shortest semiregular continued fractions

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§1. SHORTEST SEMIREGULAR CONTINUED FRACTIONS BY M. S. SRINIVASAN* (Research Student, University of Madras) Received January, 10, 1952 (Communicated by Dr. R. Vaidyanathaswamy, v.A.sc.) The shortest continued fraction.- By a semiregular continued .fraction (C. F.) of a rational number p/q, we understand a development of the form p _ _ 01 0 2 O_ --a°+ al+ a o + .... a~ or (aoO l, alO.,, a,'_~O,', a ~ where aTs are positive integers, ai+O +l> l i< r); in particular, 91 = -F 1 when a o = 0 and O,.+z-----+ 1 when a i = I. (1) Further, if we carry out the division-algorithm in such a way that each. remainder t[I Oi+l ] <.½, we get the -~ Nearest Integer Continued Fraction (N.I.C.F.). Perron has shown that there is a unique N.I.C.F. development for every rational number p/q with the possible exception of the case when the final quotient is 2. For the case a,'----=2, there are obviously two developments, since 0,'_ 1 0,. _ 0,._ 1 O,. (0 i = 4- 1). a,_l -F- 2 -- a~_l + 0,'--2 2 -- Perron has also remarked that the N.I.C.F. is shortest, in the sense that no other semiregular C.F. of p/q has a smaller number of partial quotients. * I am indebted to Dr. R. Vaidyanathaswamy and Dr. M. Venkataraman for their guidance. t Perron, Die Lehre yon Kettenbriichen. Kettenbriiche der nach ndchten ganzen, p. 168. 224

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§1.

SHORTEST SEMIREGULAR CONTINUED FRACTIONS

BY M. S. SRINIVASAN* (Research Student, University of Madras)

Received January, 10, 1952 (Communicated by Dr. R. Vaidyanathaswamy, v.A.sc.)

The shortest continued f r a c t i o n . -

By a semiregular continued .fraction (C. F.) of a rational number p/q, we understand a development of the form

p _ _ 01 0 2 O_ - - a ° + a l + a o + . . . . a~

o r

(aoO l, alO.,, a,'_~O,', a ~

where aTs are positive integers,

ai+O +l> l i < r); in particular, 91 = -F 1 when a o = 0 and O,.+z ----- + 1 when a i = I.

(1)

Further, if we carry out the division-algorithm in such a way that each. remainder

t[I Oi+l ] <.½,

we get the -~ Nearest Integer Continued Fraction (N.I.C.F.). Perron has shown that there is a unique N.I.C.F. development for every rational number p/q with the possible exception of the case when the final quotient is 2. For the case a,'----= 2, there are obviously two developments, since

0,'_ 1 0,. _ 0,._ 1 O,. ( 0 i = 4 - 1) . a,_l -F- 2 - - a~_l + 0,'--2 2 --

Perron has also remarked that the N.I.C.F. is shortest, in the sense that no other semiregular C.F. of p/q has a smaller number of partial quotients.

* I am indebted to Dr. R. Vaidyanathaswamy and Dr. M. Venkataraman for their guidance.

t Perron, Die Lehre yon Kettenbriichen. Kettenbriiche der nach ndchten ganzen, p. 168.

224

Shortest Semffegular Continued Fractions 225

In fact, there are C.F.'s for some rational numbers p/q, other than the N.I.C.F., which are shortest. For example,

11 1 1 1 1 1 1 28 3 - - 2 + 5 2 + 2 - - 6

I call these shortest continued fractions. The nature of these does not seem to have been investigated. In this note I obtain the conditions that a semi- regular C.F. may be shortest. I also show how to obtain the number of shortest C.F.'s for p/q in terms of Fibonacci numbers.

Perron shows that the singular C.F. for a rational number exists and is unique, but fails to note that it is a case of the shortest C.F. I prove this below and obtain the singular C.F. by a method different from Perron's.

THEOREM 1.--In the N.LC.F., the partial quotients 1 + , 2 -- do not occur, and conversely.

