J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
H
ANS
R
OSKAM
Prime divisors of linear recurrences and Artin’s
primitive root conjecture for number fields
Journal de Théorie des Nombres de Bordeaux, tome
13, n
o1 (2001),
p. 303-314
<http://www.numdam.org/item?id=JTNB_2001__13_1_303_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Prime divisors of
linear
recurrencesand Artin’s
primitive
root
conjecture
for number
fields
par HANS ROSKAM
RÉSUMÉ. Soit S une suite définie par une récurrence linéaire entière d’ordre k > 3. On note l’ensemble des nombres
pre-miers qui divisent au moins l’un des termes de S. Nous donnons
une
approche heuristique
duproblème
selonlequel
PS admet ou non une densiténaturelle,
et montrons que certains aspects deces
heuristiques
sont corrects. Sousl’hypothèse
d’une certainegénéralisation
de laconjecture
d’Artin pour les racinesprimi-tives, nous montrons que
PS possède
une densité asymptotique inférieure pour toute suite S"générique".
Nous donnons en illus-tration desexemples numériques.
ABSTRACT. Let S be a linear integer recurrent sequence of order k ~ 3, and define PS as the set of primes that divide at least one term of S. We
give
a heuristicapproach
to theproblem
whether PS has a naturaldensity,
and prove that part of ourheuristics is correct. Under the assumption of a
generalization
of Artin’s primitive root conjecture, we find that PS has positive lower
density
for’generic’
sequences S. Some numericalexamples
are included.
1. Introduction
An
integer
sequence ,S‘ _Ix,,1001
is said tosatisfy
a linear recurrence oforder
1~,
if there existintegers
al, a2, ... , a~ such thatIf such a sequence ,S’ does not
satisfy
a linear recurrence of order smallerthan
1~,
we say that ~S’ is a linear recurrent sequence of order l~. In this case the xn can begiven
as anexponential
expression
in the roots of the characteristicpolynomial f
=X ~ -
EZ[X]
of the recurrence.For
example,
iff
isseparable
with roots we havewhere E are determined
by
the initial valuesx
n n=lAssociated to a linear
integer
recurrent sequence ,S’ is the setPs
ofprimes
dividing
the sequence, where aprime
p is said to divide the sequence if p divides some term of the sequence. We say that ,S’ isdegenerate
if theas-sociated characteristic
polynomial
is eitherinseparable
or has two differentroots whose
quotient
is a root ofunity.
If ,S’ isdegenerate,
the setPs
ispossibly
finite. The fact thatPs
is infinite fornon-degenerate
linearre-current sequences ,S’ of order 2 is often attributed to Ward
[15].
However,
already
in 1921P61ya
[10]
proved
that for allnon-degenerate
linearrecur-rent sequences ~S’ of order k >
2,
the setPs
is infinite. Hisproof
shows thata finite set
Ps
contradicts thegrowth
rate of S.P61ya’s
method seemsinadequate
to determine the ’size’ ofPs.
Inpar-ticular,
it does not allow us to decide whetherPs
has adensity
insidethe set of all
primes.
Almost all results on the existence and value of6(PS)
are for second order linear recurrent sequences ,S’. In this second order case, the method of
proof
depends
on twoproperties
of the sequence: whether the sequence is ’torsion’[1,
4, 6,
9]
or non-torsion[13],
and whether thecharacteristic
polynomial
is reducible overQ
[1,
4, 9,
13]
or irreducible[6].
Essentially nothing
is known for sequences of orderlarger
than 2. The methods for second order sequences can be made to work in some veryspecial
cases: there existhigher
order linear recurrent sequenceshaving
a set of maximalprime
divisors ofpositive
density
[1].
Ward[16]
introduced the term maximalprime
divisors of a k-th order linear recurrent sequence for thoseprimes
that divide I~ - 1 consecutive terms of the sequence, but do not divide all terms.In this paper we focus on linear recurrent sequences ,S’ of order k > 3
whose associated characteristic
polynomial
isseparable. Unfortunately,
we are not able to prove that the setPs
ofprimes
dividing
such a sequencehas a
density.
