fibonacci advanced

8/12/2019 fibonacci advanced

1/10

ADVANCED PROBLEMS AND SOLUTIONS

Edited by

RAYMO ND E. WHI TNEY

Lock Haven State College, Lock Haven, Pennsylvania

Se nd a l l c om mun ic a t ions c onc e rn ing Adva nc e d Pr ob le m s a nd So lu tions to R a ymond E .

W hi tne y , M a the m a t i c s De p a r tm e n t , Loc k Ha ve n S ta t e C o l l e ge , Loc k Ha ve n , Pe nn s y lva n ia

17745 . Th i s de p a r tm e n t e s pe c ia l ly we lc o me s p rob le ms be l i e ve d to be new o r e x te nd ing o ld

re s u l t s . P r op os e r s s hou ld s ubm i t s o lu t ions o r o the r in fo rma t ion tha t w i l l a s s i s t t he e d i to r .

To fa c i l i t a t e t he i r c ons ide ra t ion , s o lu t ions s hou ld be s ubmi t t e d on s e p a ra t e s igne d s h e e t s

wi th in two mon ths a f t e r pub l i c a t ion of the p rob le m s .

H-195 Proposed by Verner

Hogg att, Jr.,SanJose State College,SanJose, California

Consider the array indicated below:

2

5

3

34

89

2

4

9

22

56

45

3

7

6

38

4

27

65

5

6

6

22

(i) Show tha t the row sum s ar e F , n ^ 2.

( i i) Show tha t the r is i ng diagona l sum s ar e the convolut ion of

(

F

2 n - l ^

= 0

a n d

{ ^ 2 , 2 ) } ;

= 0

,

t he ge n e ra l i z e d nu mb e rs of H a r r i s a nd S ty le s .

H-196 Proposed by J. B. Roberts, Reed College, Portland, Oregon.

(a) Let A

0

be the s e t o f i n t e g ra l pa r t s of t he pos i t i ve in t e g r a l mu l t ip l e s o f r , whe r e

1 + ^ 5

T =

413


2/10

414 ADVANCED PRO BLE MS AND SOLUTIONS [Oct .

a nd l e t A - , m = 0 , 1 , 2 , , be the s e t o f i n t e g ra l pa r t s o f t he num be r s n r

2

for n 61 A . Pr ov e tha t the col lec t ion of Z of a l l pos i t ive in teg ers is the d is jo in t

union of the A..

J

(b) Ge ner a l iz e the pro pos i t ion in (a ) .

H-197 Proposed byLawrence

Somer

Universityof Hi n o s, Urbana, Illinois.

n

s ion re l a t ions h ip :

Le t { i r } _ be the t -F ib ona c c i s e que n c e s w i th pos i t i ve e n t r i e s s a t is fy ing the re c u r -

F ind

(t)

u

n

. l i m

n

t

Vu .

Z

i

n - i

i= l

(t)

u

n+1

u

oo n

SOLUTIONS

HYPER

-TENSION

H-185 Proposedby L

Carlitz

DukeUniversity Durham, North Carolina.

Show that

(1 - 2x)

n

= ( - D

n

"

k

(

n

2

+

k

k

) (

2

k

k

) (1 - v )

n

-

k

2 F l

[-k; n + k + 1; k + 1; x]

k=0 ^ / \ /

w h e r e

2

Fi [a ,b ; c ; x ] de no te s the hyp e rge om e t r i c func tion .

SolutionbytheProposer.

We s tar t wi th the ident i ty

r s

E

(2r + 3s) (y - z) z _

r s ( r + 2s ) ^

+

.2r +3 s+l 1 - y - z

r , s = 0 y*

Now pu t y = u + v, z = v, so tha t

fr

v V

2 r

+

3 s ) L

u v

= 1

1 ;

JL J r l s l ( r + 2s ) l ,-

+

,2r +3 s+l 1 - u - 2v

r . s = 0 *

U v

'


3/10

1972] ADVANCED PRO BLEM S AND SOLUTIONS 415

The right-hand side of (*) is equal to

n=0

while the le f t -h and s ide

oo oo

E

(2r + 3s)l r s V

1

/ -x\

k

f

2r

+ 3 s + k \ , ^

r l s l ( r

+

2s)l

U V

^

{

~

1}

{ k J

( U + V )

r , s = 0 k= 0

It follow s th at

, , . ,n \" ^ , , ,k v ^ .

n

x k / 2 r + 3s + k \ (2r + 3s ) l r s

=

L - < -

1 ) r + t u

s

P +

t

( m o d

P>"

FIBONACCI IS A SQUARE

H-187 Proposed by Ira

Gessel

Harvard University, Cambridge,

Massachusetts.

