# Lucas partitions

## Full text

(1)

VOL. 21 NO. 2 (1998) 387-396

### LUCAS PARTITIONS

NEVILLEROBBINS MathematicsDepartment SanFranciscoStateUniversity

SanFrancisco,CA94132,U.S.A.

ABSTRACT. TheLucas sequenceis definedby:

2,

1,

L,-1

L,-2forn

2. Let

### r(n)

denote respectively thenumberof partitions ofnintoparts, distinct partsfrom

### (L,).

We

develop formulas thatfacilitatethe computation ofV

and

### r(n).

KEY WORDS AND PHRASES:

1991AMSSUBJECT CLASSIFICATION CODES: 11P\$1, 11P\$3,11B35.

1. INTRODUCTION

Let S denote a non-empty subset of

### N,

the setof all natural numbers. Let

### r0(n)

denoterespectivelythenumberofpartitions ofn intoparts, distinctparts, evenlymanydistinct

parts, oddly many distinct parts fromS. Define

1,

0. Let

### V(n)

have

thegeneratingfunction:

(1.1)

n=0

Let

It follows from

and

that

Furthermore,

0 for n

1. (1.3)

k=O

### tO(n).

(1.4)

REMARK. Apostol [1],p.311 and Hardy

p.255provethat

holdswhenS

### N,

but the

samereasoningappliestothe moregeneralcase. Since also

(.5)

itfollowsthat

### In

thisnote, we consider the case whereSisthesetof all

numbers,

wheren

0.

### (The

Lucasnumbersare definedby:

2,

1,L, L,_I

L,-2 ifn

### 2.)

Wewillshow howto

(2)

388

2. PRELIMINARIES NotationandDefinitions

### Fk

kthFibonaccinumber

0,

1,

ifk

kthLucasnumber

2,

1,

L_t

if k

k=l

### [z]

denotestheinteger part of therealnumber

### b]

denotesthe setofallintegers,t,such thata

t

### <

b. Inparticular,ifk

3, then

LucasIdentities

(1)

forn

2, with

2,

1 (2)

iffj

k, unlessj k-1 0

(3)

### }

isstrictly increasingifn

1

1.6" ifn

4

k+l

(7)

1

[1/2

(8)

t, where t

i--o

(9)

5F,

(10) L,+2

REMARKS.

### (1)

is the definition oftheLucas sequence (2)and

### (1).

(4)follows from(1),usinginductionandthefactthata,/9arethe rootsofu2 u 1 0. (5)and(6)follow from

### (4).

(7)through(9)may be proved usinginductiononn.

follows from

### (1)

and(9). 3. THE MAIN THEOREMS

Letnbeanaturalnumber. Wefirstaddress theissueoftherepresentability ofnas a sumofdistinct Lucas numbers. Such a representation will be calledaLucas

### representation’of

n. Ifinaddition, the summands are non-consecutiveLucas numbers,wesaythat theLucasrepresentation ofnisspecial. We will show that every natural number has a special Lucas representation. If the special Lucas

representation ofnisunique,which isusuallytheease,wecallit the minimal

### Lucas

repreentati0n ofn.

Otherwise, n has two special Lucas representations. In this case, we define the minimal Lucas

representation ofnas thespecialLueasrepresentationthatdoesnotinclude

### L0

2as asummand.

For example, 13 has the unique special (and hence minimal) Lucas representation

13 11

2

### +

L0; 12 has two special Lucas representations: 12 11

1

and 12 7

3

2

q-

### L0.

TheformeristheminimalLucas representationof12.

THEOREM1.

### Every

naturalnumber,n, hasaspecialLucasrepresentation:

n

etc.

(3.1)

where

k,+t

### >

2forall such thatI

r 1, ifr

### >

2.

PROOF. (Inductiononn) Itsufficestoconsidertheease where n:/:

### L.

Thereforethere exists

unique

3 such that

n

Letnl n-

Now

implies0

nl

(3)

that 2

r, if r

3. Thus

n], hence

n

Since

3,

implies

1, that is,k]

2. Since n

### +

rq,the conclusion nowfollows.

LEMMA 1. Let n Lj

etc.

with s

2 and ji j,+]

### _>

2 for all such that

1

### <

s 1. Let j j]. Then n

Lj+]

Furthermore, n

iff s 1

j 0, andj, j

### 1)

forall such that 1

### <

s 1.