Proof--For, in the N.I.C.F., the partial quotients at each stage is the nearest integer and therefore the remainder

0i+1 I~ ½"

Hence a i + 1 ~ 2 ; and either ai+ 1 = 2, Oi+ 2 = -~- 1, or ai+ I ~ 3. Conversely if these conditions are satisfied, the remainder at each stage is ~< ½. When the remainder is -----½ in absolute value, the final stage is reached. Hence the theorem.

§ 2. Transformation o f semiregular C.F.--

Now, given any semiregular C.F. (boOa, b~O, . . . . . b~) for p/q, we pro- ceed to show that it can be transformed into the N.I.C.F. for p[q, without increasing the number o f partial quotients. (It will follow that the N.I.C.F. for p/q is a shortest C.F. in the sense that no other semiregutar C.F. forp/q contains a smaller number of partial quotients).

For this purpose, we have, by Theorem 1 to effect the transformation in such a way that any 1 + or 2 - which occur among the partial quotients in

(boO1, b~O2, . . . . b~) (2) are removed.

226 M. S. SRINIVASAN

More preciscly, we prove that, if (2) contains the partial quotients

1 ~ l _ I 1 1 1 1 (a) ~-~ (b) _ - - 2 (c) 2- -2-k- (c') ~ 2

1

1 1 (e) 2--- a i (a 4- 4= 2 4- or 3 - ),

the 1 -k and 2 -- in these can be transformed away in such a way that there is a reduction in the number of partial quotients in cases (a), (b), (c) (c'), (d), (d') and no reduction in case (e).

Proof--Firs t ly , for the case (e), we apply the transformation

1 0 1 1 0 1 - - ( H ) a q - ~ - - x a + O - - 2 + x - - l "

Consider a C.F. in which this sequence occurs, with the comple te quotient x ~ 3 [ so that the sequence comes under (e) ] and either a > 2; or a = 2, 0 = -k 1. The identity (H) shows that the partial quotients can be trans- formed without reducing the length, so that 2 - - i s replaced by 2 -k and no new 2 -- is introduced. This transforms (e) at every stage where (e) occurs, without reducing the length.

Next for reducing (a), (b), (c), (e'), (d), (d') respectively, the following transformations, each of which reduces the number o f partial quotients by one [except (e') where the length is reduced by k] are applied : - -

Oi+ 1 1 = (ai + Oi+O (a) a i + 1 + ai+9.-~-~

0 1 0 (b) x - k ~ _ ~ = (x + O) - - 3 ;

1 0 1 1 (c) a + 2 - - 2 + x --

( c ' ) x -

which by (H) - -

Oi+ 1 x + 1 '

1 0 1 1 a-k- 0 - - ~ ~ by (H)

1 0 1 - - a + 0 - - 3 - - x + 1 by (a) .

1 0 /" 1 '~ 1 1 1 a + ~k2- - ' - - ]K_22- - 2- - x

1 0 I ' I N 1 1 1 a+2-'--~,)2---- k_21 + 2 + x - - I

. 1 1 1 1 1 1 1 ( 3 ~ ) , 1 a + 3-- 3 - 3 - 3 - b is written as~--~ ~ - .

Shortest Semiregular Continued Fractions 227

1 0 ( 1 ) 1 1 1 which by(a)-- a + ~ x . 2 ~ - - k - a i T 3 + x - - 1

. . . . . . . . . . . . . . . . . . . . . . . . " . ° . . . . . . . . . . . . . . . . . . . .

__ 1 O t ' l " ~ 1 1 1 a + 2 - - \ ~ ) o l + k + x - - 1

1 0 1 1 1 a + ~ i - T k + x - - 1

1 0 1 1 a + l + k + l + x - - I

1 0 1 a + O--k @-'2+ x - - l "

But

(d) , o ( L ) ~ ~_ ~ o ~ ( ~ ) , a + 2 - - k 2 ± x a + ~ - g--£- ~-- z,_.12-l-x

1 0 1 1 ( 1 ) 1 1 which by(H) - - a + 0 _ 2 + 2 _ 3 _ ~---- ~-224- x

_ 1 0 1 l ( 1 ) z , . - 1 1 a + O - - ) - ~ - - 2 - + ~ ~ ~ 2-4-

__ 1 0 ( 1 ) 1 1 ( 1 ) ~ 1 1 a + 0 2 3 ~- ~ ~ 2 + 2 - - ~---- 324-x

. . . . . . . . . . . . • . . . . . . . . . . . * o * ° * ° . . . . . . ° * * .

