J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
M
ICHAEL
D
RMOTA
W
OLFGANG
S
TEINER
The Zeckendorf expansion of polynomial sequences
Journal de Théorie des Nombres de Bordeaux, tome
14, n
o2 (2002),
p. 439-475
<http://www.numdam.org/item?id=JTNB_2002__14_2_439_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
The
Zeckendorf
expansion
of
polynomial
sequences
par MICHAEL DRMOTA et WOLFGANG STEINER
d6dii d Michel France d l’occasion de son anniversaire
RÉSUMÉ. Nous montrons que la fonction ’somme de chiffres’ de Zeckendorf
sz(n)
lorsque n
parcourt l’ensemble des nombrespre-miers ou bien une suite
polynomiale
d’entiers satisfait un théorèmecentral limite. Nous obtenons aussi des résultats
analogues
pourd’autres fonctions du même type. Nous montrons
également
quele
développement
de Zeckendorf et ledéveloppement
standard enbase q des entiers sont
asymptotiquement indépendants.
ABSTRACT. In the first part of the paper we prove that the
Zeck-endorf
sum-of-digits
functionsz(n)
andsimilarly
definedfunc-tions evaluated on
polynomial
sequences ofpositive integers
orprimes satisfy
a central limit theorem. We also prove that theZeckendorf
expansion
and the q-aryexpansions
ofintegers
areasymptotically
independent.
1. Introduction
Let q > 2 be an
integer.
Then a real-valued functionf
defined on thenon-negative
integers
is calledq-additive
iff
satisfieswhere E
{O, 1,... q - 11
are thedigits
in the q-aryexpansion
of the
integer
n > 0. Forexample,
thesum-of-digits
functionis a
q-additive
function. The distribution behaviour ofq-additive
functionshas been discussed
by
several authors(starting
mostprobably
with M.Manuscrit regu le 20 avril 2001.
Mend6s France
[18]
and H.Delange
[3],
see alsoCoquet
[2],
Dumont andThomas
[10, 11~,
Manstavicius[16],
and[6]
for a list of furtherreferences).
Most papers deal with the average value or the distribution of
q-additive
function. There are,
however,
also laws of the iteratedlogarithm
and moregenerally
a Strassen law for the sum ofdigits
function due to Manstavicius[17].
(It
seems to be diflicult togeneralize
such a law to the Zeckendorfsum-of-digits
function since acorresponding
Fundamental Lemmas seems tobe out of reach at the
moment,
even thegeneralization
to ajoint
law oftwo q-ary
sum-of-digits
function is notobvious,
see[8].)
The most
general
central limit theorem forq-additive
functionsf
is dueto Manstavicius
[16],
where the distribution of the valuesf (n) (0 n
N)
is considered. In this paper we are interested in the distribution of
f (P(n))
(0
nN),
where is aninteger polynomial.
Here the best knownresult is due to
Bassily
and Kitai(1).1
(Here
and in thesequel
~(x)
denotesthe distribution function of the standard normal
law.)
Theorem 1. Let
f
be aq-ddditive function
such that = 0(1)
ask -+ oo and b E
{0, ... , q - 1}.
Assume thatfor
some > 0 and let
P(n)
be apolynomial.
withinteger
coefficients, degree
rand
positive
leading
term.Then,
as N -3 00,and
where
and
This result relies on the fact that
suitably
modified centralized momentsconverge.
The main purpose of this paper is to extend this result to certain
G-ary
digital expansions.
Let a > 1 be aninteger
and the sequence G =be defined
by
the linear recurrence
-1 This theorem was only stated (and proved) for 17 =
3 .
However, a short inspection of theNow every
integer n >
0 has aunique
digital
expansion
with
integer
digits
0
aprovided
thatfor
all j >
0(which
means that16G,k-l(n)
= 0 ifEck(n)
=a).
