859-8
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) le plus grand facteur
Nous obtenons des majorations sur le nombre de
solutions de congruences de la forme n! + 2n - 1 ~ 0 (mod q) et
nous utilisons ces bornes pour montrer que
premier de
n.
lim sup P(n! + 2n -
1)/n
~
(203C02 +3)/18.
n~~
ABSTRACT. For an integer n ~ 2 we denote by P(n) the largest
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 denote by P(k) the largest prime factor
the number of distinct prime divisors of k. We also set
of h; and by
1
and
0.
P(l)
It is trivial to see that P(n! + 1) > n. Erd6s and Stewart [4] have shown
that
=
=
..
--
This bound is improved in [7] where it is shown that the above upper limit
is at least 5/2, and that it also holds for P(n! + f (~c)) with a nonzero
polynomial f (X )
Here
E
Z[X].
the method of
arguments, to show that
we use
-
,
.
[7],
-.-
which
.
we
Manuscrit reçu le 7 novembre 2003.
with
some new
,
’B.
We also estimate the total number of distinct
value of n! + 2n - 1 with 1 n X.
one
supplement
primes which divide
at least
860
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 of
independent interest.
Certainly, there is nothing special in the sequence 2n - l, and exactly
the same results can be obtained for n! + u(n) with any nonzero binary
recurrent sequence u (n) .
Finally, we note that our
with
approach
can
be used to estimate
P(n! + u(n))
arbitrary linear recurrence sequence u(n) (leading to similar, albeit
weaker, results) and with many other sequences (whose growth and the
number of zeros modulo q are controllable).
Throughout this paper, we use the Vinogradov symbols », « and A
as well as the Landau symbols 0 and o with their regular meanings. For
z > 0, log z denotes the natural logarithm of z.
an
Acknowledgments. During the preparation of this paper, F. L.
ported in part by grants SEP-CONACYT 37259-E and 37260-E,
was supported in part by ARC grant DP0211459.
and I. S.
the number of solutions of
and
2.
Bounding
some
equations
congruences
The
plays
following polynomial
an
important role
Lemma 2.1. The
has
no
our
arguments.
equation
integer solutions (n, k, m) with
Proof. One simply
Hence,
in
> 0
n > 3 and
notices that for any n > 3 and
for n > 3.
m
m
>
>
was
k > 1.
~; > 1
sup-
861
Let .~(q) denote the multiplicative order of 2 modulo an odd
For
and q > 1, we denote by
of solutions of the following congruence
and put
T(y, x, q) _ #T(y, x, q).
3.
the set
integer q >
q)
We also define
Lemma 2.2. For any prime p and integers x and y with p
we have
> x >
y+ 1
>
1,
We assume that p > 3, otherwise there is nothing to prove. Let
>
£(p) z > 1 be a parameter 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. Then
Proof.
=
...
where
and 1,f2
It is clear that
Then there exists
such that
=
«
Assume now that n E
integers 1~ and m with 0 l~
(~ - y)/z.
a nonzero
m G z, and
2n from the first and the second congruence, and then from
the first and the third congruence, we obtain
Eliminating
Now
eliminating n!,
we
derive
0 (mod p), where Fk,m(X) is given by (2.1). Because £(p) >
that for every 0 1~
has a
the polynomial
z,
nonzero coefficient modulo p and deg
m
z, thus for every 0 k
m
z there are at most z suitable values of n (since p > x > y + I > 1).
or
Fk,m(n)
-
we see
=
862
Summing
over
all admissible values of k and m,
we
derive
#T,f2 G z3 -!-
1.
inequality.
0
Therefore
Taking z
~)~~(p) 2013 1 }
=
we
obtain the desired
for any n > p with n! + 2n - 1
Thus
Obviously,
(mod p).
Lemma 2.3. For any
-
0
(mod p),
we
have 2~ - 1
integers q > 2 and x > y -~- 1 > l, we have
Assume that T(y, x, q) > 6, because otherwise there is nothing to
prove. We can also assume that q is odd. Then, by the Dirichlet principle,
there exist integers n > 4 , m > 1~ > l, satisfying the inequalities
Proof.
and such that
Arguing
as
Because
IFm,k(n)1
=
in the
proof of Lemma 2.2, we derive
0 by Lemma 2.1, we obtain
0((2.r)~). Therefore,
0
q.
(mod q).
Obviously,
=
and the result follows.
Certainly,
Lemma 2.2 is useful
Lemma 2.4. For any x the
maybe
The bound
0
large enough.
x 1/2/ log x holds for all excerpt
x.
log R/ log log R W
We remark that stronger results are
to be of help for our arguments.
seem
if ~(p) is
inequality. ~(p) >
pri7rtes p
G
only
L2/ log L concludes the proof.
known,
see
[3, 6, 9],
but
0
they do not
863
3. Main Results
Theorem 3.1. The
following
bound holds:
that the statement of the above theorem is false, we see
that there exist two constants A
(2~rz + 3)/18 and p such that the inequality P(n! + 2~ - 1) An + p holds for all integer positive n.
We let x be a large positive integer and consider the product
Proof. Assuming
Let
Q
=
P(W)
so we
have
Q
Ax + p .