1 0 ( ~ ) ~ ~ ~ 1 ==a+ O - - ~ )~-- ~ _ 2 2 + 2 - - 2 + x

_ 1 0 ( 1 k 1 1 1 - -a+O---3-- - ~ 1 2 + l q - x

1 1 1 1 1 2 + l + x 3 - - x + l

and

Hence we have,

1 0 a + 2 - -

A2

1 1 1 1 1 2 + l - - x 3-+ x --1

k 2 + x 1 0 ( 1 ) 1 1

a + 0--3-- ~ k-1 J - - x + 1

1 0 ( 1 ) 1 1 a + 2 ~ 3-2 k 2 - - x

1 O { l ' x 1 ] a + 0 - - 3 - - ~ , ) 3 -2 ~ 3 + x - - 1

228 M.S. SRINIVASAN

(d') Similarly, 1 0 ( 1 ) i 1 0 ( , ) , ,

a + ~ ~ k 2 - - a + 0 ~ 3 ~ 3-~ z,_2 2 + 2 - - ~ 2 _ 1 O f l , 1

a + O-- 3- - ~. k-13" This proves all our assertions.

Hence it incidentally follows that the shortest C.F.'s are precisely those which

(1) do not contain the partial quotient 1 + , and

(2) which contain the partial quotient 2 --, if at all, in a manner different from (b), (e), (c'), (d) and (d'). This specifies the shortest C.F.'s completely.

§ 3. The number o f shortest C.F.'s for p /q . - -

This number can be stated in terms of the N.I.C.F. for p]q rendered 1 unique, if necessary, by taking the last partial remainder in the form + ~.

0 The above proof shows that wherever the partial quotient ~ occurs

in the N.I.C.F., then by (H) it can be transformed into 2 -- without increasing the number of partial quotients. This transformation- will however intro- duce another 2 + in the case when = = 3, 0 = - -1 which could be trans- formed again by (H).

Thus the places in the N.I.C.F. which can be transformed into shortest C.F.'s are those where there are sequences of the form

1 1 1 s 1 1 l . . . . . . ,

which may be more conveniently written as

s = ~ ~: (3)a, (2)a~ (~)o, . . . . . . . . (2)at p ~ ,

where a - l - 4 = 2 + o r 3 - - ; ai, bi>~ 0; p > 2 .

Let the number of shortest C.F. equivalents to this sequence S be C (al, b~, a2, . . . .ab, bt). If there are several sequences, of the form S in the N.I.C.F., the total number of equivalent shortest C.F.'s is evidently the product of the numbers C (al, bj . . . . . bt) associated with each of the sequences.

It will be seen from what follows that this number is not affected even if the N.I.C.F. ends with the sequence; i.e., even if p does not exist.

Shortest Semiregular Continued Fractions ~29

We proceed to obtain a reduction formula to calculate tl~e number of shortest C.F. equivalents to S.

§ 4. Reduction formula. - -

THEOREM 2. - -The number o f shortest C.F. equivalents to S satisfies the reduction formula

C (al, bz, a~ . . . . . bg) =

f C (0, b 1 -- 1, a2, . . . . b t ) + (a~+ 1). C (0, bl -- 2, a2 . . . . bt) if b~ > 1.

C (a2, b2 . . . . . bt) + (al+ 1) .C (a o - - l , b, . . . . . bt) '-'f bl = 1.