Aspecial
case of these
expansions
is the Zeckendorfexpansion
where a = 1 and theG~
are the Fibonacci numbers.A function
f
is said to beG-additive,
ifAlternatively
we havewhere
fk(b)
:=f (bGk).
First we will prove the
following
theoremconcerning
the distribution ofthe sequence
f (n),
0 n N. Theproof essentially
relies on the factthat the
possible
G-ary digital
expansions
can berepresented by
a Markovchain. Note that the sequence
Gk
is alsogiven by
where a is the
positive
root of the characteristicpolynomial
of the linearrecurrence
Theorem 2. Let G be as
above, f a G-additive function
such thatfk(b) =
o
(1)
as k -3 00for
b E{0, ... , a}.
Then,
for
all q >0,
theexpected
valueof
f (n),
0 nN,
isgiven
by
Furthermore,
setwith
where
Assume
further
that there exists a constant c > 0 such thatk>0.
Then,
and
for
allpositive integers
h.(1.3)
has been shownby
Drmota[5]
forstrongly
G-additive functionsf,
i.e.
Furthermore,
it should be noted that(1.4)
provides
anasymptotic
relationfor the
variance,
too,
however,
without an error term:We will use Theorem 2 and a method similar to
Bassily
and Kitai’s toprove Theorem 3.
Theorem 3. Let
G, f
be as in Theorem 2 andP(n)
apolynomial
withand
for
allpositive
integers
h, if
we setf (P(n)) = -f (-P(n))
for
P(n)
0.Note that definition of
f (P(n))
forP(n)
0 has no influence on theresult,
because the number ofnon-negative integers
withP(n)
0 isnegligible.
Our next results concern the
indepence
of differentdigital expansions.
For
example,
in[6]
thefollowing
property
is shown.Suppose
that ql, q2 aretwo
coprime integers
andfl, f2
qm resp.q2-additive
functionssatisfying
theassumptions
of Theorem 1. Then wehave,
as N --> oo,i.e. the distribution of the
pairs
( f 1 (n), f2 (n) ),
0 nN,
can beconsid-ered as
independent.
We will extend this
property
to our moregeneral
situation.Theorem 4.
Suppose
thatfl, f2
are twofunctions satisfying
oneof
thefollowing
conditions.(i)
ql, q2 > 2 are twopositive coprime
integers
andf l, f2
ql - resp.q2-additive
functions satisfying
theassumptions
of
Theorem 1.Further-more set
Mi(N)
:=Mqt (N)
andDq; (N)
(i
=1, 2).
(ii)
q > 2 is aninteger
andfl(n)
aq-additive function satisfying
theassumptions
of
Theorem 1. a > 1 is aninteger
andf2(n)
is aG-additive
function
satisfying
theassumptions
of
Theorem 2.Further-more set
MI(N) := Mq(N), Dl(N) := Dq(N)
andMZ(N) := MG(N),
(iii)
at,a2 >
- 1 are twodifferent
integers
such that isirrational,
y 2
G = and H = the
corresponding
linear recurrentse-quences, and
fl, f2
G- resp. H-additivefunctions
satisfying
theassump-tions
of
Theorem 2. Furthermore setM1(N)
:=MG(N),
D1(N)
:=DG(N)
andM2(N)
:=MH(N),
D2(N)
:=DH(N).
Let
Pl(x),P2(x)
be twopolynomial
withinteger coefficients, degrees
rl,r2 andpositive
leading
terrra.Then,
as N - oo,and
The paper is
organized
in thefollowing
way. Section 2 is devoted to theproof
of Theorem 2. Section 3provides
aplan
of theproof
of Theorem 3.Sections 4-6 collect some
preliminaries
which are needed for theproof
ofTheorem 3 in Section 7.
Finally,
theproof
of Theorem 4 ispresented
inSection 8.
2. Proof of Theorem 2
Our aim is to
study
the distribution behaviour off (~c),
0 nN,
i.e.the random variable
YN
definedby
If we define
(k,N
by
and
by
then we
obviously
havei.e.
