On a problem of Erdős involving the largest prime factor of $n$

0.17 MB·English

by  Aleksandar Ivić

4
0
0
2 ON A PROBLEM OF ERDO˝S INVOLVING

n
a THE LARGEST PRIME FACTOR OF n
J

8

] Aleksandar Ivic´
T
N
.
h Abstract. Let P(n) denote the largest prime factor of an integer n(≥2), and let
at N(x) denotethe numberof natural numbersnsuchthat 2≤n≤x, and n doesnot
m divide P(n)!. We prove that
[
x logt
2 N(x)=x 2+O log2x/logx 2 ρ(logx/logt) t2 dt,
v (cid:18) (cid:18)q (cid:19)(cid:19)Z
6
5 where ρ(u) is the Dickman-deBruijnfunction. In termsof elementaryfunctionswe
0 have
1 N(x)=xexp − 2logxlog2x(1+O(log3x/log2x)) ,
1
3 thereby sharpening and cornrecpting recent results of K. Ford and J.-oM. De Koninck
0
and N. Doyon.
/
h
t
a
m
1. Introduction and statement of results
:
v
i Let P(n) denote the largest prime factor of an integer n ( 2). In 1991 P. Erd˝os
X
≥
[5] proposed the following problem: prove that N(x) = o(x) (x ), where N(x)
r
a → ∞
denotes the number of natural numbers n such that 2 n x, and n does not
≤ ≤
divide P(n)!. This problem is connected to the so-called Smarandache function
S(n), the smallest integer k such that n k!.
|
Erd˝os’s assertion was shown to be true by I. Kastanas [13], and S. Akbik [1]
proved later that N(x) xexp( 1√logx) holds. K. Ford [8] proposed an as-
≪ −4
ymptotic formula for N(x). His Theorem 1 states that
√π(1+log2)
(1.1) N(x) (logxlog x)3/4x1 1/u0ρ(u ) (x ).
∼ 23/4 2 − 0 → ∞
In this formula one should correct the constant, as will be show in Section 4. Here,
as usual, log x logx and log x = log(log x) for k 2. The function ρ(u) is
1 ≡ k k−1 ≥
1991 Mathematics Subject Classification. 11N25, 11N37.
Key words and phrases. The largest prime factor of n, Dickman-de Bruijn function.
Typeset by AMS-TEX
2 A. Ivi´c
the continuous solution to the difference delay equation uρ (u) = ρ(u 1) with
′
− −
the initial condition ρ(u) = 1 for 0 u 1 and is commonly called the Dickman
≤ ≤
(or Dickman-de Bruijn) function. The function u = u (x) is implicitly defined by
0 0
the equation
2
logx = u x1/u0 1 ,
0
−
(cid:16) (cid:17)
so that
1/2 2
2logx log x log2 log x
(1.2) u (x) = 1 3 + +O 3 .
0
log x − 2log x 2log x log x
(cid:18) 2 (cid:19) ( 2 2 (cid:18) 2 (cid:19) !)
The asymptotic formula (1.1) and the explicit expression (1.2) enabled Ford (op.
cit.) in Corollary 2 to deduce that
(1.3) N(x) = xexp (√2+o(1)) logxlog x (x ).
− 2 → ∞
n p o
Unaware of Ford’s work, J.-M. De Koninck and N. Doyon [3] recently tackled
this problem again. In [3] they published the result that
(1.4) N(x) = xexp (2+o(1)) logxlog x (x ).
− 2 → ∞
n p o
Unfortunately, (1.4) is not true (it contradicts (1.3)). Namely the argument in
[3] leading to their (3.8) does not hold, and the crucial parameter u = logx/logy
in the formula for Ψ(x,y) has to be evaluated more carefully. A straightforward
modification of their proof leads then again to (1.3).
The aim of this note is to sharpen (1.1) and (1.3), and to provide asymptotic
formulas for sums of Sr(n) and 1/Sr(n) when r R is fixed. The results are
∈
contained in the following
THEOREM 1.