PROOF. Using(2)andourhypothesis,wehave

### r+L-2(,_])

where 1ifji=0 andj=2i-1

r

0 otherwise Thuswehave:

=1 =1 i=0 i=0

Now (8)impliesn

Lj+I

Ifs 1

js 0, and jz j-

for all suchthat

1

### <

s-1, then the weak inequalities in

### (3.2)

may be replaced with equalities, which yields

n

### .

Conversely, if n Lj+I

### j,

then the weak inequalities in

### (3.2)

become equalities. Thisimpliess 1

(andhenceji j-

for all suchthat

1 <i<s-l,andj=O.

2.

liff(i,j)

or

PROOF. Suppose

1. Ifj 0,then

3,so 2. Ifj 1and

### <

j,then 0. Now suppose

j

1. Then

and

imply 1

### L-2.

Therefore L,-2 1,so 3and j 2. Theconversefollowsbydirect substitution.

3. If

n

### (3.3)

then this special

### Lueas

representation ofnisunique.

PROOF. Let k be the least indexsuch that the special representation

is notunique. By

inspection, k

### >

4. Thusnhasasecond specialLueasrepresentation:

n

etc.

Lj (3 4)

with ji ji+l

### >

2for all such that1

### <

s 1. In fact, (2)implies

### >

2. Letj jl. Now (3.4)

implies

n, so

### L.

If2[j,thenLemma implies

Lj+] 1, so

Butthen

### L+I,

an impossibility. If21j,thenLemma implies

Lj+]

1. Since k

4,Lemma2 implies

1. Therefore

1, sothat

Since

we must have

hence k j

1.

yields

n

etc.

hence L_2

etc.

### By

definition ofk,we musthaves 2,

### J2

k 2. Butthenj] ./2 1,an

impossibility.

THEOREM2. Letnhavetwodistinct specialLueasrepresentations:

n

etc.

with k,+]

2 for all

such that 1

r-1;

n

n

etc.

with ji- j+l

2 forall

such that 1

### <

s-1. (3.6) Assume also that j=jl<kl. Then k]=2s-l, kn=k=l, and ji=2(s-i) for all with l<i<s.

(4)

ROBBINS

musthaven

1

L1, sok,z

1.

andLemma implyj,

forall

suchthatI

### <

s.

THEOREM3. Letnhave the specialLucasrepresentation:

n

etc.

### Lk,.

(3.7)

This specialLucasrepresentationisuniqueunless

1 and

2h

1for some h

### >

1, in which casenhasasecond specialLucasrepresemation:

n

etc.

etc.

### (3.8)

PROOF. Ifr 1, then the specialLucasrepresentation(3.7)isunique byLemma3 Ifr

### _>

2,

supposethatnhasasecond specialLucasrepresentation:

n

etc.

### (3.9)

Again, by Lcmma3,s

2. Ifjl

### <

kl, thenthe conclusion follows fromTheorem2,withr 2 and h s. Now supposthatj, k/forall suchthat 1

u 1(forsomeu

2),butj,,

k,. Let

u--1

m n-

etc.

also

u-1

m n

etc.

### L3,.

=1

Now Theorem 2 implies j,+,

### 2(s-u-i)

for all such that 0<_

s-u, u=r-1,

k,

1

### +

1, k, 1. The conclusionnowfollows from Theorem 2,with

Combining the results of Theorems 1, 2, and 3,wehave:

THEOREM4.

### Every

naturalnumber,n, has auniqueminimalLucasrepresentation:

n

### (3

10)

where (i) k,-ki+l >_2 for all such that l_<i_<r-1, ifr_>2; (ii) fir_>2 and k=0, then

k_

3.

### LEMMA

4. Letnhave theminimalLucasrepresentation given by

### (3.10)

inTheorem4above. Then

k

n

fir

### _>

2.

PROOF. (Inductionon

Clearly,

### <

n, soitsufficestoshowthat

+. Letr 2, so

n

If

### k2 _>

1, then by hypothesis,

2, so

implies

Thus n

If k=0, then by

and (2), we have

so

Ifr

3, letnl n

etc.

### L,.

Clearly,this is a minimalLucas representationofn, so

byinductionhypothesis,wehavenl

hencen

Since 1

1

1by

hypothesis,

implies

Thereforen

### L_

The three following theoremspermitthe computation of

and

THEOREM5.

ifn

### _>

0.