However,
for alarge
class of linear recurrent sequences we can prove thatPs
haspositive
lowerdensity,
if we assume ageneralization
of Artin’s
conjecture
onprimitive
roots.The paper is
organized
as follows. In section 2 wegive
a different char-acterization of the setPs .
This characterization is used in section3,
togive
a heuristicapproach
to theproblem
whether the setPs
contains a set ofpositive
density.
Section 3 also contains a theoremstating
thatpart
ofthe heuristics is correct. This
theorem,
combined with ageneralization
ofdensity
for’generic’
sequences S. Afterproving
the theorem in section4,
we
give
some numericalexamples
in section 5.2. Reformulation of the
problem
Let S = be an
integer
sequencesatisfying
the linear recurrenceof order > 2 with
separable
characteristicpolynomial
The sequence ,S’ is an element of
V f,
the set of all sequences ofalgebraic
numbers
satisfying
the linear recurrence with characteristicpolynomial f.
By
defining
addition and scalarmultiplication
of sequencescomponent-wise,
the setV f
becomes aQ-vectorspace.
Asf
isseparable,
the sequencesform a
Q-basis
forV f ,
and there exist cl , ... , Ck EQ
such that xn -
Ciai
for allpositive integers
n. The absolute Galoisgroup
GQ
ofQ
permutes
thea2’s
and actstrivially
on theintegers
xn.Using
that the form a basis forV f,
we find thatthe set is
GQ-invariant.
Moreover,
each a EGQ
induces the samepermutation
on theNow let a denote the residue class of X in the free
Q-algebra
A-Ca ~X~~ / ( f )
of rank k.By
the Chinese remaindertheorem,
thec2’s
determinea
unique
c E A such thatc(ai) =
c2 for i =1, ... ,
k. Because of the lastremark of the
previous
paragraph,
GQ
actstrivially
on c. We conclude thatthere exists a
unique
c E(~ ~X ~ / ( f )
such thatDenote the set of
primes
that divide the sequence Sby
Ps.
Aprime
p isan element of
Ps if
andonly
if there exists aninteger n
such thatIn order to
interchange
thetrace-map
and the reduction modulopZ,
wechoose an
integer
d ~
0 such that cd EZ (X ~ / ( f )
and define R =Z[~].
The
(maximal)
primes
of thering
R are of the formpR,
with p a rationalprime
notdividing
d. In thefollowing
wedisregard
thefinitely
manyprimes
dividing
d. This restriction has no effect on the existence or value of thedensity
ofPs.
The elements can are contained in C~ =
R~X~/( f ),
a freeR-algebra
ofrank k. As the trace is stable under base
change
and theQ-algebras
A and 0 0pQ
areisomorphic,
we can writeequation
(2)
asTo compute xn modulo a
prime p
ofR,
we tensor the trace-map Tr :We find that a
prime
p of R divides the sequence if andonly
if there exists aninteger n
such thatwhere the bar means reduction modulo In other
words,
if H denotesthe kernel of the trace map Tr :
Fp ,
we have thefollowing
criterionfor a
prime
p of R to divide the sequence:This characterization hints to
why
the determination ofPS
is difficult. Thegroup H is an additive
subgroup
of thering
whereas,
for almostall
primes,
c(a)
is a coset of themultiplicative subgroup
(a)
C(O/pO)* .
Such a mixture of an additive and
multiplicative
structure isnotoriously
difficult to
study.
3. Heuristic
approach
In this section we let p range over the
primes
of R andpredict
the’probability’ that p
divides a fixed linear recurrent sequence S. If this’probability’
ispositive,
weexpect Ps
to be a set ofprimes
ofpositive
density
inside the set of allprimes.
By
theequivalence
(4)
of section2,
the likelihood that p divides S willdepend
on the size of thesubgroup
(a)
C(O/pO)* .