3
log x
(1.5) N(x) = xexp √2L(x) 1+g (x)+O 3 ,
0
− log x
( (cid:18) 2 (cid:19) !!)
where
(1.6) L = L(x) = logxloglogx,
and, for r > 1, p
−
log x+log(1+r) 2 log2 2
g (x) = 3 − − 1+
r
2log x log x
(1.7) 2 (cid:18) 3 (cid:19)
(log x+log(1+r) log2)2
3 − ,
− 8log2x
2
On a problem of Erdo˝s 3
THEOREM 2.
log x x logx logt
(1.8) N(x) = x 2+O 2 ρ dt.
s logx !!Z2 (cid:18)logt(cid:19) t2
THEOREM 3. Let r > 0 be a fixed number. Then we have
2
1 log x
(1.9) = xexp √2rL(x) 1+g (x)+O 3 ,
2 n x Sr(n) (− r−1 (cid:18)log2x(cid:19) !!)
≤X≤
where L(x) and g (x) are given by (1.6) and (1.7), respectively. Moreover, for
r
any fixed integer J 1 there exist computable constants a = ζ(r +1)/(r +1),
1,r
≥
a ... ,a such that
2,r J,r
J a xr+1
(1.10) Sr(n) = xr+1 j,r +O .
logj x logJ+1x
2 n x j=1 (cid:18) (cid:19)
≤X≤ X
The asymptotic formula (1.8) is sharper than (1.5)–(1.7), since it is a true as-
N(x)
ymptotic formula, while (1.5)–(1.7) is actually an asymptotic formula for log .
x
On the other hand, the right-hand side of (1.5) is given in terms of elementary
functions, while (1.8)contains the (non-elementary) function ρ(u). Thus is seemed
in place to give both (1.5) and (1.8), especially since the proof of (1.5) requires
only the more elementary analysis which follows the work of De Koninck–Doyon
[3].
The asymptotic formula (1.9)seems to be new, while (1.10) sharpens the results
of S. Finch [7], who obtained (1.10) in the case J = 1.
Acknowledgement. I thank Prof. G. Tenenbaum for pointing out formula
(2.6) to me, and indicating how it improves the error term O(log x/log x) in the
3 2
first version of Th. 2 (coming from (2.5)), to O( log x/logx). I also thank the
2
referee for valuable remarks.
p
2. The necessary results of Ψ(x,y)
The proofs of both Theorem 1 and Theorem 2 depend on results on the function
(2.1) Ψ(x,y) = 1 (2 y x),
≤ ≤
n x,P(n) y
≤ X ≤
4 A. Ivi´c
which represents the number of n not exceeding x, all of whose prime factors do
not exceed y. All the results that follow can be found e.g., in [9], [10] and [14].
We have the elementary bound
logx
(2.2) Ψ(x,y) xexp (2 y x)
≪ −2logy ≤ ≤
(cid:18) (cid:19)
and the more refined asymptotic formula
log(u+1)
Ψ(x,y) = xρ(u) 1+O
(2.3) logy
(cid:18) (cid:18) (cid:19)(cid:19)
u = logx/logy, exp((log x)5/3+ε) y x.
2 ≤ ≤
Note that the Dickman–de Bruijn function ρ(u) admits an asymptotic expansion,
as u , which in a simplified form reads
→ ∞
log u
ρ(u) = exp u logu+log u 1+O 2 .
− 2 − logu
(cid:26) (cid:18) (cid:18) (cid:19)(cid:19)(cid:27)
Let, for given u > 1, ξ = ξ(u) be defined by uξ = eξ 1, so that
−
2
log u log u
(2.4) ξ(u) = logu+log u+ 2 +O 2 .
2 logu logu
!
(cid:18) (cid:19)
If u > 2, v 1u, then we have the asymptotic formula
| | ≤ 2
(2.5) ρ(u v) = ρ(u)exp vξ(u)+O((1+v2)/u) .
−
(cid:8) (cid:9)
This result (see e.g., Hildebrand–Tenenbaum [10]) for bounded v (see (3.3.9) of
| |
de la Bret`eche–Tenenbaum [2]) yields the asymptotic formula
1
(2.6) ρ(u v) = ρ(u)evξ(u) 1+O ( v v ).
0
− · u | | ≤
(cid:26) (cid:18) (cid:19)(cid:27)
3. Proof of Theorem 1
We turn now first to the proof of Theorem 1. The idea is to obtain upper
and lower bounds of the form given by (1.5)–(1.7). For the former, note that if
P2(n) n, then n is counted by N(x), since P2(n) cannot divide P(n)!. Therefore,
|
if T (x) denotes the number of natural numbers n x such that P2(n) n, then
0
≤ |
N(x) T (x). The desired lower bound follows then from the formula for T (x),
0 0
≥
which is a special case of the formula for (r 0 is a given constant)
≥
1
(3.1) T (x) := ,
r Pr(n)
2 n x,P2(n)n
≤ ≤X |
On a problem of Erdo˝s 5
obtained by C. Pomerance and the author [12], and sharpened by the author in
[11]. Namely for T (x) it was shown that
r
3
log x
(3.2) T (x) = xexp (2r +2)1/2L(x) 1+g (x)+O 3 ,
r r
− log x
( (cid:18) 2 (cid:19) !!)