Pro b le m: Show tha t

a

pos i t i ve in t e ge r

n is a

F i b o n a c c i n u m b e r

if

and only

if

e i the r

5n

2

+

4 o r

5n

2

- 4 is a

s q u a r e .

Solution by the Proposer.

L e t

F

0

= 0, F

4

= 1, F - = F

+

F _-

be the F ibona c c i s e r i e s a nd L

0

= 2 , =

1 ,

L

,- = L

+

L

n

be the Lu cas ser ie s . I t i s wel l known tha t

r+ 1

r r - 1

(1) ( -I)"

+

F*

r

=

F

r + 1

F

r

_

1

(2)

L

r

=

F

r + 1

+

F

r

_

x

Sub t ra c t ing fou r t ime s the f i r s t f rom the s qua re of the secon d equat io n, we have

whe nc e

L

2

_

4 ( - l )

r

-

4 F

2

= (F _,, - F -

)

2

= F

2

,

r

r r+1 r - 1 r

5 F

2

+

4 ( - l )

r

= L

2

.

r

r


6/10

418 ADVANCED PR OB L EM S

AND

SOLUTIONS

[Oct.

T h u s if n is a F i b o n ac c i n u m b e r , e i t h e r 5n

2

+ 4 o r 5n

2

- 4 is a s q u a r e .

I have

two

p r o o f s

of the

c o n v e r s e .

F i r s t

Proof. We us e the

t h e o r e m ( H a r dy

and

W r i g h t ,

An

In t roduc t ion

to the

T h e o r y

of

N u m b e r s ,

p. 153)

tha t

if p and q ar e

i n t e g e r s ,

x is a

r e a l n u m b e r ,

and

| (p /q)

- x | 2, so by the two t h e o r e m s

quoted above , k /n = F - / F for s o m e r, and s inc e bo th f ra c t ions a re r e d u c e d , n = F .

Second Proof. A s s u m e 5n

2

4 = m

2

. T h e n m

2

- 5n

2

= 4, so

m + \/ 5n m - N/5n _

n

"

2 * 2

l

'

and s ince

m and n

ha ve

the

s a m e p a r i t y ,

m

+ \

r

5n , m - V5n

_ _ and

a r e i n t e g e r s

in Q ( N / 5 ) ,

w h e r e

Q is the

r a t i o n a l s ,

and

s inc e the i r p roduc t

is 1,

the y

are

u n i t s . It is

wel l known (Hardy

and

W r igh t ,

p. 221)

t h a t

the

on ly in t e g ra l un i t s

of

Q(\/I>)

are

of the fo rm

x~

, w h e r e x = (1 + \/"5)/2.

T h e n

we

have

(m

+

^ 5 n ) / 2

= x

r

= \ (x

r

+ y

r

) + f L ^ ^ L N/5 ,

2

L ^ J

w h e r e y = -1 /x . Now x. + y = L and


7/10

1972] ADVANCED PROB LEM S AND SOLUTIONS 419

(

X

r

- y

r

) / ^ 5 = F

r

(Hardy and Wri ght , p . 148) . Thu s

| ( m + N/5n) = | ( L

r

+ ^ 5 F

r

> ,

so n = F .

S U M S E R IE S

H-189 Proposedby L Carlitz DukeUniversity Durham, NorthCarolina Corrected).

Show that

(2r + 3s ) l (a - b y )

r

b

S

y

r + 2 s

^ r l s ( r + 2 s ) l

H

_, ,2r +3 s+l , ,

2

r 9 S = 0

' (1 + ay) 1 - ay - by

2

Solution

b y

theProposer.

P u t

i - ar

so tha t

1 - ax - bx

2

A

m=0

m

x

m

1 = E *

(1 - ax - bx

2

)(l - y)

n

J

m , n = 0

v

ni n

x

y

R e p la c ing y by x" y th i s be c o me s

1 V ^

m

~

n n

z /

J

G x y

m , n = 0

1 - ax - bx

2

)( l - x y)

Hence tha t par t of the expans ion of

(1 - ax - bx

2

) ( l - x V )

tha t i s independen t of x is equal to


8/10

420 ADVANCED PRO BLE MS AND SOLUTIONS [Oct .

(*)

1 - ay - by

2

On the oth er hand, s ince

(1 - ax - bx

2

)( l - x~ y) = (1 + ay) - x(a - by) - bx

2

- x " y ,

we have

(1 - a x - b x ^ U - x - V ) "

1

= [ ^ a - b ^ + b x ^ + x - V l

v T S t

E

(r + s + t ) l (a - by) b y r+ 2s -t

r ls l t l , - ^

x

r + s + t + l

x

r,s,t=0

( 1 + a

^

The pa rt of th is sum tha t i s independent of x is obta ined by taking t = r + 2s . We ge t

E

( 2 r + 3 s) l ( a - b y ) V y

r 2 s

r

T

.sI(r + 2s)l ,.. , v2 r+3 s+i

r , s = 0

( 1 + a y )

Since th is i s equal to (*) , we have proved the s ta ted ident i ty .

