Journal of Inequalities in Pure and
Applied Mathematics
ERDŐS-TURÁN TYPE INEQUALITIES
LAURENŢIU PANAITOPOL
University of Bucharest
Faculty of Mathematics
14 Academiei St.
RO–70109 Bucharest
Romania.
EMail: pan@al.math.unibuc.ro
volume 4, issue 1, article 23,
2003.
Received 10 January, 2003;
accepted 19 February, 2003.
Communicated by: J. Sándor
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
Denoting by (rn )n≥1 the increasing sequence of the numbers pα with p prime
and α ≥ 2 integer, we prove that rn+1 − 2rn + rn−1 is positive for infinitely many
values of n and negative also for infinitely many values of n. We prove similar
1
1
properties for rn2 − rn−1 rn+1 and rn−1
− r2n + rn+1
as well.
Erdős-Turán Type Inequalities
2000 Mathematics Subject Classification: 11A25, 11N05.
Key words: Powers of prime numbers, Inequalities, Erdős-Turán theorems.
Laurenţiu Panaitopol
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
On the Difference rn+1 − rn . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Erdős-Turán Type Properties . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
4
8
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1.
Introduction
Let (rn )n≥0 be the increasing sequence of the powers of prime numbers (pα with
p prime and α ≥ 2 integer). Thus, we have r1 = 4, r2 = 8, r3 = 9, r4 = 16,
etc. Properties of the sequence (rn )n≥1 were studied in [5] and [3].
Denote by pn the n-th prime number. In [1], Erdős and Turán proved that
pn+1 − 2pn + pn−1 is positive for infinitely many values of n and negative also
for infinitely many values of n. Until now, no answer is known for the following
question raised by Erdős and Turán: Do there exist infinitely many numbers n
such that
pn+1 − 2pn + pn−1 = 0?
2
Erdős
and Turán also
proved that each of the sequences (pn − pn−1 pn+1 )n≥2 and
1
1
− p2n + pn+1
has infinitely many positive terms and infinitely many
pn−1
n≥2
negative ones.
Denoting by (qn )n≥1 the increasing sequence of the powers of prime numbers, the author proved in [4] that the value of qn+1 − 2qn + qn−1 changes its
sign infinitely many times.
In the present paper, we raise similar problems for the sequence (rn )n≥1 . We
need a few preliminary properties, which will be proved in the next section.
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2.
On the Difference rn+1 − rn
Property 2.1. We have
(2.1)
lim sup(rn+1 − rn ) = ∞.
n→∞
Proof. Let m ≥ 4. We show that, among the numbers
√
m! + 2, m! + 3, . . . , m! + [ m],
there is no term of the sequence (rn )n≥1 .
√
Assume that there exists an integer a such that 2 ≤ a ≤ [ m] and
(2.2)
m! + a = pi
where p is prime and i ≥ 2.
The relation (2.2) can also be written in the form
m!
a
+ 1 = pi , whence a = pj with 1 ≤ j ≤ i.
a
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It follows that
m!
m!
+ 1 = pi−j , hence j is not divisible by p.
j
p
p
If ep (n) is Legendre’s function, we have ep (m) = j, that is,
∞ X
m
(2.3)
= j.
ps
s=1
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√
√
Since a ≤ m, it follows that pj ≤ m, that is, m ≥ p2j , and then (2.3)
implies that
2j 2j 2j p
p
p
j≥
+
+ · · · + 2j
2
p
p
p
2j−1
2j−2
=p
+p
+ ··· + p + 1
≥ 22j−1 + 22j−2 + · · · + 2 + 1
= 22j − 1.
Since for j ≥ 1 we have 22j − 1 > j, we obtained a contradiction.
Since our assumption turned out to be false, it follows that for every m ≥ 4
there exists k = k(m) such that
√
rk ≤ m! + 1 and rk+1 ≥ m! + [ m] + 1,
√
whence rk+1 − rk ≥ [ m], and finally
lim sup(rn+1 − rn ) = ∞,
n→∞
(2.4)
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and the proof ends.
We now denote an =
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rn+1 −rn
n log2 n
and recall that, in [2], H. Meier proved that
pn+1 − pn
lim inf
< 0.248.
n→∞
log n
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In connection with this result, we prove:
Property 2.2. We have
(2.5)
lim inf an < 0.496.
n→∞
Proof. We consider the indices m such that
pm+1 − pm
< 0.248.
log m
Both the numbers p2m and p2m+1 occur in the sequence (rn )n≥1 , that is, p2m = rk
and p2m+1 = rh , with k = k(m), h = h(m) and h ≥ k + 1. In [5], it was proved
that, for m ≥ 1783, we have
(2.6)
p2m ≥ rm > m2 log2 m.