where L(x)and g (x) aregiven by (1.6)and (1.7). Wenote that Erd˝os, Pomerance
r
and the author in [6] investigated the related problem of the sum of reciprocals of
P(n). They proved that (this result is also sharpened in [11])
(3.3)
1 1 x log x x logx 1
= Ψ ,p = x 1+O 2 ρ dt.
2 n x P(n) p x p (cid:18)p (cid:19) s logx !!Z2 (cid:18)logt(cid:19) t2
≤X≤ X≤
The same argument leads without difficulty to
x log x x logx logt
(3.4) T (x) = Ψ ,p = x 1+O 2 ρ dt.
0 p x (cid:18)p2 (cid:19) s logx !!Z2 (cid:18)logt(cid:19) t2
X≤
It remains yet to deal with the upper bound for N(x). As in [3], we shall use
the inequality (p denotes primes)
x
(3.5) N(x) Ψ ,pr .
≤ pr
2≤r≤lXogx/log2 p≤Xx1/r (cid:18) (cid:19)
To see that (3.5) holds note that, if n is counted by N(x), then n must be divis-
ible by pr which does not divide P(n)!. The condition r 2 is necessary, since
≥
squarefree numbers q (> 1) divide P(q)!. Moreover, pr P(n) cannot hold, since
≤
otherwise the numbers p,2p,... ,rp would all divide P(n)!, and so would pr, which
is contrary to our assumption. Thus n = prm(r 2), P(m) P(n) < pr, which
≥ ≤
easily gives (3.5).
We restrict now the ranges of r and p on the right-hand of (3.5) to the ones
which will yield the largest contribution.
The contribution of r > 3L is trivially
xp r x2 3L = xexp( 3log2 L),
− −
≪ ≪ − ·
r>3L p
X X
which is negligible, since 3log2 > √2. Similarly, the contribution of p > exp(2L)
is negligible, and also for p > exp((logx)1/3) and r > (logx)1/6log x the contri-
2
bution is, for some C > 0,
xp r xexp( C logx log x),
≪ − ≪ − · 2
r>(logxX)1/6log2xp>exp(X(logx)1/3) p
6 A. Ivi´c
which is negligible.
With (3.5) and (2.3) we see that in the range
L := exp((logx)1/3) < p exp(1L), r (logx)1/6log x
1 ≤ 4 ≤ 2
we have
log(x/pr) logx rlogp
u = = −
logpr logp+logr
logx+O((logx)2/3+ε) logx
= = 1+O (logx)ε 1/3 .
−
logp+O(log x) logp
2
(cid:16) (cid:16) (cid:17)(cid:17)
Therefore by (2.3) we obtain, as x ,
→ ∞
x x logx logx
Ψ ,pr exp (1+o(1)) log ,
pr ≤ pr − logp logp
(cid:18) (cid:19) (cid:18) (cid:18) (cid:19)(cid:19)
which gives
x
Ψ ,pr
pr
L1<p≤Xexp(14L) Xr≥2 (cid:18) (cid:19)
1 logx logx
x max exp (1+o(1)) log
≪ L1<p≤Xexp(41L) p2 L1<p≤exp(14L) (cid:18)− logp (cid:18)logp(cid:19)(cid:19)
xexp( (2+o(1))L),
≪ −
which is negligible. The contribution of p L ,r > (logx)1/6log x is easily seen
≤ 1 2
to be also negligible. This means that the main contribution to the right-hand
side of (3.5) comes from the range
(3.6) exp(1L) p exp(2L), 2 r (logx)1/6log x,
4 ≤ ≤ ≤ ≤ 2
or more precisely, for some C > √2,
N(x) xexp( CL)
≪ −
x
(3.7)
+ Ψ ,pr .
pr
2≤r≤(logXx)1/6log2x exp(14L)X≤p≤exp(2L) (cid:18) (cid:19)
By trivial estimation it transpires that the contribution of r 8 in (3.7) will be
≥
negligible, so that we may write
(3.8) N(x) xexp( CL)+Lmax (x),
c,r
≪ − c>0 T
2 r 7
≤X≤
On a problem of Erdo˝s 7
where, for a given constant c > 0 and r N,
∈
x
(x) := Ψ ,pr ,
Tc,r pr
exp(cL 1)<p exp(cL) (cid:18) (cid:19)
− X≤
since (again by (2.3)) the contribution of c c , c a large positive constant, is
0 0
≥
negligible. If p is in the range indicated by (x), then
c,r
T
logxp r logx log x
u = − = 1+O 2 ,
logpr cL s logx !!
log x
logu = log x logc logL+O 2 .
