On the largest prime factor of n! + 2n- 1
par FLORIAN LUCA et IGOR E. SHPARLINSKI
RÉSUMÉ. Pour un entier n > 2, notons
P(n)
leplus grand
facteur premier de n. Nous obtenons des majorations sur le nombre desolutions de congruences de la forme n! + 2n - 1 ~ 0
(mod q)
etnous utilisons ces bornes pour montrer que
lim sup
P(n!
+ 2n -1)/n
~(203C02 +3)/18.
n~~
ABSTRACT. For an integer n ~ 2 we denote
by P(n)
thelargest
prime factor of n. We obtain several upper bounds on the number of solutions of congruences of the form n! + 2n - 1 ~ 0(mod q)
and use these bounds to show that
lim sup P(n! -2n- 1)/n ~ (203C02 +3)/18.
n~~
1. Introduction
For any
positive integer k
> 1 we denoteby P(k)
thelargest prime
factorof h; and
by
the number of distinctprime
divisors of k. We also setP(l)
= 1 and = 0.It is trivial to see that
P(n! + 1)
> n. Erd6s and Stewart[4]
have shownthat
.. --
This bound is
improved
in[7]
where it is shown that the above upper limit is at least5/2,
and that it also holds forP(n!
+f (~c))
with a nonzeropolynomial f (X )
EZ[X].
Here we use the method of
[7],
which wesupplement
with some new arguments, to show that- , . -.- . , ’B.
We also estimate the total number of distinct
primes
which divide at leastone value of n! + 2n - 1 with 1 n X.
Manuscrit reçu le 7 novembre 2003.
These results are based on several new
elements,
such as bounds for the number of solutions of congruences with n! + 2n -1,
which could be ofindependent
interest.Certainly,
there isnothing special
in the sequence2n - l,
andexactly
the same results can be obtained for n! +
u(n)
with any nonzerobinary
recurrent sequence
u (n) .
Finally,
we note that ourapproach
can be used to estimateP(n!
+u(n))
with an
arbitrary
linear recurrence sequenceu(n) (leading
tosimilar,
albeitweaker, results)
and with many other sequences(whose growth
and thenumber of zeros
modulo q
arecontrollable).
Throughout
this paper, we use theVinogradov symbols
», « and Aas well as the Landau
symbols 0
and o with theirregular meanings.
Forz >
0, log z
denotes the naturallogarithm
of z.Acknowledgments. During
thepreparation
of this paper, F. L. was sup-ported
inpart by grants
SEP-CONACYT 37259-E and37260-E,
and I. S.was
supported
in partby
ARC grant DP0211459.2.
Bounding
the number of solutions of someequations
andcongruences The
following polynomial
plays
animportant
role in ourarguments.
Lemma 2.1. The
equation
has no
integer
solutions(n, k, m)
with n > 3 and m > k > 1.Proof.
Onesimply
notices that for any n > 3 and m > ~; > 1Hence,
> 0 for n > 3.Let
.~(q)
denote themultiplicative
order of 2 modulo an oddinteger q >
3.For
and q > 1,
we denoteby q)
the setof solutions of the
following
congruenceand
put T(y, x, q) _ #T(y,
x,q).
We also defineLemma 2.2. For any
prime
p andintegers x
and y with p > x >y+ 1
>1,
we have
Proof.
We assume thatp > 3,
otherwise there isnothing
to prove. Let£(p)
> z > 1 be aparameter
to be chosen later.Let y + 1 ni ... nt x be the
complete
list of t =T(y,
x,p)
solutions to the congruence n! + 2n - 1 - 0
(mod p),
y + 1 n x. Thenwhere
and
1,f2
=It is clear that «
(~ - y)/z.
Assume now that n EThen there exists a nonzero
integers
1~ and m with 0 l~m G z,
andsuch that
Eliminating
2n from the first and the second congruence, and then from the first and the third congruence, we obtainNow
eliminating n!,
we deriveor
Fk,m(n)
- 0(mod p),
whereFk,m(X)
isgiven by (2.1).
