁱ⁺¹

3 +

n

X

i=0

(2n+ 2i)!

(2i)!(2n−2i)!(3s)²ⁱ, p_6n+5 = 2

n

X

i=0

(2n+ 2i+ 2)!

(2i+ 1)!(2n−2i)!

(3s)²ⁱ⁺¹

3 ,

p6n+6 =−

n

X

i=0

(2n+ 2i+ 2)!

(2i+ 1)!(2n−2i)!

(3s)²ⁱ⁺¹

3 +

n+1

X

i=0

(2n+ 2i+ 2)!

(2i)!(2n−2i+ 2)!(3s)²ⁱ, p_6n+7 =

n

X

i=0

(2n+ 2i+ 2)!

(2i+ 1)!(2n−2i)!

(3s)²ⁱ⁺¹

3 +

n+1

X

i=0

(2n+ 2i+ 2)!

(2i)!(2n−2i+ 2)!(3s)²ⁱ, and

q6n+2 = 3

2n

X

k=0

(−1)^k (2n+k)!

k!(2n−k)!(3s)^k, q_6n+3 =

2n+1

X

k=0

(−1)^k+1(4n−k+ 2)(2n+k)!

k!(2n−k+ 1)! (3s)^k, q_6n+4 =

2n+1

X

k=0

(−1)^k(2n−2k+ 1)(2n+k)!

k!(2n−k+ 1)! (3s)^k, q_6n+5 =

2n+1

X

k=0

(−1)^k+1 (2n+k+ 1)!

k!(2n−k+ 1)!(3s)^k, q6n+6 =

2n+2

X

k=0

(−1)^k(4n+k+ 4)(2n+k+ 1)!

k!(2n−k+ 2)! (3s)^k, q6n+7 = 2

2n+2

X

k=0

(−1)^k(n+k+ 1)(2n+k+ 1)!

k!(2n−k+ 2)! (3s)^k.

Proof of Theorem3. Replacenby 2nandsby 3sin Lemma 1 (9). Then by the combinatorial expressions of p_6n+2 and q_6n+2 in Lemma 4 we have

− 1 (3s)²ⁿ⁺¹

Z 1 0

x²ⁿ(x−1)²ⁿ

(2n)! e^1/(3s)dx=

2n

X

k=0

(2n+k)!

k!(2n−k)!(3s)^k+ +

2n

X

k=0

(−1)^k (2n+k)!

k!(2n−k)!(3s)^k−e^1/(3s)+ 1

3 ·3

2n

X

k=0

(−1)^k (2n+k)!

k!(2n−k)!(3s)^k=

=p_6n+2−e^1/(3s)+ 1 3 q_6n+2, which is (18).

Replacenby 2n+ 1 andsby 3sand divide both sides by 3 in Lemma 1 (9). Then, using the expressions of p6n+5 andq6n+5 in Lemma 4, we get (21).

Replacenby 2nand sby 3sin (10) and (11) in order to get (19). Then e^1/(3s)+ 1

3 q_6n+3+ 1 3(3s)²ⁿ⁺¹

Z 1 0

(x+ 1)x²ⁿ(x−1)²ⁿ

(2n)! e^x/(3s)dx=

= e^1/(3s)+ 1

3 ·2(2n+ 1)

2n+1

X

k=0

(−1)^k+1 (2n+k)!

k!(2n−k+ 1)!(3s)^k+

+ 2

3(3s)²ⁿ⁺¹ Z 1

0

x²ⁿ⁺¹(x−1)²ⁿ

(2n)! e^x/(3s)dx−

−e^1/(3s)+ 1 3

2n

X

k=0

(−1)^k(2n+k+ 1)!

k!(2n−k)! (3s)^k+1−

− 1 3(3s)²ⁿ⁺¹

Z 1 0

x²ⁿ(x−1)²ⁿ⁺¹

(2n)! e^x/(3s)dx=

= 2 3

2n

X

k=0

(2n+k+ 1)!

k!(2n−k)! (3s)^k+1+2

3(2n+ 1)

2n+1

X

k=0

(−1)^k+1 (2n+k)!

k!(2n−k+ 1)!(3s)^k−

−2n+ 1 3

2n+1

X

k=0

(2n+k)!

k!(2n−k+ 1)!(3s)^k−1 3

2n

X

k=0

(−1)^k(2n+k+ 1)!

k!(2n−k)! (3s)^k+1 =

= 1 3

n

X

i=0

(2n+ 2i+ 1)!

(2i)!(2n−2i)!(3s)²ⁱ⁺¹+2n+ 1 3

n

X

i=0

(2n+ 2i+ 1)!

(2i+ 1)!(2n−2i)!(3s)²ⁱ⁺¹+ +

n

X

i=0

(2i)(2n+ 2i)!

(2i)!(2n−2i+1)!(3s)²ⁱ−(2n+1)

n

X

i=0

(2n+ 2i)!

(2i)!(2n−2i+1)!(3s)²ⁱ=p_6n+3.

In a similar manner, replacenby 2nandsby 3sin (9) and (11) in order to get (20), replace n by 2n+ 1 and s by 3sin (10) and (11) in order to get (22), and replacenby 2n+ 1 andsby 3sin (10) and (11) in order to get (23).

Theorems 4 and 5 can be similarly proved using the corresponding com- binatorial expressions from Lemma 4. If we let p_n/q_n be the nth convergent of the continued fraction expansion of (e^1/(3s+1)+ 1)/3 then, for example,

p18n+2= 2

3n

X

i=0

(6n+ 2i)!

(2i)!(6n−2i)!(3s+ 1)²ⁱ, q_18n+2= 3

6n

X

k=0

(−1)^k(6n+k)!

k!(6n−k)! (3s+ 1)^k, p18n+3=−5

3

3n

X

i=0

(6n+ 2i)!