In the extreme case that agenerates
(O/pO)* ,
the intersectionc(a)
nH isnon-empty
and p divides the sequence. To make theimportance
of the size of(a)
moreexplicit,
we fix aprime
p of R such that ca is a unitmodulo
pO.
As thetrace-map
issurjective,
its kernel H has index p. Theprobability
for arandomly
chosen xEO/pO
to have non-zero traceis
1-!.
The elements in the cosete(1i)
c are notrandomly
chosenp
elements of
0/ pO.
However,
if we let p range over theprimes
ofR,
it seemsplausible
that the additive andmultiplicative
structure of thering
are
’independent’.
Therefore,
we assume that as pvaries,
the’probability’
that an element in the coset
e(1i)
has non-zero trace is1 - p.
As a furthersimplification,
we assume the’probability’
that all elements ine(1i)
have non-zerotrace,
to beequal
to(1 - p)#~(a).
Note that theseassumptions
are wrong for k = 1. In this case, the
trace-map
is theidentity,
and none ofthe elements in =
F*
has zero trace. With the aboveassumptions
and theequivalence
(4),
we conclude that the’probability’
that aprime
pdoes not divide a fixed linear recurrent sequence is
approximated
by
In order to obtain a set of
primes
ofpositive
density
that do divide the sequence, we need a set ofpositive
density, consisting
ofprimes
p forwhich this
probability
is’small’,
or,equivalently,
for which the order of 1i E is’large’.
For an
integer
a ~
0,
the order of a EF*
divides p - 1. In1927,
Emil Artinconjectured
that if a is notequal
to -1 or a square, the set ofprimes
p for which the order of a EF*
equals
p - 1 haspositive
density.
Moregenerally,
weexpect
that for each h EZ>i ,
the setTh
ofprimes
p for which the order of a E F* is(p-1)/h,
has adensity
b(Th).
Note that8(Th)
is notpositive
for all a and h EZ>l;
if a is a square, there are no oddprimes
pfor which a E
F*
has order p - 1.However,
we doexpect
theequality
8(Th)
= 1 to hold. In otherwords,
weexpect
that for ’most’primes
p, the order of a modulo p is ’almost’ maximal. These
conjectures
arestill open, but have been
proved
under theassumption
of thegeneralized
Riemannhypothesis
[5,
7,
14].
To
adapt
Artin’sconjecture
to oursituation,
we first need an upperbound for the order of a E The trivial upper bound for this order is the
exponent
e(p)
of(C~/pC~)*.
The value ofe(p)
as function ofthe
prime
pdepends
on thesplitting
type
of the characteristicpolynomial
f
modulo p. Moreprecisely,
forprimes
p notdividing
the discriminant off,
we havee(p)
=lcmd(pd -
1),
with dranging
over thedegrees
of theirreducible factors of
f
EFp[X].
Now,
let T be the set ofprimes
with a fixedsplitting
type.
We say that a, or,equivalently,
f ,
satisfies thegeneralized
Artinconjecture
forprimitive
roots, if thefollowing
holds.For each h E the set
Th
of
primes
p E T
for
which the orderof
d E ise(p)/h,
has adensity
6(Th).
Moreover,
we have~h 18(Th) _
8(T) .
Artin’s
original
conjecture
excludes theintegers
~1. In ourgeneral
setting,
there are more a’s for which the above
conjecture
does not hold.However,
weexpect
theconjecture
to hold for’generic’
a. As it is not ourprime
interest to
classify
theexceptional
a’s,
we do notspecify
what we meanby
‘generic’.
Instead wegive
anexample
of anon-generic
a. Assumef
does not factor overQ
as aproduct
of linearpolynomials.
By
Chebotarev’sdensity
theorem,
the set T ofprimes
p for whichf
does notsplit completely
into linear factors modulo p, has apositive density.
Forp E T,
theexponent
e(p)
is at leastp2 -1.
