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RESEARCH NOTES
ON A DENSITY PROBLEM OF ERDÖS
SAFWAN AKBIK
(Received 13 April 1998 and in revised form 10 June 1998)
Abstract.For a positive integern, letP(n)denotes the largest prime divisor ofnand define the set:(x)== {n≤x:ndoes not divideP(n)!}. Paul Erdös has proposed that
|S| =o(x)asx→ ∞, where|S|is the number ofn∈S. This was proved by Ilias Kastanas.
In this paper we will show the stronger result that|S| =O(xe−1/4 logx).
Keywords and phrases. Number theory, primes, factorial, density, divisibility.
1991 Mathematics Subject Classification. 11B05, 11N25.
Introduction. For a positive integern, letP(n)denote the largest prime divisor of nand define the set
(x)==
n≤x:ndoes not divideP(n)!}. (1) Paul Erdös [1] proposed that|S| =o(x)asx→ ∞, where|S|is the number ofn∈S. A solution [3] was provided by Ilias Kastanas. There was also [3] a claim of proving that
|S(x)| =O(x/logx). In this paper, we show the stronger result.
Theorem. For some constanta >0, we have
|S| =O xe−a√
logx
. (2)
In fact,a=1/4suffices.
Lemma1. Letν(n)be the number of distinct prime divisors ofn. Define S1=
n≤x:ν(n) >4kloglogx, k≥1
. (3)
Then
|S1| =O x
(logx)k
(4) uniformly ink.
Proof. It is well known [2] that if d(m) is the number of divisors of m, then
m≤xd(m)=O(xlogx). Sinced(m)≥2ν(m), O(xlogx)≥
m≤x
2ν(m)≥
m∈S1
2ν(m)≥
m∈S1
(logx)(4log2)k≥ |S1|(logx)k+1, (5)
and the lemma follows.
656 SAFWAN AKBIK
Lemma2. LetC(x)=C=(logx)k, wherek=k(x)will be chosen later. Define S2=
n≤x:p2|nfor some primep > C
. (6)
Then
|S2| =O x
(logx)k
. (7)
Proof. Sincen∈S2if and only ifn=tp2for someC < p≤√xand somet≤x/p2,
|S2| =
C<p≤√ x
x p2
≤
C<p≤√ x
x p2=O
x C
=O x
(logx)k
. (8)
The first big O in (8) follows since
p>C1/p2 ≤ ∞
[C]du/u2 = 1/[C] = O(1/C), forC≥1.
Lemma3. Let S3=
n≤x:pα|nfor someα≥Tand some primeP≤C
, (9)
whereT=2logC. Then
|S3| =O x
(logx)k
. (10)
Proof.
|S3| =
p≤Cα≥T
x pα
≤2
p≤C
x
pT ≤2 x 2T p
4
p2=Ox 2T
=O x
C2log2
≤O x
C
=O x
(logx)k
.
(11)
The first inequality in (11) is valid because
α≥T1/pα≤1/pT+1/pT+1+ ··· ≤2/pT and the second is valid becausepT=2T(p/2)T≥2T(p/2)2forT≥2.
Lemma4. LetS(x)=S−(S1∪S2∪S3), then, for any n∈S, we have
P(n)≤2CT . (12)
Proof. Letn∈S. Thenn∈Sand sondoes not divideP(n)!. There exists a prime p0dividingnsuch that
νp0(n) > νp0
P(n)!
, (13)
whereνp(m)denotes the largest integertsuch thatptdividesm. Sinceνp0(P(n)!)≥ 1, (13) implies thatνp0(n)≥2. Sincen∈S2, it follows thatp0≤C. Alson∈S3, so thatT≥νp0(n). Hence,
T≥νp0(n) > νp0
P(n)!
≥P(n)
2p0 ≥P(n)
2C (14)
which implies (12). Note that the third inequality in (14) is true because νp(m!)≥ m/2pfor anymand anyp|m. This is because
νp(m!)≥ m
p
>m
p −1≥ m
2p, p=m, (15)
and[m/p]=1≥m/2pifp=m.
Proof of the Theorem. For n∈S, we haven∈S1∪S2∪S3. Thus, νp(n)≤ 4kloglogx. Also, ifpαis any prime power dividingn, then one of the following two possibilities must occur:
(a) p≤Candα≤T, (b) p > Candα=0 or 1.
Case (a) generates at mostC(T+1)≤2CT prime powers. For Case (b), the number of prime powerspαwithp > C andα≤1 is at mostP(n). By Lemma 4, this is at most 2CT. Hence, the number of possible prime powerspαthat divide ann∈Sis at most 4CT. But such anncan be the product of at most 4kloglogxdistinct prime powers.
Therefore,
|S| ≤(4CT )4kloglogx=(8ClogC)4kloglogx
=e4kloglogx(log8+kloglogx+logk+logloglogx)≤e8k2(loglogx)2 (16) since log8+log(kloglogx)≤kloglogx, forkloglogx≥4.
Choosing
k=1 4·
logx
loglogx, (17)
(16) gives|S| ≤e(1/2)logx=x1/2. Hence, S=O
xe−(1/4)√
logx
. (18)
From (17), we havex/(logx)k=xe−1/4√
logx. Lemmas 1, 2, and 3 imply that
|Si| =O xe−1/4√
logx
, i=1,2,3. (19)
Finally,S=S∪[S∩(S1∪S2∪S3)]. Hence, (18) and (19) yield
|S| ≤ |S|+|S1|+|S2|+|S3| =O
xe−(1/4)√
logx
, (20)
and (2) follows witha=1/4.
Remark. Ifπ(x)is the number of prime integers that are less than or equal tox, an early version of the prime numbers theorem asserts that
π(x)= x
2
du logu+O
xe−a√
logx
, (21)
for some constanta. Although the bigOterms in (19) and (2) are similar, there is no apparent relationship between the PNT and (2).
658 SAFWAN AKBIK References
[1] P. Erdös,A Proposed Problem, Amer. Math. Monthly98(1991), 965.
[2] G. H. Hardy and E. M. Wright,An Introduction to the Theory of Numbers, 5th ed., The Claren- don Press, Oxford University Press, New York, 1979. MR 81i:10002. Zbl 423.10001.
[3] I. Kastanas,The smallest factorial that is a multiple ofn, Amer. Math. Monthly101(1994), 179.
Akbik: Department of Mathematics, Hofstra University, Hempstead, NY11550, USA
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