I T S A M O D W O R L D

H-190 ProposedbyH. H.Ferns Victoria, ritish Columbia.

P r o v e t h e f o l l o w i n g

2

r

F = n (mod 5)

2

r

L = 1 (mod 5) ,

whe re F a nd L a r e the n F ibona c c i a nd n Luc a s nu m be rs , r e s p e c t iv e ly , a nd r i s

the lea s t res idu e of n - 1 (mod 4) .

Solutionb ytheProposer.

In an unpub l ished pape r by the pr op os er , i t i s shown tha t

^\ E

2k

n

+1

)

5

k

k=0


9/10

1 9 7

2 ] ADVANCED PROB LEM S AND SOLUTIONS 421

Hence

IT]

k = l

x

'

Thus

5

k

.

2

n 1

F

n

= n (mod 5)

Let n - 1 = 4m + r , wh ere 0 < r < 4 . Then

But

Hence

2 F = 2 F = 2 2 F = n (mod 5) .

2

4 m

= ( 2

4

}

m ^

1 ( m o d 5 )

^

To p rove

use

2

r

F = n (mod 5) .

2

r

L = 1 (mo d 5) ,

M

k= 0

%

'

(which is der ived in the same paper) and proceed as above*

JU S T S O MA N Y T W O S

H-192 Proposed byRonaldAlter UniversityofKentucky Lexington, Kentucky.

I f

3n+l

c

n

3=0

M1+J.

. s( J H

prove tha t

c

=

2

6 n + 3

. N

s

(N odd , n >0 ) .

n

olution

by

the

Proposer

In the s eque nce

b

k

= b

k - l "

3 b

k - 2 '

( k

"

l j b

i =

b

2 = D >

it is ea sy to show tha t 2 is the hi gh est pow er of 2 tha t div ide s b. if and only if k = 3

(mod 6) , Als o , by der iv in g the ap pro pr ia te Bine t for mu la , i t fo l lows tha t


10/10

422 ADVANCED PROB LEM S AND SOLUTIONS [Oct .

Thus

^ i m - c - ^ i - -

k ^ 1

2

j=o

The de s i re d re s u l t fo l lows by obs e rv ing

6n+3 6n+2 n *

Ed i t o r i a l No te : P l e a s e s ubmi t s o lu t ions fo r a ny o f t he p rob le m p r opo s a l s . W e ne e d f re s h

blood

A GOLDEN SECTION SEARCH PROBLEM

R E X H . S H U D D E

Waval Post Graduate School, Monterey, California

Afte r t i r i ng of u s ing num e rou s qua d ra t i c func t ions a s ob je c t ive func t ions fo r e xa m ple s

in my ma the ma t i c a l p rog ra m min g c ou rs e , I pos e d the fo llowing p ro b le m fo r

myself:

De s ign

a un imoda l funct ion ove r t he (0,1) i n t e rva l wh ic h i s c onc a ve , h a s a ma x im um in the in t e r io r

o f (0 ,1 ) , a nd i s no t a qua d ra t i c func t ion . The pu rp os e w a s to de m ons t ra t e num e r i c a l ly the

golden sec t ion search.*

My f i r s t thoughts we re to add two funct ions which ar e concav e o ver the (0 ,1) in te rva l

with the pr op er ty tha t one goes to -o a t 0 and the o th er goes to -o a t 1 . My two in i t ia l

c ho ic e s we r e log x a nd l / (x - 1 ) . The go lde n s e c t ion s e a rc h s t a r t s a t t he two po in t s x

t

=

1 - (1/0) and x

2

= l/(p wh ere 0 = (1 +

N / 5 ) / 2 .

Af t e r s e a rc h ing w i th 8 po in t s , I no t i c e d tha t

the in te rva l of un cer ta in ty s t i l l conta in ed the f i r s t se ar ch point so I thought i t abou t t im e to

f ind the loca t io n of the ma xim um an alyt i ca l ly . I wa s dumfounded to d is co ve r tha t i f I con t in-

ued indefin i te ly wi th the se ar ch m y in ter va l of un cer ta in ty would s t i l l conta in the in i t ia l sea rc h

point .

* Doug la s J . W i lde , Op t imum Se e k ing M e tho ds , P re n t i c e Ha l l , Inc . (1964) .

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