(2i)!(6n−2i)!(3s+1)²ⁱ+1 3

3n

X

i=0

(6n+ 2i+ 2)!

(2i+1)!(6n−2i)!(3s+1)²ⁱ⁺¹, q_18n+3=

6n+1

X

k=0

(−1)^k−1(18n−2k+ 3)(6n+k)!

k!(6n−k+ 1)! (3s+ 1)^k, p_18n+4= 1

3

3n

X

i=0

(6n+ 2i)!

(2i)!(6n−2i)!(3s+ 1)²ⁱ+1 3

3n

X

i=0

(6n+ 2i+ 2)!

(2i+1)!(6n−2i)!(3s+ 1)²ⁱ⁺¹, q18n+4=

6n

X

k=0

(−1)^k(6n+k+ 1)!

k!(6n−k)! (3s+ 1)^k+1, p_18n+5= 2

3n

X

i=0

(6n+ 2i+ 2)!

(2i+ 1)!(6n−2i)!(3s+ 1)²ⁱ⁺¹, q18n+5= 3

6n+1

X

k=0

(−1)^k−1(6n+k+ 1)!

k!(6n−k+ 1)! (3s+ 1)^k.

If we letpn/qnbe thenth convergent of the continued fraction expansion of (e^1/(3s+2)+ 1)/3 then, for example,

p_18n+15=−

3n+2

X

i=0

(6n+ 2i+ 4)!

(2i)!(6n−2i+ 4)!(3s+ 2)²ⁱ+ +1

3

3n+2

X

i=0

(6n+ 2i+ 6)!

(2i+ 1)!(6n−2i+ 4)!(3s+ 2)²ⁱ⁺¹, q18n+15=

6n+5

X

k=0

(−1)^k−1(12n−k+ 10)(6n+k+ 4)!

k!(6n−k+ 5)! (3s+ 2)^k,

p18n+16=

3n+2

X

i=0

(6n+ 2i+ 4)!

(2i)!(6n−2i+ 4)!(3s+ 2)²ⁱ+ +1

3

3n+2

X

i=0

(6n+ 2i+ 6)!

(2i+ 1)!(6n−2i+ 4)!(3s+ 2)²ⁱ⁺¹, q_18n+16=

6n+5

X

k=0

(−1)^k(6n−2k+ 5)(6n+k+ 4)!

k!(6n−k+ 5)! (3s+ 2)^k, p_18n+17= 2

3

3n+2

X

i=0

(6n+ 2i+ 6)!

(2i+ 1)!(6n−2i+ 4)!(3s+ 2)²ⁱ⁺¹, q18n+17=

6n+5

X

k=0

(−1)^k−1(6n+k+ 5)!

k!(6n−k+ 5)! (3s+ 2)^k, p_18n+18=

3n+3

X

i=0

(6n+ 2i+ 6)!

(2i)!(6n−2i+ 6)!(3s+ 2)²ⁱ−

−1 3

3n+2

X

i=0

(6n+ 2i+ 6)!

(2i+ 1)!(6n−2i+ 4)!(3s+ 2)²ⁱ⁺¹, q_18n+18=

6n+6

X

k=0

(−1)^k(12n+k+ 12)(6n+k+ 5)!

k!(6n−k+ 6)! (3s+ 2)^k, p_18n+19=

3n+3

X

i=0

(6n+ 2i+ 6)!

(2i)!(6n−2i+ 6)!(3s+ 2)²ⁱ+ +1

3

3n+2

X

i=0

(6n+ 2i+ 6)!

(2i+ 1)!(6n−2i+ 4)!(3s+ 2)²ⁱ⁺¹, q18n+19=

6n+6

X

k=0

(−1)^k(6n+ 2k+ 6)(6n+k+ 5)!

k!(6n−k+ 6)! (3s+ 2)^k.

Acknowledgements.This research was partially supported by the Grant-in-Aid for Scientific Research (C) (No. 18540006), the Japan Society for the Promotion of Science.

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[6] T. Komatsu,Some combinatorial properties of the leaping convergents. In: B. Landman, M. Nathanson, J. Neˇsetˇril, R. Nowakowski and C. Pomerance (Eds.), Combinatorial Number Theory, Proceedings of the Integers Conference 2005 in Celebration of the 70th Birthday of Ronald Graham, held at the university of West Georgia, Carrollton, Georgia, USA, October 27–30, 2005, pp. 315–325. Walter de Gruyter, Berlin, 2007.

[7] T. Komatsu,A proof of the continued fraction expansion ofe^2/s. Integers7(1) (2007), A30, 8 pp. (electronic)

[8] T. Komatsu,Leaping convergents of Hurwitz continued fractions. In: Diophantine Anal- ysis and Related Fields(DARF2007/2008), AIP Conf. Proc.976, pp. 130–143. Amer.

Inst. Phys., Melville, NY, 2008.

[9] T. Komatsu, A diophantine approximation ofe^1/s in terms of integrals. To appear in Tokyo J. Math.

[10] T. Osler, A proof of the continued fraction expansion ofe^1/M. Amer. Math. Monthly 113(2006), 62–66.

[11] O. Perron, Die Lehre von den Kettenbr¨uchen, 2nd Ed. Chelsea Publishing Co., New York, 1950.

[12] D.S. Thakur,Patterns of continued fractions for the analogues ofeand related numbers in the function field case. J. Number Theory66(1997), 129–147.

Received 16 August 2008 Hirosaki University

Revised 13 January 2009 Graduate School of Science and Technology Hirosaki, 036-8561, Japan

komatsu@cc.hirosaki-u.ac.jp