YN
is a(weighted)
sum ofTherefore,
we will first have ade-tailed look at It turns out that constitutes an almost
stationary
Markov
chain,
as the next lemma shows. We want to mention that thisfact is also a consequence of results from Dumont and Thomas
[10, 11].
Inour case this is a
quite simple
observation. Therefore we decided topresent
a short
proof
of thisfact,
too. Thisprocedure
issimpler
and shorter thanintroducing
the notation of[10,
11]
and tospecialize
afterwards.Lemma 1.
For fixed j,
the random variablesform
a Markovchain with
where
with initial states
and
Rerraark. The matrices are no transition matrices of a Markov process,
but
they
describe transition matrices in view of the relations(2.1)-(2.3).
However,
it turned out to be easier to work with 3 x 3-matrices instead of(a
+1)
x(a
+I )-matrices.
Proof.
A sequence ofnon-negative integers
is aG-ary digital
expan-sion of an
integer
n, if andonly
a for all i >0,
= 0 if Ei = a andEi
~
0only
for a finite number of i(cf.
e.g. Grabner andTichy
[13]).
Letbe the set of
G-ary digital expansions
for n Thenand it can be
easily
seen that(2.1)
holds. For k =0,
even =a]
isequal
to =1].
We have
because we can take a block
(o, E1, ... ,
fj-l)
of the set on the left side of theequation,
shift it to theleft,
set = 0 andget
a one-onecorrespondence
Since the other
probabilities
b],
1 b a,
areequal,
we haveNow we show that we have a Markov chain.
where the third
equation
is validonly
if(bo, ... , bk+i) E Bk+2.
Otherwisethe
probability
is 0(for
bk+i
=a,
bk 0
0,
(bo, ... ,
or undefined(for (bo,
... ,Bk+ 1).
If theprobability
isdefined,
we thus haveSimilarly
to(2.1),
(2.2)
and(2.3)
are easy to see. Henceand the transition from to is
entirely
determinedby
(2.4).
0Corollary
1. Theprobability
distributionof
isgiven by
with
Proof.
Let P be the matrix obtainedby neglecting
the C~(Q2J-k»)
terms inthe matrix The
eigenvalues
of P are1, 0
and theeigenvector
to the
eigenvalue
1 withLemma 1
suggests
toapproximate
thedigital
distributionby
astation-ary Markov chain
(Xk, k >
0),
with(stationary)
probability
distributionPr[Xk = b~ =
pb, 0 b a, and transition matrixP,
i.e.The next lemma shows how we can
quantify
thisapproximation
for finitedimensional distributions.
Lemma 2. For every h > 1 and
integers
0kl
k2
...kh j
wehave
and
consequently
Since
we
just
have toapply
(2.6)
andCorollary
1 and the lemma follows. D The case ofgeneral
N is very similar.Lemma 3. The
probability
distributionof
Ç,k,N
for
Gj
NGj+l
withj
> k isgiven
by
for
all b E{0, ... , a}.
Furthermore,
thejoint
distributionfor
given
by
Therefore
otherwise
where we have used
A similar
reasoning
can be done for thejoint
distribution,
e.g. we havej:
otherwise
Thus,
we canproceed
in the same way. 0We now turn to the derivation of
EN
=EYN ,
i.e. to theproof
of(1.2),
the first
part
of Theorem 2. Sincefor N the
expected
value ofYN
isgiven by
and q
> 0 is asufficiently
small number(to
be chosen in thesequel).
Furthermore,
we havewhich
implies
It seems that the variance
Var YN
cannot be treated in a similar(easy)
way.
Therefore,
we use some additionalassumptions
andpresent
aproof
of
(1.4)
together
with the distributional result(1.3).
The above calculation indicates that we
just
have to concentrate ondig-its with A k B
(defined
in(2.9)).