PROOF. (Inductionon

### )

Thestatement istreebyinspectionif n 0or1. Ifn

### _>

2, andifL,is partitionedintoseveral distinct parts, then(7)implies thatthelargest partmustbeL,_. Therefore, by

(1),we have

1

1

### [ n]

(byinductionhypothesis)

### 2)].

(The "1" in

(5)

LUCASPARTITIONS

PROOF. (Induction onn) The statement is truebyinspection if n 0 or 1. Ifn

### _>

2, then reasoning as in the proof of Theorem 5, we have

### 2)]

by Theorem 5 and induction hypothesis.

THEOREM7.

0 ifn 2,3(mod4)

-1 ifn 0,1(mod

### 4)"

PROOF. Thisfollows from

### (1.6)

andfrom Theorems5and6.

Having settledthecase wherenis aLucas number,wenow considerthecasewherenis asum of

twoor more distinct, non-consecutiveLucasnumbers. Then, byTheorem 4,nhasaunique minimal

Lucasrepresentation:

n

(3.

k-1

wherer

2,

### ki+l >_

2for all suchthat

r 1, andif

0, then k_

### _>

3. Alternatively,wecouldwrite:

n

cjLj

### (3.12)

where(i)c, 1; (ii) % 0 or1 for alljsuchthat 0

j

### <_

8 1; (iii)%-lcj 0for alljsuch that

1

j

8;(iv)if

### co

1,thenc2 0.

Ifweomitthe conditions(iii)and(iv),then(3.12)correspondstoaLucasrepresentation ofn. The %willbe called the digits of the representation.

Referring again to (3.11), let

>0,n2

### =n-L

>0. Given any Lucas

representation of n, define the initial segment as the first

### k

-/c2digits;definethe terminal se_arnentas the remaining digits.

### In

the minimal Lucas representation of n, the initial segment consists ofa 1

followedby

### M

1O’s,andcorrespondstothe minimalLucasrepresentationof

### ,

whilethe

terminalsegment correspondstotheminimalLucasrepresentation ofnl. Lucasrepresentations ofnmay be obtainedasfollows:

T_vpeI. Arbitrary combinations of Lucas representations of the integers correspondingtotheinitial andterminalsegments in the minimalLucasrepresentation of n,namely

and

### n.

Clearly, the number of

### Type

Lucas representations ofnis

### Typ_e

II. Supposethatin a non-minimalLucasrepresentation of n, theinitial segment endsin10,

whilethe terminal segmentstartswith0. Ifthisblock ofdigits, consisting of100, isreplacedby 011,

then a newLucasrepresentation ofnis obtained. Anecessaryconditionfor theexistenceof

### Type

II Lucasrepresentationsisthat

### Type

III. In the minimal Lucas representation of n, if

1 and k_ 2h

1 for some

/z

### >

1, then by Theorem 3, anew Lucasrepresentation ofrtis obtainedbyreplacing

by

etc.

### L0.

Thethreefollowing theorems enable ustocompute

and

THEOREM8. If k

1, then

k

### +

1.

PROOF. Let n=/nt+

1

nl

so the number of

### Type

Lucas

representations ofnis

### []

k. Since/-,1 hasnoLucasrepresentationbutthe

minimalone,thereareno

### Type

II Lucasrepresentations ofn. Byhypothesis and Theorem 3,there is a

unique

### Type

III Lucasrepresentation ofn. Therefore

k

1. THEOREM9. Ifk

1, then

k;

k

### +

1.

(6)

N.

Lucasrepresentation ofn hasj

### +

1 terms, and thus contributes to

### + 11

iffj is odd. The

conclusion nowfollows.

THEOREM10. If k

1, then

1;

### 1)

2. PROOF. Thisfollowsfrom

### (1.6)

and from Theorems8and9.

InTheorems 11, 12, and 13 below, wedevelop formulas for

and

### a(rt)

inthe case where n

Lk,n

### +

1. Inorder todo so, we mustbe ableto count the number of

### Type

II Lucasrepresentations ofn. Wethereforeneed to determine the numberofLucasrepresentations ofn thatdonotincludethe largest possibleLucasnumberas apart. Thisquestionisaddressed byLemma5.

### LEMMA

5. Letnhavethe minimalLucasrepresentation:

n

etc.