Because£(p)
>z, we see that for every 0 1~ the
polynomial
has anonzero coefficient modulo p and
deg
= m z, thus for every 0 km z there are at most z suitable values of n
(since p > x > y + I > 1).
Summing
over all admissible values of k and m, we derive#T,f2 G z3 -!-
1.Therefore
Taking z
=~)~~(p) 2013 1 } we obtain the desired inequality.
0
Obviously,
for any n > p
with n! + 2n - 1 - 0 (mod p),
we have 2~ - 1
(mod p).
ThusLemma 2.3. For any
integers q >
2 andx > y -~- 1 > l, we
haveProof.
Assume thatT(y,
x,q) > 6,
because otherwise there isnothing
toprove. We can also assume
that q
is odd.Then, by
the Dirichletprinciple,
there exist
integers n > 4 , m > 1~ > l, satisfying
theinequalities
and such that
Arguing
as in theproof
of Lemma2.2,
we derive 0(mod q).
Because 0
by
Lemma2.1,
we obtain q.Obviously, IFm,k(n)1
= =0((2.r)~). Therefore,
and the result follows. 0
Certainly,
Lemma 2.2 is usefulonly
if~(p)
islarge enough.
Lemma 2.4. For any x the
inequality. ~(p) > x 1/2/ log x
holdsfor
allexcerpt
maybe pri7rtes p
x.The bound G
log R/ log log
R WL2/ log L
concludes theproof.
0We remark that
stronger
results areknown,
see[3, 6, 9],
butthey
do notseem to be of
help
for ourarguments.
3. Main Results Theorem 3.1. The
following
bound holds:Proof. Assuming
that the statement of the above theorem isfalse,
we seethat there exist two constants A
(2~rz
+3)/18 and p
such that the in-equality P(n!
+ 2~ -1)
An + p holds for allinteger positive
n.We let x be a
large positive integer
and consider theproduct
Let
Q
=P(W)
so we haveQ
Ax + p .Obviously,
For a
prime
p, we denoteby
sp thelargest
powerof p dividing
at least oneof the nonzero
integers
of the form n! + 2n - 1 for n x. We also denoteby
rp thep-adic
order of W .Hence,
and
therefore, by (3.1)
and(3.2),
we deduceWe let .M be the set of all
possible pairs (p, s)
which occur on the left handside of
(3.3),
thatis,
and so
(3.3)
can be written asAs
usual,
we use7r(y)
to denote the number of and recall thatby
the Prime Number Theorem we have7r(y) = (1
+o(I))y/logy.
Now we introduce subsets
93, E4 E A4,
whichpossibly overlap,
andwhose contribution to the sums on the left hand side of
(3.4)
isAfter
this,
westudy
the contribution of theremaining
set G.~ Let
Si
be the set ofpairs (p, s)
E .M withp x/ log
~.By
Lemma2.3,
we have
because
obviously
sp « :~log
x.9 Let
E2
be the set ofpairs (p, s)
E .M with s >x/(log ~)2. Again by
Lemma
2.3,
andby
theinequality
we have
because
Q
=O(r) by
ourassumption.
~ Let
E3
be the set ofpairs (p, s)
E A4 with~’(p) xll2l log
x.Again
by
Lemmas 2.3 and2.4,
andby
theinequality
sp W we have~ Let
E4
be the set ofpairs (p, s)
EA4)(Si
UE3)
with sxl/4. By
Lemma 2.2 and
by (2.2),
we haveWe now
put L
=.M~ U£2
UE3 U ~4).
The above
estimates, together
with(3.4),
show thatThe
properties
of thepairs (p, s)
E G can be summarized asIn what
follows,
werepeatedly
use the above bounds.We now remark that because
by
ourassumption P(n! -f- 2" -1)
An -~ ~we see that =
T( L(p - p) /A ] , ,x,pS).