Now assume in addition thatak
isrational,
for somefixed k E The order of a E is at most
k(P-1),
soTh
is finite for all h E We find that¿h=18(Th)
= 0 isstrictly
less than6(T),
and the above
conjecture
does not hold.Now we return to our linear recurrent sequence S of order k > 2. Assume
that the associated characteristic
polynomial
of S satisfies thegeneralized
Artinconjecture.
For thefollowing,
wedistinguish
two cases; eitherf splits
into distinct linear factors modulo p, or not. In the former case we say that
p is a
splitting
prime
or that psplits
completely.
First we take for T the set of
primes
thatsplit
completely.
The set Tis
isomorphic
toF*,
and hasexponent
e(p) =
p - 1.By
assumption
there exists h E such that the setTh
ofprimes p E T
for which a E has order(p - 1 ) /h,
has apositive
density.
According
to our
heuristics,
the’probability’
that p ETh
does not divide the sequence is(1 2013 i)(P")/.
For p - oo thisexpression
converge to whichp
lies
strictly
between 0 and 1.Therefore,
weexpect
thatTh
contains twosubsets of
positive
lowerdensity,
oneconsisting
ofprimes
that do dividethe sequence, and one
consisting
ofprimes
that do not divide the sequence.As a consequence, it seems
plausible
that the set ofprimes
in T that divide ,S’ has apositive
lowerdensity, strictly
less than 1. Note thatalthough
theabove formula for the
’probability’
doesgive
us an indication of what toexpect,
it does notpretend
togive
aprecise
value for thedensity
of the setof
primes
inTh
that divide ,S’.Therefore,
our heuristics do notpredict
theexistence of the
density
of the set ofprimes
in T that divide S.Now,
let T be the set ofprimes
that do notsplit completely.
For h E we defineTh
as the set ofprimes p E T
for which the order of a E(O/P 0)*
isequal
toe(p) /h.
Asf
does notsplit completely
modulop E T,
we havee(p) ~
p2 -1.
Using
our heuristicformula,
weexpect
the’probability’
thatp E
Th
does not divide S to be at most(1 2013 ) )/.
Thisexpression
p
converges
exponentially
to0,
for p - oo.Therefore,
for all h E the subset ofTh
ofprimes
that do not divide the sequence, should have zerodensity.
According
to ourgeneralization
of Artin’sconjecture,
thedensity
of
Uh5:nTh gets
arbitrarily
close to6(T),
for n -~ oo. As aconclusion,
weexpect
the set ofprimes
that do notsplit completely
and that do not divide the sequence to have zerodensity.
Up
to now, we discussed linear recurrent sequences of any order k > 2.The theorem below and the data in section 5
support
the above heuristics for linear recurrent sequences of order k > 3.However,
for aspecific
lin-ear recurrent sequence S of order 2 with irreduciblepolynomial,
the aboveheuristics are
proved
to be false. Let ,S’ = be the linear recurrentsequence with irreducible characterisitic
polynomial f
= X 2 -
5X + 7 and initial values xl - 1 and X2 = 2. Let a and a be the roots off
and de-noteby
0 thering
ofintegers
ofQ(a).
Lagarias
[6]
showed that aprime
p divides ~’ if and
only
if the order ofa/a
E is divisibleby
3. This condition isequivalent
with certainspitting
conditions for p in certainnumber fields.
Using
Chebotarev’sdensity
theorem,
Lagarias
showed that the set ofprimes
that do notsplit completely
inK/Q
and that do not divide ~S’ haspositive density. According
to ourheuristics,
this set should have zerodensity.
Let ,S’ be a linear recurrent sequence of 3. Recall from section
2 that a denotes the residue class of X in the
ring
0 =d is some fixed
integer.
Define for each h EZ>i
the setTh
={P :
f
is irreducible modulo p, and[((O/pO)* : (1i)]
=~}.