The reason is that we obtainuniform estimates for this range. The
following
lemma is a directconse-quence of Lemmata 2 and 3. Note that it is not necessary to assume that
ki, ... ,
kh
are ordered and thatthey
are distinct.Lemma 4. For
every h >
1 andfor
every A > 0 we haveuniformly for
allintegers
(where
A,
B aredefined
in(2.9)
with anarbitrary
1/ >0)
andbl, b2, ... , bh
E10, 1,
... ,al,
whereThis observation causes that we have to truncate the
given
functionf (n)
and have to consider
In order to finish the
proof
of Theorem 2 it is(luckily)
enough
to proveThis is due to the
following
lemma and(2.11).
for
E ~8if
andonly if
for
all x ~ DLFurthermore,
if for
all h > 0then we also have
and
conversely.
Proof.
We consider the three(sequences of)
random variablesSuppose
first that thelimiting
distribution ofXN
is Gaussian and that all moments converge. Sinceant the same is true for
YN.
Further,
we know thatThus,
itimmediately
follows that thelimiting
distribution ofZN
is the sameas that of
YN
and that all moments ofZN
converge to the same limits asthe moments of
YN .
It is also clear that the converse
implications
are valid. Thiscompletes
the
proof
of Lemma 5. 0converge to the
corresponding
moments of the normal law. We will do this in twosteps.
First we prove a central limit theorem(with
convergenceof
moments)
for the exact Markov process and then we compare thesemoments to those
of7(n),
i.e.(1.4).
Obviously
theproof
(1.3)
of Theorem 2 iscompleted
then.The next lemma
provides
a central limit theorem forE
fk(Xk),
whereXk
is thestationary
Markov process definedby
(2.5).
Lemma 6.
Suppose
that there exists a constant c > 0 such that(2)
> cfor all j >
0. Then we have’
and the sums
of
the random variablesfk(Xk)
satisfy
a central limit theorem.More
precisely
and
for
all h > 0 wehave,
as N ~ oo,Proof.
Let(which
does notdepend
onk)
denote the transition function of the Markovchain
(Xk, k > 0)
andits
ergodicity
coefficient. If thefk
areinjective
onf 0, ... , a},
then0)
is a Markov chain withergodicity
coefficient(3
and weget,
by
Lemma 2 of Dobru0161in[4]
and with= Qk2 >
c,If some of the
fk
are notinjective,
weget
the same resultby
considering
in-jective
functions Ik
which tend tofk.
SinceD(N)2
=and = Var
EB k=A
fk (Xk)
this proves(2.11)
if{3
ispositive.
Suppose ,Q
= 0. Then there exist Xl, X2 E{0, ... , a}
and a set A suchthat
P(xl,
A)
= 0 andP(X2, A)
=1,
becauseP(x, A)
attainsjust finitely
many values. We have
P(x,
{0})
> 0 for all x.Hence,
if 0 EA,
weget
a contradiction to
P(xi, A)
= 0and,
if0 ft A,
weget
a contradiction toFor each h >
2,
the moments Eare jointly
bounded because offk(b)
= O(1).
Hence,
if thefk
areinjective,
all conditions of Theorem 4 ofLifsic
[15]
are satisfied and we have convergence of(absolute)
moments tothose of the normal distribution. An
inspection
of Lifsic’proof
showsthat,
as
above,
this is valid fornon-injective
fk
too. DNow we are able to compare the moments of
f (n)
and E
Lemma 7. For every h > 1 and
every A
> 0 we haveProof.
We haveand
By
Lemmata 4 and6,
theseexpressions
areequal
up to an error termO
((log N)h/2-À).
Since A can be chosenarbitrarily,
the lemma isproved.
0
3. Plan of the Proof of Theorem 3
We set
M,
D and7
as in Theorem 2 with theonly
differenceLemma 5 shows that it is
enough
to proveand
In
fact,
we prove that the centralized momentsand
converge
(for
N -~oo)
by comparing
them toAh(N’’).