Thus, putting
rp = we obtainTherefore, (3.6)
where
and
To estimate
U,
we observethat, by
Lemma2.3,
Furthermore,
Hence
We now estimate V. For an
integer a >
1 we let7~~
be the set ofprimes p Q
withThus,
=R(p)p.
Accordingly,
let,Ca
be the subset ofpairs (p, s)
E L for which p EPa.
We see that if
(p, s)
E L and n x, thenp2
> n, and therefore thep-adic
order of n! is
I - I
For
pEP a
we also haveClearly, if n >
p thenordp(n! -t- 2’’~ - 1)
> 0only
for n = 0(mod f(p)).
Because
p~(p)
»x3/2 j (log x)2
> x we seethat,
x,Therefore,
for n ap - 1 and n - 0(mod f (p)),
we haveThus,
On the other
hand,
for(a
+l)p,
we haveordp(n!)
>n/p -
I > a.Hence,
for n - 0(mod L’(p)),
we deriveAs we have mentioned
ordp(rt!
+ 2n -1)
= 0 for every n > p with rc = 0(mod R(p)). Thus, T((a
+I)p,
= 0 for(p, s)
ELa.
For a =
l, 2, ...,
let us defineWe then have
where
For every a >
1,
and(p, s)
EL,,
as we have seen,We now need the
bound,
which is a modified version of
(2.2). Indeed,
if = :~ thenXa,p
= x,and we count solutions in an
empty
interval. If = ap - 1(the
otheralternative),
we thenreplace
the congruence moduloP’ by
the congruence modulo p and remark that because n > p we have n!+2~-1 - 2n-1(mod p)
and(3.8)
is now immediate.We use
(3.8)
forx1~2~(log ~)z >
s >x1~4,
and Lemma 2.3 for >s >
x1/2/ (log x)2. Simple calculations lead to the bound
We now have
Thus, putting everything together,
andtaking
into account that the setsP",
I 0152 =l, 2, ...,
aredisjoint,
we deriveHence
Substituting (3.7)
and(3.9)
in(3.6),
andusing (3.5),
we derivewhich contradicts the
assumption A (2~r2
+3)/18,
and thus finishes theproof.
0Theorem 3.2. For any
sufficiently large
x, weProof.
In the notation of theproof
of Theorem3.1,
we derive from(3.2)
and Lemma
2.3,
thatObviously
sp «xlogx/logp,
therefore rpThus,
forany
prime
number p,which
together
with(3.1)
finishes theproof.
D4. Remarks
We recall the result of
Fouvry [5],
which asserts thatP(p - 1) >
holds for a set of
primes
p ofpositive
relativedensity (see
also[1, 2]
for thisand several more related
results). By
Lemma2.4,
thisimmediately implies
that
~(p) > p~v68
for a set ofprimes
p ofpositive
relativedensity. Using
this fact in our
arguments,
one caneasily
derive thatactually
However,
the results of[5],
or other similar results like the ones from[1, 2],
do not
give
any effective bound on the relativedensity
of the set ofprimes
with
P(p-1) > pO.668 ,
and thus cannot be used toget
anexplicit
numericalimprovement
of Theorem 3.1.We also remark
that,
as in[7],
one can use lower bounds on linear forms inp-adic logarithms
to obtain an "individual" lower bound onP(n!+2n-1).
The
ABC-conjecture
can also used in the same way as in[8]
forP(n!
+1).
References
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Florian LUCA
Instituto de Matemdticas
Universidad Nacional Autonoma de Mexico C.P. 58089, Morelia, Michoacan, Mexico
E-mail : fluca@matmor.unam.mx
Igor E. SHPARLINSKI
Department of Computing Macquarie University Sydney, NSW 2109, Australia E-mail : igoreics .mq. edu. au