Since pm ∼ m log m, it follows that rk ∼ k 2 log2 k. But rk = p2m , hence
k log k ∼ m log m. One can show without difficulty that k(m) ∼ m. It then
follows that
√
√
√
√
rk+1 − rk
rh − rk
pm+1 − pm
<
=
.
log k
log k
log k
Since log k ∼ log m, we get
√
√
rk+1 − rk
pm+1 − pm
≤ lim inf
< 0.248.
lim inf
m→∞
k→∞
log k
log m
√
√
Since rk ∼ k log k and rk+1 ∼ (k + 1) log(k + 1) ∼ k log k, it follows that
rk+1 − rk
lim inf
< 0.496,
k→∞
k log2 k
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where k = k(m). Consequently,
lim inf
n→∞
rn+1 − rn
< 0.496.
n log2 n
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3.
Erdős-Turán Type Properties
For k ≥ 2 we denote
Rk = rk+1 − 2rk + rk−1 ,
and prove
Property 3.1. There exist infinitely many values of n such that
Rn > 0,
Erdős-Turán Type Inequalities
and also infinitely many ones such that
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Rn < 0.
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Pm
Proof. Denoting Sm = k=2 Rk , we have Sm = rm+1 − rm − r2 + 1. By (2.1)
we have lim sup Sm = ∞, hence Rn > 0 for infinitely many values of n.
m→∞
P
Denoting σm = m
k=2 kRk , we have
rm
2
2
σm = m(rm+1 − rm ) − rm − r2 + 2r1 = m log m am − 2 2
.
m log m
2
2
Since rm ∼ m log m, we get by (2.5) that lim inf σm = −∞, hence Rn < 0
m→∞
for infinitely many values of n.
For k ≥ 2, denoting ρk =
1
rk−1
−
2
rk
+
1
rk+1
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, we have
J. Ineq. Pure and Appl. Math. 4(1) Art. 23, 2003
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Property 3.2. There exist infinitely many values of n such that
ρn > 0,
and also infinitely many ones such that
ρn < 0.
m
P
0
Proof. For α > 3, denoting Sm
(α) =
k α ρk , we get
k=2
0
(α) = −
Sm
mα (rm+1 − rm ) mα − (m − 1)α
−
rm rm+1
rm
m−1
X k α − 2(k − 1)α + (k − 2)α
+
+ O(1).
rk
k=2
We have
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2
2
rk ∼ k log k,
k − (k − 1)α ∼ αk α−1 ,
k α − 2(k − 1)α + (k − 2)α ∼ α(α − 1)k α−2 ,
α
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whence
mα (rm+1 − rm )
mα−3 am
∼
,
rm rm+1
log2 m
mα − (m − 1)α
αmα−3
∼
,
rm
log2 m
α(α − 1)k α−4
k α − 2(k − 1)α + (k − 2)α
∼
.
rk
log2 k
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Since
m−1
X
k=2
k α−4
(α − 3)mα−3
∼
,
log2 m
log2 m
it follows that
0
Sm
(α) ∼
mα−3
· − am − α + α(α − 1)(α − 3) .
2
log m
0
Then lim Sm
(3.1) = −∞, and thus there exist infinitely many values of n
m→∞
such that ρn < 0.
0
On the other hand, we have by (2.5) that lim sup Sm
(4) = ∞, which shows
Erdős-Turán Type Inequalities
Laurenţiu Panaitopol
m→∞
that there exist infinitely many values of n such that ρn > 0.
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A consequence of Properties 3.1 and 3.2 is the following.
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Property 3.3. There exist infinitely many values of n such that
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rn−1 rn+1 > rn2 ,
and also infinitely many ones such that
rn−1 rn+1 < rn2 .
√
n−1
Proof. If rn > rn+1 +r
, then rn > rn−1 rn+1 . On the other hand, if
2
1
1
+ rn+1
, then
rn−1
1
1
√
rn < 2
+
< rn−1 rn+1 ,
rn−1 rn+1
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and then the desired conclusion follows by Properties 3.1 and 3.2.
Open problem. Do there exist infinitely many values of n such that
rn+1 − 2rn + rn−1 = 0?
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References
[1] P. ERDŐS AND P. TURÁN, On some new question on the distribution of
prime numbers, Bull. Amer. Math. Soc., 54(4) (1948), 371–378.
[2] H. MEIER, Small difference between prime numbers, Michigan Math. J.,
35 (1988), 324–344.
[3] G. MINCU, An asymptotic expansion, (in press).
[4] L. PANAITOPOL, Some of the properties of the sequence of powers of
prime numbers, Rocky Mountain J. Math., 31(4) (2001), 1407–1415.
[5] L. PANAITOPOL, The sequence of the powers of prime numbers revisited,
Math. Reports, 5(55)(1) (2003), (in press).
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