P r o o f - - W e divide the shortest (~.F. equivalents to S into classes / ' ~ ( i = 0; 1, 2, . . . . a l + 1) respectively, given b y : - -

(i) / 'o ~- the class of equivalents beginning with a not followed by 3; i.e., of the form

(a) ~ 2 ~ (3) ~_13 followed by M = the transformations o f

(2)a1-~ (3-),,,....p, when b~ > 1 ;

or (b) c~ 2 ] (3)~-t 4 followed by N = the transformations of

(3)~,-z (2) b~ . . . . p when bl = 1.

(ii) / ' i (i = 1, 2 . . . . . a1 -- 1) = the class of equivalents beginning with O)i not followed by 3; i.e., of the form

(a) ct (3)i 2 2 (3-)~_i_x3 followed by M when b~ > 1,

or (b) c~ (3-)i 2 ~. (3)~-i-1 4 followed by N when bl = 1.

(iii) ~,~ = the class of equivalents beginning with = (3)~ not followed by 3; i.e., of the form

(a) = (3)~ 2 followed by transformations of (2)6_~ (3)~, (2)6. . . . . p when b~ > 1,

or (b) ~ (3-)~2 followed by transformations of (3)a, (2)~. . . . . p when b~ = !.

class of equivalents not beginning with =; i.e., of (iv) F~1+1----- the the form

(a) ( ~ )

or (b)

(3-),3 followed by M when b 1 > 1

2 (3)a~ 4 followed by N when bt = 1.

230 M . S . SRINIVASAN

We obtain the forms stated above by application of (H) to the class F~ with cL ----- 2, 0 = 1 in succession, by which ~ 2 2 becomes 2~3 and 3 2 ~ becomes 2 ~ . H shows that if the first 2 in (3-)~, (2)~ is transformed by (H), the preceding ] will be changed..

The number o f shortest C.F. equivalents in Pal is C (o, bx - - 1, a2, . . . . b t ) when bl > 1, and C (a2, bs . . . . . bt) when bi = 1.

The number of shortest C.F. equivalents in each of the other classes is

= C (0, bt - - 2, as . . . . . bt) when bl > 1 and = C ( a s - - l , b., . . . . . bt) when bl = 1 .

Hence the total number o f shortest C.F. equivalents in S is

C (0, bt - - I, as, . . . . bt) + (at + 1). C (o, bl --2, as, . . . . b t ) when bt > 1 and

C (as, b2, . . . . bt) + (al 27 t) C (a s - - 1, bs, . . . . bt) when bt = 1. Corollary.--The shortest C.F. in which 2~= is not preceded by k occurs

only in the class F 0 when a+ is the first quotient in S. I f ~ is the first quotient, the shortest C.F. of the above type occurs'only in the class F~+t. (This will be used to prove the uniqueness of the singular C.F., in Theorem 3.)

By successive application of the reduction formula, the evaluation of C (al, b l , . . . , bt) is obtained finally in terms of C (% fl). Now we find C (a, fl) explicitly in terms of Fibonacci numbers. We have,

C (a,/3) ---- C (0, fl -- 1) + (a + 1). C (0, ~q -- 2) and C ( 0 , / ~ ) = C ( 0 , f l - - 1 ) + C ( o , f l . - -2) .

Now C ( 0 , 0 ) = the number of shortest C.F. 's in 1 I when ~ > 2, ~ + p p > 2, which = 1. Also C(0 , 1 ) = the number of shortest C.F. 's in

1 1 1 cL + 2 + p by definition, which = 2; these two being evidently given by"

'1 1 1 1 1 1 c~ 4- 2 + p ~ 4- 1 z F 2 - - p + 1"

Now write C (0, fl) --- f ( f l + 1) so that

f ( 1 ) - - l ; f(2)-----2; and f ( f l + 1 ) = f ( f l ) + f ( B - - 1 ) . Hence,

f ( 3 ) = 3, f ( 4 ) = 5, f ( 5 ) = 8 . . . . . f (O) = l, f ( - - l ) = 0 , f ( - - 2) = l, f ( - - 3) = - -1 . . . .

Shortest Semiregular Continued Fractions 231

are the Fibonacci numbers. Thus, the C-function may be considered as a generalisation of Fibonacci numbers.