2 − − s logx !
Therefore, for 2 r 7,
≤ ≤
x logx log x
(x) exp 1+O 2
Tc,r ≪ pr − cL s logx ×
exp(cL−1X)<p≤exp(cL) n (cid:16) (cid:16) (cid:17)(cid:17)
(1 log x+ 1 log x+O(1))
× 2 2 2 3
1 o log x
xexp cL(r 1) L 1+O 3 ,
≪ − − − 2c log x
(cid:26) (cid:18) (cid:18) 2 (cid:19)(cid:19)(cid:27)
so that the largest contribution is for r = 2. The function c+ 1/(2c) attains its
minimal value √2 when c = 1/√2, hence
log x
(x) xexp √2L 1+O 3 ,
c,r
T ≪ − log x
(cid:26) (cid:18) (cid:18) 2 (cid:19)(cid:19)(cid:27)
and we obtain
log x
(3.9) N(x) xexp 2logxlog x 1+O 3 .
≤ − 2 log x
(cid:26) (cid:18) (cid:18) 2 (cid:19)(cid:19)(cid:27)
p
This is somewhat weaker than the result implied by Theorem 1. But we saw
that the main contribution to (3.7) comes from (x), which is quite similar
c,2
T
to the main sum appearing in the estimation (lower bound) of T (x) (cf. [12]).
0
The only difference is that in the case of (x) we have Ψ(xp 2,2p), while in
c,2 −
T
the case of T (x) the terms Ψ(xp 2,p) appear. In the relevant range for p (i.e.,
0 −
exp(1L) p exp(2L)) this will not make much difference, so that using the
4 ≤ ≤
arguments [12] we can obtain for N(x) the upper bound implied by the expression
standing in (1.5)–(1.7). It is only for the sake of clearness of exposition that we
gave all the details for the slightly weaker upper bound (3.9). This discussion
completes the proof of Theorem 1.
8 A. Ivi´c
4. Proof of Theorem 2
For the proof of Theorem 2 we need an expression that is sharper than (3.5),
which is only an upper bound. If n is counted by N(x) then
a) Either P2(n) divides n, and the contribution of such n is counted by T (x)
0
(cf. (3.1)), or
b) The number n is of the form n = pqbk,p = P(n),b 2,P(k) < p,P(k) = q,
≥ 6
where henceforth q (as well as p) will denote primes. Since n does not divide
P(n)!, then (similarly as in Section 3) it follows that qb > P(n). Also, again as
in Section 3, it is found that the contribution of p exp(L/4) = Y is negligible.
1
≤
The contribution of p > exp(2L) = Y is also negligible. Namely in this case, since
2
q2 n,q > P(n)/b exp(2L)/logx, the contribution is clearly
| ≫
x p 2 xexp( 3L).
−
≪ ≪ −2
p exp(2L)/logx
≫ X
Therefore, ifb) holds, weneedonlyconsidernumbers nforwhich thereisanumber
b 2 and a prime q (p/b,p), such that n = pqbk,p = P(n),P(k) < p,P(k) = q.
≥ ∈ 6
The contribution of P(k) = q can be easily seen to be negligible, as can be also
shown for the one pertaining to P(k) = p. Using the technique of Section 3 (or
of [8]) it is seen that the main contribution comes from the sum with b = 2. If
b = 2, namely if n = pq2k, p/2 < q < p and q2 k with some prime p/2 < q < p,
1| 1
then n is counted at least twice, but it is seen that the contribution of such n is
xexp( 3L). Therefore the contribution to N(x) of integers satisfying b) equals
≪ −2
x
(4.1) O(xexp( 3L))+ Ψ ,p .