Let

n-

0. Let

### T(n)

denote the number ofLucas representationsofnthat donotinclude

as apart; let

### rE(n), to(n)

denote respectively the number of such representations consisting of evenly,oddly manyparts. Then

Tg(n)

### ).

(3.3)

(3.14)

(3.5)

PROOF. It followsfromthe definitions of

that

### T(n)

isthe number of

Lucasrepresentationsofnthat doinclude

as apart;

### To(n)

are respectively

thenumber ofsuchrepresentations consisting of evenly, oddlymanyparts. Ifn

let

n

etc.

Lj,

s

### 2)

(3.16)

be aLucasrepresentation ofnthatincludes

as apart.

### fit

follows fromLemma4that

### L

isthe

largest part.) Correspondingto

### (3.16),

there isaLueas representationof n"

etc.

### (3.17)

This correspondence is clearly a bijection, so that

### r(n)-T(n)=r(n).

Furthermore, the number of parts in

and

### (3.17)

differ in parity. Therefore

and

Theconclusions

### (3.13),

(3.14), (3.15)nowfollowifn

Ifn

### L,

so that nl 0, then dearly no other Lueas representation ofn includes

### L

as a part. Therefore

Furthermore,

1

### r(n).

TaORM11. nhave the minimaLucas representation:

n

etc.

(3.18)

where(i)r

2;(ii)if

1, then

Letnl n

n2

n

0. Then

if

if

### )"

PROOF. Byhypothesis, therearenoType IH Lucas representationsofn. Asmentionedearlier,

thenumberof

### Type

Lucas representationsofnis

thenthere

areno

### Type

II Lueas representationsof n,sothat

If

### 21(kl- k),

then the number of Type II Lucas representations of n is the number ofLucas

(7)

LUCAS PARTITIONS 393

I. If

2, then

### [

PROOF. (Inductionon n) Byinspection, the conclusionis trueif2

n

5. Ifn

6, then

3

L,-1

L,-3

### 3).

NowTheorem 11 implies

### 3). By

inductionhypothesis,wehave

and

Therefore

COROLLARY:2. Ifn

1,then

n.

PROOF. If m

### _>

1, then via (5), we have

### L1).

Now Theorem 11 implies

2m. again, via

we have

### 3).

NowCorollary implies

### [(4m- 2)]

2m-1. REMARK. Corollaries and2imply (independently) that the function

### r(n)

isasurjeetion fromN toN.

THEOREM12. Letn

n

1. Then

ifk

if

if

if

PROOF.

### By

hypothesisandTheorem 3, any

### Type

HI Lucasrepresentation ofnmustarisefroma

corresponding

### Type

HI Lucasrepresentationofnl n

### Lk.

Thusitsufficesto count the

### Type

and IILucasrepresentations ofnconsisting ofevenlymany parts.

### Type

Lucasrepresentationofnwith evenly manyparts will occurwheneverthe initial and terminal segmentsagreein parity. Therefore the numberof such representationsisgiven by:

k,

k2

If,2

If

then no

### Type II

Lucas representations of n can arise. In particular, if

### 3(rood 4),

by simplifying the last formula, we obtain

Similarly,if

weobtain

### k,2),

we wish to countthe numberof

### Type II

Lucasrepresentationson nthat have evenly manyterms. Each such representation originates fromaLucasrepresentation ofnwhose initial

has

### k2)

terms,

and whose terminal

### segment’s

number of terms therefore differs in parity from

If

### 4),

then the number of

### Type

II Lucas representationsofnis

by Lemma5.

### In

thiscase, the number of

### Type

Lucasrepresentations ofnis

Thusweobtain:

If

### 4),

thenthe number of

### Type II

Lucasrepresentationsofnis

### rE(n2),

byLernma5. Inthiscase,the number of

### Type

Lucasrepresentations ofnis

### k2)r(n).

Thereforewe obtain:

1

THEOREM13. Letn

Lk,n

## -

/1. Then

if

0

if

1

if

2

0 if

3

(8)

394

6. Ifn

1, then

1

1

0

if n 1

ifn 2

ifn 0,3

### 4)

PROOF. (Induction on

### n)

The conclusionholdsby inspection if 1

n

4 Ifn

5, then

1

q-

-b

-b

### 1).

Now Theorem 13 implies

### 1),

sotheconclusionfollows from theinductionhypothesis.