We see that

C (al, az . . . . . ai_x, O, ai+ 1 . . . . . a~_,A-----

C (al, as, . . . . , ai_l + ai+l . . . . . a2,,). As a particular case C (2, 3) = 9; this may be verified by writing down

106 whose N.I.C.F. is explicitly the nine shortest C.F.'s for 2 ~

1 1 1 1 1 1 0 + 3 ~ - - 3 - - 2 4 2 + 2 + 3

The nine shortest C.F.'s are : - - 1 1 1 1 1 1

(i) 0 + 3 _ _ 3 _ _ 2 4 2 4 2 4 3 1 1 1 1 1 1

(ii) 0 + 2 + 2 _ ~ - _ ~ + 2 + 3

1 1 1 1 1 1 (iii) O + 3 - - 2 + 2 - - 4 - - 2 - - 4

1 1 1 1 1 1 (iv) 0 + ~ 2 _ 3 + 2 + 3

1 1 1 1 1 1 (v) 0 + ~ + 2 _ 3 _ 4 _ 2 _ 4

1 1 1 1 1 1 (vi) 0 + 3 ~ 3 - - 2 - - 3 + 3

1 1 1 1 I 1 (vii) 0 4 ~ - - _ ~ 2 + ~ 2 74

1 1 1 1 1 ! (viii) 1 2 - - 3 - - 3 - - 3 + 2 + 3 1 ! 1 1 1 1

( i x ) 1 . . . . 2 - - 3 - - 3 - - 4 - - 2 - - 4 "

§5. A ~emi-regu!ar C.F.

Singular continued .fractions.--

ao + 01 0 2 O~

a l+ a2+ . . . . a~. is said to be singular if

ai>~ 2;! (ai + 0 ) > / 2 ; o~ = ~: 1 (o< i ~ r).

232 M . S . Sm~IVASAN

- - 1 This implies that 1 -4- and ~ do not occur as partial quotients. Hence

no sequence of forms (a), (b), (c), (c'), (d), (d') occurs. Therefore by our previous remark, the singular C.F. is a shortest C.F.

THEORFM 3.--Zlm singular C.F. for p/q exists and is unique. Proof--By examining the classes F o, Fi, Fal , and Fal+l of shortest C.F.

equivalents to S, we see that the condition for singularity can be satisfied only in the class F 0 or F~1+I according as a or ~ is the first quotient in S. (Theorem 2 corollary). The next sequence (of the form S) in this class contains the singular C.F. in the corresponding F o or Fal+l class of that sequence. Proceeding in this manner, with every successive sequence, we will see. finally that the singular C.F. for p/q exists, and is unique.

The singular C.F. can be written down, given the sequence S, applying the following results : - -

(1) I f S = ~ (2),~p then S.C.F. = a (2),~p,

(2) If S = a (3-),, (2),~p then S.C.F. = a22 (3),,-1 3 (2)~_~p,

(3) I f S = g (2),~p then S.C.F. = (a -- 1) ~3 (2),~_~p,

(4) I f S = ~ (3-),, (2),~p then S.C.F. = (a - - 1) ~ (3)m 3 (2),~_2 p,

(5) I f S = ~ (~),, 2p then S.C.F. = ~ 2 2 (3),,_t (p + 1),

(6) I f S = ~ (3),, 2p then S.C.F. = (a -- 1) ~ (3-)~ (p + I).

Another method for obtaining the S.C.F. (Singular C.F.) is by direct transformation of the N.I.C.F. of p/q. In the N.I.C.F. 2 -- does not occur.

- -1 ~ -1 We transform ~ into - - by (H), wherever the former occurs in the

N.I.C.F. The C.F. finally obtained contains same number of partial quotients as the N.I.C.F. and satisfies the conditions for singularity. This is the singular C.F. of p/q.

- -1 I f the N.I.C.F. does not contain 2--+' it is itself the singular C.F. If

of the N.I.C.F. is -- 1, it is transformed to ; 1 ; for, the the last term singv~-

- - 1 lar C.F, does not contain ~ as last term,