−2 pq2
Y1<Xp≤Y2p/2X<q<p (cid:18) (cid:19)
This is different from Ford [8, eq. (2)], who had Ψ x ,q instead of the correct
pq2
Ψ x ,p . This oversight does affect his result (1.1)(cid:16), which(cid:17)is not correct since the
pq2
con(cid:16)stant(cid:17)is not the right one (1+log2 should be replaced by 2), but the relation
(1.3) remains valid, since both expressions with the Ψ-function are of the same
order of magnitude. It follows that our starting relation for the proof of Theorem
2 takes the shape
x
(4.2) N(x) = T (x)+O(xexp( 3L))+ Ψ ,p .
0 −2 pq2
Y1<Xp≤Y2p/2X<q<p (cid:18) (cid:19)
By using the asymptotic formula (2.3) for Ψ(x,y), the prime number theorem in
the standard form (see e.g., [14])
x dt
(4.3) π(x) = 1 = +O(xexp( logx)),
logt −
p x Z2
X≤ p
On a problem of Erdo˝s 9
we write the sum in (4.2) as a Stieltjes integral and integrate by parts. We set for
brevity
log x
R := 2 .
s logx
Then we see that the sum in question equals
Y2 1 t 1 logx logy
(1+O(R))x ρ 2 1 dydt
logt ty2logy logt − logt −
ZY1 Zt/2 (cid:18) (cid:19)
Y2 1 1 1 logx
= (1+O(R))x ρ 2z 1 dzdt,
tlogt tzz logt − −
ZY1 Z1 log2/logt (cid:18) (cid:19)
−
on making the substitution logy = zlogt. Since both z and 1/z equal 1+O(R)
in the relevant range, by the use of (2.5) we see that our sum becomes
Y2 e2ξ(logx/logt 1) logx 1
(1+O(R))x − ρ 1 t zdzdt
−
tlogt logt −
(4.4) ZY1 (cid:18) (cid:19)Z1−log2/logt
Y2 e2ξ(logx/logt 1) logx
−
= x(1+O(R)) ρ 1 dt.
t2log2t logt −
ZY1 (cid:18) (cid:19)
Now we use (2.6) (with u 3 replacing u, v = 2), setting u = logx/logt,
− −
Y t Y , to obtain that
1 2
≤ ≤
e2ξ(u 1)ρ(u 1) = e2ξ(u 1)ρ((u 3)+2)
− −
− −
= e2ξ(u 1) 2ξ(u 3)ρ(u 3) (1+O(R)).
− − −
− ·
Since ξ (u) 1/u (u ) we have
′
∼ → ∞
1
e2ξ(u 1) 2ξ(u 3) = eO(1/u) = 1+O = 1+O(R).
− − −
u
(cid:18) (cid:19)
Therefore the last integral in (4.4) equals
Y2 logx dt
(4.5) (1+O(R)) ρ 3 .
logt − t2log2t
ZY1 (cid:18) (cid:19)
With the change of variable
logx dt du
u = , =
logt tlog2t −logx
10 A. Ivi´c
it follows that (4.5) becomes
1 y2 logx
S := (1+O(R)) ρ(u 3)exp du,
logx − − u
Zy1 (cid:18) (cid:19)
logx logx
y := 1 , y := 4 .
1 2slog2x 2 slog2x
But since uρ (u) = ρ(u 1), integrating by parts we obtain
′
− −
1 y2 logx
S = (1+O(R)) uρ (u 2)exp du,
′
logx − − − u
Zy1 (cid:18) (cid:19)
1 y2 logx logx
= (1+O(R)) ρ(u 2) 1+ exp du,
logx − u − u
Zy1 (cid:18) (cid:19) (cid:18) (cid:19)
y2 1 logx
= (1+O(R)) (u 1)ρ (u 1) exp du,
′
− − − u − u
Zy1 (cid:18) (cid:19) (cid:18) (cid:19)
y2 logx logx
= (1+O(R)) ρ(u 1) exp du,
− u2 − u
Zy1 (cid:18) (cid:19) (cid:18) (cid:19)
y2 1 logx
= logx(1+O(R)) ρ (u) exp du,
′
− u − u
Zy1 (cid:18) (cid:19) (cid:18) (cid:19)
y2 logx logx
= logx(1+O(R)) ρ(u) exp du,
u3 − u
Zy1 (cid:18) (cid:19) (cid:18) (cid:19)
where we used several times that u 1/R in the range of integration, so that
≍
lower order terms could be absorbed by the O(R)–term. Making again the change
of variable logx/logt = u, we obtain
Y2 logx logt T (x)
0
S = (1+O(R)) ρ dt = (1+O(R)) ,
logt t2 x
ZY1 (cid:18) (cid:19)
where (3.4) was used. This proves (1.8).