By proceeding
as inthe
proof
of Lemma 7 andby using
thefollowing
lemma,
it follows that foreach fixed
integer
h >0,
Bh(N) -
0 andCh(N) -
0as N -> oo.
(Of
course, this proves Theorem 3. Wejust
have toreplace
Lemma 4
by
thefollowing
property.)
Lemma 8
(Main
Lemma).
LetP(n)
be aninteger polynomials of degree
r > 1 and
positive leading
term. Thenfor
every h > 1 andfor
every A > 0we have
and
uniformly for
allintegers
It turns out that this lemma can be
proved similarly
to that ofBassily
and Kitai
[1],
i.e. withhelp
ofexponential
sums. Theonly difficulty
isto
get
a nice condition forextracting
thedigits
ek (n)
withoutusing
greedy
tiling
of the unit square. Section 5provides
proper estimates forexponen-tial sums. These are the two main
ingredients
of theproof
which is thencompleted
in Sections 6 and 7.4.
Tilings
The aim of this section is to
provide
propertilings
of theplane
corre-sponding
to ourdigital expansions
in order toget
ananalogue
to q-aryexpansions
where we haveif
~x~
denotes the fractionalpart
of x.For our
expansions,
we will have to take into account the values ofI I I B
and
By taking
just
one value intoaccount,
there areoverlaps
and we cannotget
something
like(4.1)
or(4.2).
Proposition
1. LetAb,
0 b a, denoterectangles
in theplane
R2
defined
as the convex hullof
thefollowing
corners:Then these
rectangles
induce aperiodic tiling of
theplane
withperiods
Zx Z,
i.e.
they
constitute apartition
of
the unit square modulo 1. Theirslopes
are
(a,
1), (20131,0:)
and their areas are = Pb, b =0,...,
a, with Pb asin
Corollary
1.Furthermore, if
ek(n)
= b thenEssentially,
thisproposition
says that there is ananalogue
to(4.1)
forG-ary
expansions
with a small error of order 0(a-k)
for the k-thdigit.
We want to remark that Farinole
[12]
considered a very similarquestion.
Remark. The
rectangles Ab
modulo 1 constitute a Markovpartition
of theExample.
Beforeproving
theproposition,
we illustrate theexample
a = 3:which looks like follows in
R /Z:
Proof of Proposition
1.Suppose
that n isgiven by
n Then wewith the abbreviations
where we have used
(1.1)
and that is aninteger
for.
Similarly
weget
By R6nyi
[19],
we know thatgraphically)
implies
(lexico-Hence,
if Ek a,then y
is boundedby
and
by
if Ek = a.
Similarly, x
is boundedby
for all Ek,
by
for E~ = 0 and
by
for f-k > 0.
If we
put
these limits into we obtain thegiven
corners for
Ab.
It is now an easy exercise that(the
interiorsof)
theserectangles
arepairwisely disjoint
(and
situated as in theexample)
and thatthey
induce aperiodic tiling
in91
withperiods 7~2.
05.
Exponential
SumsIn order to prove the Main Lemma we have to
study exponential
sumsof the form ~.. ,
and
with
integers
i as usual.Lemma 9. Let I
(log
N)6
for
alli, j
andbe
integers
withI
for arbitrary
constants 6 >0, 1/
> 0.Then, if
0,
for
all 77’
"1.Proof.
Clearly
we haveFor the lower
bound,
we first remark thatak
isgiven
by
where the sequence is defined
by
G’ =
0,
Gi -
1 andG) =
+
for j >
2. Therefore we havewith and We have
if A00orBi4O and
becauseG~
isgiven by
(cf.
(1.1)).
HenceThe next two lemmata are
adapted
from Lemma 6.2 and Theorem 10 ofLemma 10. Let
P(n)
be apolynomials of degree
r withleading coefficient
/3.