7. Ifn

1, then

### 1)".

PROOF. The conclusion holds by inspection if 1

4. If

5, then

1

### 1).

NowTheorem 13implies

### 1).

The conclusion nowfollows from Lemma6.

8. Ifj

n

L_3- 1, then

O.

PROOF.

### n)

by(1),so theconclusionfollows from the hypothesis and Theorem13.

LEMMA9. If0

n

1, then

### a(n).

PROOF. By (9),wehave

### n).

The conclusion nowfollows from the hypothesis and Theorem13.

THEOREM14. Ifnbelongsto

wherek

2, andm

1 n, then(i)

and(ii)

### a(n).

PROOF. Itis easilyseenthat mbelongsto

iffndoes. Now

### (7)

implies thereis abijection

betweenthe partitions of m,nrespectivelyinto distinctLueasparts. Thus

### r(n).

Furthermore,

sincethe leftsideof

### +

2 terms, itfollows that under this bijectiort, corresponding partitions of

m andn will have numbers of parts that agreeordisagreeinparity accordinglyas k is even orodd. Therefore

COROLLARY3. If k

1, then

1;

### 2)

2. PROOF. Thisfollows from Theorems 10 and14.

THEOREMIS. Ilk

3,then

2.

PROOF. If 1

3

k, let k.i

nE

Thus

z.2

z,3

Nowz,3

### E(L+-I)-E(SF-I).

But Lemma 9 implies

so

Thus

implies

L-2, soZk,3

### +

I. Now

Lernma 9 implies z,3

1. Also, zk,2

n

### 1}

0

by Lemma 8. Now zk,a

n

### 1}.

Theorem 14 implies that Z,l

n

Thus we have:

### 1)),

fromwhichthe conclusion follows.

TtlEOREM16. If k

2andif

n

1, then

2-

1 if

n

2

if

1

n

5F, 1

1 if

n

1

PROOF. If

1

n

### 5F-

1, then Lemma \$ implies

If

n-SF n-SF

n

1, then

j)[

[a(j)l

### 5Fk)

byLemma 9. Also, Lemmas 8 and 9 imply

1

so

1

### E(2Lk)+ E(n-5Fk).

Finally, ifL+I_<n_<2L-2, let m=2L-n, so that 2_<m_<Lk_2. We must show that

### Now

Lemmas 8 and 9 imply

and

(9)

LUCAS PARTITIONS

when2

m

### _< L-2.

This is trivially truewhenm 2. If 3

m

L-2,then

3

m

3, so that by

### (I), (9),

Lemma9and Theorem 14, we have

Therefore

rn-1 m-1 rn-2

### E(2Lk-m)

]a(2Lk-j)] ]aCj-1)]

### E(m-2),

so wearedone.

3=2 3=2 =I

TIIEOREM17. lira 0.

PROOF. Ifk

2,

6

### I}.

Itsufficestoshow that lira

0. Ifn

### e

Ii, then by Theorem 16,wehave:

I. Since

### E(n)

isnon-decreasing and

### n-SFk _<

Li-2, it follows that

### E(n)< E(2L)+E(L_2)+I.

By Theorem 16, we have

I,so weobtain

2, hence

Since n

### <

Lk+l, We get E(n)

### <

E(L+l)L+ Jr" 2E(L.) SO that

L+

tk

Since

### E(n)

is non-decreasing and

### Lk

tendstoinfinitywithk, itsufficestoshow that lira

0. In fact,since

1

### 1),

itsufficestoshow that

Ifk_>l,

### letck=E(Lk-1).

Thusc=0, c2=c3=2,

### By

Theorem 15, we have: ck+2--Ck+l-}-2ck-2+2. Let the

### {ck}

have the generating function:

c,z

### .

Usingthemethodof[2], p.337-350,weobtain:

2 2z3

z

k=l

2z

2z2

Thereforeck

where the

### bi

areconstants, and the ti are the roots of the equation: x

### 2x2+

2x- 2 0. Using Cardan’s formula, if u

v=

then

t2

Thus

1.138,

b3, and

2

Now0

But

### (6)

impliesthat the fightsideof thelastinequality tendsto 0asktendstoinfinity. Therefore lim

0,we aredone.

THEOREM15.

### a(n)

assumes each of thevalues 0, -1-1, +/-2infinitely often.

PROOF. Theorem7 implies

### a(n)

0, 1 infinitelyoften,whileTheorem 10implies

### a(n)

2

infinitely often.