Theorem. Let S be a linear recurrent sequence of order k >
3,
and define for eachpositive integer
h the setTh
as above. For everypositive integer
h,
all butfinitely
manyprimes
inTh
divide S.We will prove this theorem in the next section. The main idea of the
proof
is to relate the
primes
p ETh
that divide S to the existence ofFp-rational
points
on someprojective
algebraic
variety
defined overFp.
Forlarge
p,the existence of these
points
isguaranteed by
a result ofLang
and Weil.The theorem does not
imply
thatPs
contains a set ofpositive
density.
There are twoproblems.
First of all the set T ofprimes p
forwhich f
EFp[X]
is irreduciblemight
beempty,
forexample
iff
is reducible overQ.
Iff
isirreducible,
a necessary and sufficient condition for T to havepositive
density,
is the existence of ak-cycle
in the Galois group off . Namely,
suppose
f
is irreducible modulo theprime
p. Thedecomposition
group ofa
prime
above p in a normal closure ofQ[X]/(f)
actstransitively
on the roots off ,
and therefore contains ak-cycle.
The fact that the existence of ak-cycle
is sufficient for T to havepositive
density
follows from Chebotarev’sdensity
theorem.Much more difficult is the
question
whether there exists h EZ>i
such thatTh
haspositive
density.
An affirmative answer to thisquestion
ispre-cisely
thegeneralization
of Artin’sconjecture
that we discussed above. Inorder to conclude that
Ps
haspositive
lowerdensity,
we need thefollowing
precise
version of thisconjecture.
Generalized Artin
conjecture:
Letf
EZ[X]
be an irreducible monicpolynomial,
a a root off,
and define K =Q (a)
withring
ofintegers O K .
Furthermore,
let T be the set ofprimes
that are inert in and definefor each h E
Z>l,
the setTh
ofprimes p E T
for which thesubgroup
(1i)
C has index h.Then
Th
has a naturaldensity
6(Th)
for all h EMoreover,
we havethe
equality
=8(T).
As we saw
before,
although
ithappens
that theconjecture
fails to hold forspecific
a, weexpect
theconjecture
to hold for’generic’
a. Based oncom-puter
calculations forquadratic
polynomials,
Brown and Zassenhaus madea similar
conjecture
[2].
Under some mildassumptions
on thepolynomial
f , they conjectured
that the setTl
haspositive
density.
The
generalized
Artinconjecture
has beenproved
forintegers
a 0
(0, + 1 ) ,
or,equivalently,
for linearpolynomials f ,
under theassumption
of theproved
[12]
thatpart
of theconjecture
holds for irreduciblequadratic
poly-nomials
f .
Unfortunately,
we are not able to prove thegeneralized
Artinconjecture
for asingle polynomial
ofdegree
at least3,
not even under theassumption
of thegeneralized
Riemannhypothesis.
Corollary.
Let S be a linear recurrent sequence oforder k >
3. Assumethe associated characteristic
polynomial
satisfies thegeneralized
Artincon-jecture,
and contains ak-cycle
in its Galois group. ThenPs
contains a set ofpositive
density
Proof.
Let a be a root of the characteristicpolynomial f ,
and letOK
be thering
ofintegers
of the number field K =Q(a).
For allprimes
p notdividing
d,
thering
O/pO
isisomorphic
toBy
theassumption
on
f ,
we conclude thatTh
has adensity
6(Th)
for allpositive integers
h.According
to thetheorem,
thedensity
of the setTS h
ofprimes
p ETh
that divide S isequal
to6(Th) .
Denote the upperdensity
and lowerdensity
ofTs
=T fl PS
by
S(TS)
and6(TS)
respectively.
Using
the aboveremarks,
wefind:
r’V"’B I"V’B
We conclude that the set
Ts
has adensity
and moreover we have6(T) .
As the Galois group off
contains ak-cycle,
thisdensity
ispositive.
D
4. Proof of the theorem
As in section
2,
we choose aninteger
d #
0 such that xn =where 0 =
R (X ~ / ( f )
is a freealgebra
over R =Z[~j,
and c e O isuniquely
determined
by
Fix h EZ>i
and define the setLet p
ETh
be aprime
notdividing
d and write 1i for the reduction of amodulo
pO.