For every To >0,
we have a T > 0 such thatimplies
Lemma 11. Let
P(n)
be as in Lemmas 10. For every To >0,
we have aT > 0 such that
implies
as N -4- 00.
Note that we can
apply
these two lemmas for/3
=S/(a
+1)
withS 0
0for any choice of T > 0 since
Lemma 10 can be deduced for r > 12 from Theorem I in
Chapter
VI ofVinogradov
[20]
because ofif P
E(q, +-,].
Forgeneral
r, the two lemmata can beproved by replacing
qby -
in theproofs
of Lemma 6.2 and Theorem 10 of Hua andusing
thefollowing
lemma.2
Lemma 12.
where
llxll
=min((x),1 - (~)).
Proof.
In each of the intervals[mo,
(m
+1)0)
and(1 -
(m
+1)R,1 -
mo],
0
rn 2 (~~,
we have at most one Therefore6. The
Boundary
of theTilings
Lemma 13. Let
P(x)
be anarbitrary polynomial of degree
r and A > 0.Set
where
(8A6
denotes theboundary of
Ab.)
Let(log
N)n
klogo
NT -(log N)n
for
some(fixed) 77
> 0 and A anarbitrary
positive
constant.Then,
uni-formly
ink,
we haveProof.
We usediscrepancies
to prove this lemma. Theisotropic
discrepancy
JN
of thepoints
(xl,l, x1,2), ... ,
(xN,l, xN,2)
inJR2
is definedby
where the supremum is taken over all convex subsets C of
’B’2
=R2
/Z~.
Itcan be estimated
by
the normaldiscrepancy
DN
which is definedby
where the supremum is taken over all 2-dimensional intervals I of
~2:
(see
Theorem 1.12 of Drmota andTichy
[9]).
To
get
an estimate forDN
we use thefollowing
version of Erdös-where M is an
arbitrary
positive integer
(and -1
=+oo) (cf.
Theorem 1.21of
[9]).
-
-We set and M =
(log
N)2À.
Then wehave,
sinceUb(A)
is the union of 4 convex subsets and the conditions ofLemmata 9 and 10
hold,
Similarly
weget,
with Lemma11,
We can choose To > 2A and the
inequalities
areproved.
D7. Proof of Main Lemma
For b E
{0, ... , a}
letWb(X,y)
be a functionperiodic
mod1,
definedexplicitly
in[0, 1]
x[0, 1]
by
Its Fourier
expansion
isgiven by
where
V(Ab)
denotes the set of vertices of therectangle Ab
andthe set of vertices
adjacent
to(xl,x2) E V(Ab) (cf.
Drmota[7],
Lemma1).
uniformly
for all(mI,m2),
where the constantsimplied
by « only depend
on
Ab
and +i
:= miI
m2,+2
:= mi . onAb
and mi := mi + := m2 -For
(small) A
> 0 we consider the functionThe Fourier
expansion
given by
of this function isif (ml, m2) =1= (0, 0)
and Hence and as 0.It is clear that 0 1 for every
pair
(zi , z2)
and thatWe define
and
We set
for A
greater
than the error termsFurthermore,
setand let be the set of vectors M =
(mi i,
ml,2,... , 7 Mh,l
mh,2)
withinteger
entriesThen we have
where
and
for all
i, j,
Lemmata 9 and 10provide
if 0. Lemma 11
provides
a similar result forprimes.
Since
(mi,l,
mi,2 )
e(mZ,1, mZ,2 )
is,
up to aconstant,
anorthogonal
trans-formation,
we haveFor the M with for some
i, j ,
weget
similarly
if we set Therefore we have
(and
a similarexpression
forE2).
Since the main termdepends
onA,
wewant to
replace
TM
by
Hence we have to estimate the
difference ’
]
By
(7.3),
we haveFirst assume
I
for alli, j.
Then we obtain from(7.6)
andand it remains to estimate the sum of the
TM
andTM
with(log
N)a~2
for somei, j
whichsatisfy
MV =0,
i.e.This is done
by
thefollowing
lemma,
whereonly
one of theequations
isneeded.