### By

Theorems 13 and 7,

1. Therefore

### a(n)=

1

infinitely often. Finally, with k

2, let n

5

### L1.

Now Theorem 13 implies

2. Therefore

### a(n)

2infinitelyoften.

THEOREM19.

2for alln.

PROOF. If

### la(n)l >

3for somen,letnbe the least such imeger.

### By

Theorems7and 10,n

### +

1. Letn havetheminimalLucasrepresentation:

n

etc.

where r

2; let ni nt-1

for1

r, with

0.

### By

hypothesis and Theorem 13,we musthave:

with

0

Theorem13implies

if/

0

0 if

1

if

2

if

3

Therefore

with

_=2

### (mod4).

IfweapplyTheorem 13repeatedly,

we eventually get

### a(0)).

Now Theorem 7 implies

### <

1,contrary tohypothesis.

(10)

Table

0 0 51 6 3 0 30 6308

0 -1 52 6 2 -2 32 6877

2 0 -1 2 2 53 6 3 0 32 7491

3 2 0 2 3 54 8 4 0 32 8155

4 2 0 2 5 55 5 3 33 8862

5 2 2 2 4 6 56 5 3 34 9622

6 2 0 4 9 57 7 3 -l 35 10438

7 3 -1 5 12 58 6 3 0 35 11316

8 2 0 5 16 59 6 3 0 35 12247

9 2 0 5 20 60 6 3 0 35 13249

lO 3 2 6 26 61 6 3 0 35 14319

ll 3 -I 7 33 62 6 3 0 35 15464

12 3 -1 8 41 63 6 3 0 35 16678

13 3 2 9 50 64 6 3 0 35 17981

14 4 2 0 9 62 65 7 4 36 19369

15 3 2 10 75 66 5 2 -1 37 20845

16 3 -1 11 90 67 5 2 -1 38 22413

17 3 -1 12 107 68 8 4 0 38 24089

18 4 2 0 12 129 69 6 3 0 38 25868

19 3 2 13 151 70 6 4 2 40 27754

20 3 2 14 178 71 6 3 0 40 29759

21 5 2 -1 15 208 72 7 3 -1 41 31893

22 4 2 0 15 244 73 4 2 0 41 34149

23 4 2 0 15 281 74 4 2 0 41 36541

24 4 2 0 15 326 75 5 3 42 39078

25 5 3 16 375 76 7 3 -1 43 41771

26 3 -1 17 431 77 5 2 -1 44 44609

27 3 -1 18 491 78 5 3 45 47619

28 4 2 0 18 561 79 8 4 0 45 50802

29 4 2 0 18 638 80 7 4 46 54170

30 4 3 2 20 723 81 7 3 -1 47 57715

31 4 2 0 20 816 82 7 3 -1 48 61471

32 6 3 0 20 922 83 9 5 49 65434

33 5 2 -1 21 1037 84 6 3 0 49 69613

34 5 2 -1 22 1163 85 6 3 0 49 74013

35 5 3 23 1302 86 7 3 -1 50 78664

36 6 3 0 23 1458 87 8 4 0 50 83561

37 4 2 0 23 1624 88 8 5 2 52 88715

38 4 2 0 23 1808 89 8 4 0 52 94140

39 6 3 0 23 2009 90 6 3 0 52 99862

40 5 3 24 2231 91 7 3 -1 53 105871

41 5 2 -1 25 2467 92 7 3 -1 54 112190

42 5 2 -1 26 2729 93 7 4 55 118835

43 6 3 0 26 3012 94 8 4 0 55 125830

44 4 2 0 26 3321 95 6 3 0 55 133160

45 4 3 2 28 3651 96 6 3 0 55 140867

46 4 2 0 28 4014 97 10 5 0 55 148958

47 5 2 -1 29 4406 98 8 4 0 55 157456

48 4 2 0 29 4828 99 8 4 0 55 166353

49 4 2 0 29 5282 100 8 4 0 55 175400

50 7 4 30 5777

REFERENCES

### T.M.,

IntroductiontoAnalyticNumberTheory, Springer Verlag,1976.

### [2]

GRAHAM, R.L.,KNU’Ud, D.E. andPATASHNIK,O.,

### HARDY,

G.H. andWRIGHT, E.M., Introduction tothe Theory

### of

Numbers, OxfordUniversity

4thEd.,1960.

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## References

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