Obviously,
For a prime p, we denote by sp the largest power of p
of the nonzero integers of the form n! + 2n - 1 for n
by rp the p-adic order of W . Hence,
and
therefore, by (3.1)
and
We let .M be the set of all
side of (3.3), that is,
and
so
(3.3)
can
(3.2),
we
x.
deduce
possible pairs (p, s)
be written
at least one
We also denote
dividing
which
occur on
the left hand
as
As usual, we use 7r(y) to denote the number of
and recall
that by the Prime Number Theorem we have 7r(y) = (1 + o(I))y/logy.
Now we introduce subsets
93, E4 E A4, which possibly overlap, and
whose contribution to the sums on the left hand side of (3.4) is
After this, we study the contribution of the remaining set G.
864
~
Let
we
9
Si be the set of pairs
have
(p, s)
E
.M with p
x/ log ~. By Lemma 2.3,
because obviously sp « :~ log x.
Let E2 be the set of pairs (p, s) E .M with s >
Lemma 2.3, and by the inequality
we
x/(log ~)2. Again by
have
because Q
O(r) by our assumption.
Let E3 be the set of pairs (p, s) E A4 with ~’(p)
by Lemmas 2.3 and 2.4, and by the inequality sp W
=
~
xll2l log x.
Again
we
have
865
~
Let E4 be the set of pairs (p, s) E
Lemma 2.2 and by (2.2), we have
We now put L
.M~
The above estimates,
=
The
U£2
U
together
A4)(Si
U
E3)
with s
E3 U ~4).
with
(3.4),
properties of the pairs (p, s) E G
can
show that
be summarized
In what follows, we repeatedly use the above bounds.
We now remark that because by our assumption P(n! -fwe see that
T( L(p - p) /A ] , ,x,pS).
we obtain
Thus, putting rp
=
=
Therefore,
(3.6)
where
and
xl/4. By
as
2" -1)
An -~ ~
866
To estimate
U,
we
observe
that, by
Lemma
2.3,
Furthermore,
Hence
We
Q
p
estimate V. For
with
now
an
integer a >
1
we
let 7~~ be the set of primes
Thus,
R(p)p.
Accordingly, let ,Ca be the subset of pairs (p, s)
=
We see that if
order of n! is
(p, s)
E L and
n
E L for which p E Pa.
then
x,
p2 > n, and therefore the p-adic
I
For
pEP a
we
Clearly, if n >
Thus,
I
also have
ordp(n! -t- 2’’~ - 1) > 0 only for n =
p~(p) » x3/2 j (log x)2 > x we see that,
p then
Because
Therefore,
-
for n
ap - 1 and
n -
0
(mod f (p)),
we
have
0
(mod f(p)).
x,
867
On the other hand, for
Hence, for n - 0 (mod L’(p)),
As
we
have mentioned
(a + l)p,
we
ordp(rt! + 2n - 1)
(mod R(p)). Thus, T((a + I)p,
For
a
=
l, 2, ..., let
us
we
have
ordp(n!)
>
n/p -
I >
a.
derive
=
=
0 for
0 for every n > p with
(p, s)
E
rc =
0
La.
define
We then have
where
For every a >
We
now
1, and (p, s)
need the
E
L,,
as we
have seen,
bound,
which is a modified version of (2.2). Indeed, if
:~ then Xa,p
x,
and we count solutions in an empty interval. If
ap - 1 (the other
alternative), we then replace the congruence modulo P’ by the congruence
modulo p and remark that because n >
p we have n!+2~-1 - 2n-1
and
is
now
immediate.
(mod p)
(3.8)
We use (3.8) for
s > x1~4, and Lemma 2.3 for
>
s >
calculations
lead
to
the
bound
Simple
=
=
x1~2~(log ~)z >
x1/2/ (log x)2.
We
now
have
=
868
Thus, putting everything together, and taking
P", 0152 l, 2, ..., are disjoint, we derive
into account that the sets
=
I
Hence
Substituting (3.7)
and
which contradicts the
(3.9)
in
(3.6),
and
assumption A
using (3.5),
(2~r2 + 3)/18,
we
derive
and thus finishes the
0
proof.
Theorem 3.2. For any
sufficiently large
Proof. In the notation of the
and Lemma 2.3, that
Obviously sp « xlogx/logp,
any prime number p,
proof of
x,
we
Theorem
therefore rp
3.1,
we
derive from
(3.2)
Thus,
for
869
which
together
with
(3.1)
finishes the
proof.
D
4. Remarks
We recall the result of Fouvry [5], which asserts that P(p - 1) >
holds for a set of primes p of positive relative density (see also [1, 2] for this
and several more related results). By Lemma 2.4, this immediately implies
that ~(p) > p~v68 for a set of primes p of positive relative density. Using
this fact in our arguments, one can easily derive that actually
However, the results of [5], or other similar results like the ones from [1, 2],
do not give any effective bound on the relative density of the set of primes
with P(p-1) > pO.668 , and thus cannot be used to get an explicit numerical
improvement of Theorem 3.1.
We also remark that, as in [7], one can use lower bounds on linear forms
in p-adic logarithms to obtain an "individual" lower bound on P(n!+2n-1).
The ABC-conjecture can also used in the same way as in [8] for P(n! + 1).
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870
Florian LUCA
Instituto de Matemdticas
Universidad Nacional Autonoma de Mexico
C.P. 58089, Morelia, Michoacan, Mexico
E-mail : [email protected]
Igor E. SHPARLINSKI
Department of Computing
Macquarie University
Sydney, NSW 2109, Australia
E-mail : igoreics .mq. edu. au