Lemma 14. We have
where
E’
denotes the sum over allinteger
solutions(ml, ... , mH)
of
thelinear
equation
(with integers -yi :A 0)
such thatI mi
I
>for
some i. The constantimplied by
« does notdepend
on the ~.Proof.
First we remark that mi = 0 for some i reduces theproblem
to asmaller one. For H = 1
(as
well as for H =2),
the lemma is trivial. Hencewe assume H > 1 and 0 for all i.
For every choice of
(mi,..., I MH- 1),
let m H be thecorresponding
solution of(7.10).
First we sum up over all choices withand obtain
If we consider
only
Imil I
>(log
N)b~2
for some i H -1,
we have thusIt remains to estimate the sum over the choices
(ml’...’ MH-1)
with- - _...i ,9
-Then we have
We
split
thepossible
range of112m21
intoFor
J2,
we obtainSumming
up over all such(ml, ... ,
mH)
with(log
for somei,
weget
Thus it suffices to consider m2 with
1’2m21 E
12
from now on. Thisimplies
with
We
split
thepossible
range of|y3m3|
intoand
13
=(0,2~imi)] B
J3.
Similarly
to(7.12),
we obtainand the sum over these
(7~1,...
rrcH)
can be estimated as in(7.13).
Forall other m3, we have
We can
proceed
inductively
and in theonly remaining
case we would haveWe
apply
Lemma 14 for(7.7)
with H = 2h - 1.Multiplying
each termof the sum in
(7.9)
by
min(l,
(where
mh 2 is determinedby
(7.8)),
gives
and the same estimate for
TM .
Hence
where
Together
with(7.5),
we obtainif we choose To = 2A and 6 =
8(h -
The result does not
depend
on the choice of thepolynomial
P(n).
If weset
P(n)
= n, Lemma 4implies
Similarly
weget
Remark. In the case h = 1 we have MV = 0
only
for(ml, m2) _ (0, 0)
and8. Proof of Theorem 4
In order to prove
independence
of differentdigital expansions
we canproceed
essentially along
the same lines as for theproof
of Theorem 3. Wejust
have toreplace
the Main Lemma(Lemma 8)
by
thefollowing
three(main)
lemmas(corresponding
to the threeparts
of Theorem4)
whichand the
corresponding
statement forprimes.
Therefore the twodimensional moments converge to those of the twodimensional normal law andTheo-rem 4 is
proved.
Lemma 15. Let ql, q2 be two
positive coprime integers
andPi (x),
P2 (x)
two
integer
polynomials of degrees
r, resp. r2 withpositive
leading
terms.Then
for
everyhl, h2 >
1 andfor every A
> 0 we haveand
uniformly for
allintegers
Lemma 16. Let
q >
2 and a > 1 be twointegers
andPl(x),
P2(x)
twointeger
polynomials of degrees
rl resp. r2 withpositive
leading
terms. Thenfor
everyhl,
h2 >
1 andfor every A
> 0 we haveuniformly for
allintegers
Lemma 17. Let
al, a2 >
1 be twointegers
such that is irrationaland let G = and H = denote the
corresponding
second orderrecurrent sequences.
Furthermore,
letPl(x), P2 (x)
be twointegers
polyno-mials
of degrees
rl resp. r2 withpositive
leading
terms. Thenfor
everyhl, h2 >
1 andfor
every A > 0 we haveand
uniformly for
allintegers
The
proofs
of these lemmas runalong
the same lines as theprevious
Main Lemma
(compare
also with[1]
and[6]).
We have to consider sums ofthe
type
with
Especially,
r2, then theproof
isstraightforward
and very similar tothat of
Proposition
1 in[6].
The reason is that there are no cancellationsin the
leading
coefficient of thepolynomial
MlVlPi(n) +M2Y2P2(n)
andconsequently
one candirectly apply
Lemmata 10 and 11 in order to estimatethe
corresponding exponential
sums.Therefore we concentrate on the case rl = r2. Here we have to
adapt
certain
properties.