By assumption
f
is irreducible modulo p, and =is a field of order
pk.
As iscyclic,
thesubgroup generated by
1iconsists of the h-th powers in
((O/pO)*,
and we findTogether
with theequivalence
(4)
in section2,
thisyields
thefollowing
characterization:We will show that if p is
large enough,
then theright
hand side of(5)
is satisfied. Thisimplies
that all butfinitely
manyprimes
inTh
divide,S’,
and the theorem isproved.
By extending
scalars,
we find thatO[X,,..., Xk]
is a freeJR[Xi,...,
Xk]-algebra.
Define thehomogeneous polynomial
Note that
Fh
isindependent
of theprime
p. The coefficientof
Xl
inFh
is the h-th term of the sequence S. If this coefficient is zero,then all
primes
divide S and the theorem isproved. Therefore,
we assumethat is non-zero, so
Fh
ishomogeneous
ofdegree
h. LetVh
Cbe the
projective
algebraic
set definedby
F~.
The reductionVh
ofVh
modulo p isgiven
by
thepolynomial
For j?i,... x~ E
Fp,
theequation
on theright
hand side of theequivalence
(5)
becomesFh(xl, ... ,
Xk)
=0,
so we arrive atwhere
Vh(Fp)
denotes the set ofFp-rational
points
ofVh.
In order to prove that
Vh
is anon-singular
projective variety,
we letWh
cPk-1 (FP)
be theprojective
algebraic
set definedby
If p
f h
andc ~
0,
then thepartial
derivatives ofGh
do not vanishsi-multaneously on
andWh
is a smoothprojective
hypersurface
of dimension k - 2. As k is at least
3,
the intersection of two differentirreducible
components
ofWh
isnon-empty
by
[3,
Theorem 7.2 on page48~.
However,
thepoints
of such an intersection aresingular
points
ofWh.
Because
Wh
issmooth,
this proves thatWh
isabsolutely
irreducible. Note that if kequals 2,
thenWh (Fp)
consists of hpoints,
and is notabsolutely
irreducible.The varieties
Wh and Vh
areisomorphic
overNamely,
we canThe determinant
of 0
isequal
to times the Vandermonde determinant of1iP, ... ,
As d EOlpO
is ofdegree
k overFp,
the determinantis non-zero
and 0
is anisomorphism.
We conclude that for all p ETh
that do not dividecdh,
thealgebraic
setVh
is asmooth,
absolutely
irreduciblehypersurface
inPk-1(Fp)
ofdegree
h.The result of
Lang
and Weil[8,
Theorem1]
yields
thefollowing
estimate for the number of rationalpoints
ofVh:
there exists a constantC,
depending
on h and k
only,
such that for allprimes p
ETh
notdividing
cdh we haveNote that instead of
[8]
one can also use theWeil-conjectures
to obtainthis estimate. We conclude that
if p E Th is large enough,
thenVh (Fp)
isnot
empty
and(6)
implies
that p divides ~S’. This finishes theproof
of the theorem.5. Numerical
examples
In the table
below,
we have collected some numerical data. For each ofthe five recurrences, there are two recurrent sequences. For each of these
sequences, we determined how many of the 5133
primes
up to 5-104
dividethe sequence and ordered them
by
theirsplitting
type.
Forexample,
for 853primes
up to 5 -104,
thepolynomial f
=X~ - 2X 2
+ X - 6 EFp[X]
splits
into three linear factors. Out of these 853primes,
293 divide thesequence defined
by f
and the initial conditions xi =1, x2 -
-2 andX3 = 5. A ’mixed’
splitting
type just
means thatf
EFp[X]
factors asthe
product
of a linear an aquadratic polynomial.
The first 6 sequenceshave an irreducible characteristic
polynomial
with Galois group63,
thesymmetric
group on 3 elements. For the next 2 sequences, thepolynomial
is irreducible with
cyclic
Galois group. The characteristicpolynomial
for the last two sequences factors overQ
as aproduct
of a linear and aquadratic
factor.
For each of these sequences, the conclusions that we drew on heuristic
arguments
in section3,
seem to be correct. Almost allprimes
that do notsplit completely,
divide the sequence, and out of theprimes
thatsplit
completely,
apositive proportion
does divide the sequence and apositive
proportion
does not.The first two characteristic
polynomials
are related. If a is a root off
=X~-2X~+X-6,
then~3
is a rootof 9
=X 3 - 254X2 - 3311X - 46~56.
In other
words,
the third and the fourth sequence aresubsequences
of thefirst and the
second,
respectively.
Therefore,
the first sequence has at leastas many
prime
divisors as the third sequence. The heuristicalarguments
in section 3
explain why
the first sequence hasconsiderably
moresplitting
prime
divisors than the third sequence, andwhy
both sequences have almost the same number ofnon-splitting
prime
divisors. First we note that as aand
fl
defineisomorphic
numberfields,
thesplitting
type
off
modulo p isthe same as that
of g
modulo p. Our heuristical modelpredicts
that for anylinear recurrent sequence S of order k >
3,
almost allprimes
that do notsplit completely
should divide ,S’. Inparticular,
almost allprimes
that do notsplit completely
divide both the first and the third sequence. Now let p be asplitting
prime,
so the’probability’
that p does not divide either of thesequences is
positive.
If the order of a E(0/ pO) *
is notrelatively
prime
to6,
the order =as
is smaller than that of a.According
to ourheuristics,
it is then more
likely
for p to divide the first sequence, than to divide the third. Forexample,
there are no oddsplitting
primes
p for which the orderof
fl
E(0 /pO) *
is p - l. On the otherhand,
out of the set ofprimes
thatsplit
completely,
theprimes
p for which the order of a E(0 /pO) *
isp -1,
have the
largest ’probability’
to divide the first sequence.Therefore,
the number ofsplitting
primes
dividing
the first sequence should belarger
than the number ofsplitting
primes
dividing
the third sequence.We conclude with a few observations. As we can see from the fifth and the
sixth sequence, the number of
splitting
primes
dividing
a linear recurrentseems to
dependent
notonly
on the characteristicpolynomial,
but also onthe initial conditions. This can not be
explained by
our heuristics.Our theorem is in fact
empty
for the fifth and sixth sequence.Namely,
let a be a root of f
= X 3 - X2 - X -1,
andlet p
be aprime
for whichf
isirreducible modulo p. As a has norm
1,
the index(1i)]
is at leastp-1,
and the setTh
is finite for all h E Thisargument
also shows that thegeneralized
Artinconjecture
does not hold forf .
However,
the datasuggest
that our heuristical conclusions on the number ofprime
divisorsis still correct. This can be
explained by
theassumption
that for ’most’primes
p, the order of a E is still ’aslarge
aspossible’.
To be moreprecise,
if weadapt
our heuristicsby replacing
e(p)
by
theexponent
e(p)
of the kernel of the norm N :(C~/pC7)* -~
Fp,
we can draw the sameconclusions as in section 3. For a
generalization
of Artin’sconjecture
forunits in real
quadratic
fields,
we refer to[11].
Finally
we note that further numericalexperiments
seem tosuggest
that for the sequences ,5’ in thetable,
thesplitting
primes
dividing
,S’ have adensity.
As anexample,
from the table we see that34%
of the 853splitting
primes
up to5 ~ 104
divide the first sequence. Wecomputed
that out of thefirst 853
splitting
primes
greater
than 5 -104,
thispercentage
is35%.
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[16] M. WARD, The maximal prime divisors of linear recurrences. Can. J. Math. 6 (1954), 455-462 Hans ROSKAM Mathematisch Instituut Universiteit Leiden Postbus 9512, 2300 RA Leiden The Netherlands E-mail : roskamQmath.leidenuniv.nl