Lemma 18.
Suppose
that ql,q2 >
2 arecoprime integers
and Cl, C2, rpos-itive
integers.
Forarbitrary
(but
fixed)
integers
hl,
h2
let(1 ~ j ~
hip, 2
E{1,2})
besatisfying
0 mod q
andI
(log N)6,
where J > 0 is anygiven
constant. Set’
Then,
for
we
uniformly
havefor
allgiven
01/’
1/, where q =max{ql, q2}.
This lemma is
implicitly
contained in theproof
ofProposition
2 of[6],
the statement of which is that of Lemma 15 for r = 1.
However,
by using
Lemmata
10,
11(which
have not been used in thisgenerality
in[6])
and18,
Lemma 15 follows asProposition
2 of[6].
Lemma 19. Let
q >
2 and a > 1 be twointegers
and cl , c2, rpositive
integers.
Forarbitrary
(but fixed)
integers
hl,
h2,
let(1 j h 1)
beintegers
satisfying
fl
0 mod q andI
(log N)a
and letM~2)
(1
ih2, j
E{I, 2})
beintegers satisfying
(log N) 6,
where ð > 0 is anygiven
constant. Letand
Then,
for
and
for
we
uniformly
havefor all given 0 n
Proof.
The upper bound is trivial.Thus,
we concentrate on the lowerbound. We
have,
with(5.1)
and +1)
=Gka
+Gk-1,
with
integers
7~’n(l),
m~2~
and therefore S = 0 if andonly
if theequa-tions
hold. Since
(Gk, Gk+I)
= 1 for allk,
weobtain q hl
IClm 1
and hence(for
sufficiently
large
~ ~)
which is notpossible
form(l) 0
0 mod q.Hence we may assume S
~
0. In order toget
a lower bound forS,
weuse Baker’s theorem
(see
[21])
saying
that for non-zeroalgebraic
numbers0:1, ~2? - " ? an and
integers bl,
b2,... , bn
we have eitheror
where
and real numbers
~1~2?-"
with whereh(.)
denotes the absolute
logarithmic height.
Set - = +
h2 -
1).
Then there exists aninteger
K with 0 Kh,
+h2 -
2 such that for allj, t
So fix K with this
property.
First supposeThen we have
log
I
and weI., " can
apply
Baker’s theorem forI and obtain
for a certain constant C > 0. Of course, this
implies
for some constant c > 0 and all T > 0.
Otherwise we have some Sl, S2 such that for all
J - J
, Here we
get
by
Baker’stheorem,
as
above,
and can estimate S - S
by
Lemma 20. Let
al, a2 >
1 be twointegers
such that isirrational,
detG =
(Gj)
and H =(Hj)
denote thecorresponding
second order recurrentsequences and Cl, C2, r be
positive
integers.
For
arbitrary
(but fixed)
integers hl, h2
let(1 j
hi,
j,
i E{1, 2})
beintegers satisfying
1
(log
N)6
(where
6 > 0 is anygiven
constant)
such thatand
Then,
for
we
uniformly
havefor
allgiven
0q’
q, where a =max { 0:1,
a2l
-Proof.
Again
we can concentrate on the lower bound and haveThe
assumption
thata2+4
is irrational ensures0:2 %
Hence S is zero if andonly
if theequations
we
get
m~l~ -
0 and thusm~ = m~2~
= =,S’1
=82
= 0.Hence ,S’
~
0 and the lower bound is obtainedsimilarly
to Lemma 19. DAcknowledgement.
The authors aregrateful
to an anonymous refereefor his careful
reading
of aprevious
version of this paper and for manyvaluable
suggestions
toimprove
thepresentation
and theproofs.
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Wiedner Hauptstra8e 8-10/113
A-1040 Wien
Austria