Journal of Inequalities in Pure and
Applied Mathematics
AN ASYMPTOTIC EXPANSION
GABRIEL MINCU
Universitatea Bucureşti
Facultatea de Matematică
Str. Academiei nr. 14, RO-70109,
Bucureşti, România.
E-Mail: gamin@al.math.unibuc.ro
volume 4, issue 2, article 30,
2003.
Received 24 December, 2002;
accepted 12 May, 2003.
Communicated by: L. Tóth
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c
2000
Victoria University
ISSN (electronic): 1443-5756
153-02
Quit
Abstract
In this paper we study the asymptotic behaviour of the sequence (rn )n of the
√
powers of primes. Calculations also yield the evaluation rn − pn = o logns n
for every positive integer s, pn denoting the n-th prime.
2000 Mathematics Subject Classification: 11N05, 11N37
Key words: Powers of primes, Inequalities, Asymptotic behaviour
Contents
1
2
3
4
5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
On the Asymptotic Behaviour of π̃ . . . . . . . . . . . . . . . . . . . . . . 6
Initial Estimates for rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Computing the Coefficients of the Polynomial Rm . . . . . . . . . 14
An Asymptotic Expansion
Gabriel Mincu
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
1.
Introduction
One denotes by:
• pn the n-th prime
• rn the n-th number (in increasing order) which can be written as a power
pm , m ≥ 2, of a prime p.
• π(x) the number of prime numbers not exceeding x.
An Asymptotic Expansion
• π̃(x) the number of prime powers pm , m ≥ 2, not exceeding x.
Gabriel Mincu
The asymptotic equivalences
π(x) ∼
(1.1)
x
log x
Title Page
Contents
and
JJ
J
pn ∼ n logn
(1.2)
are well known.
M. Cipolla [1] proves the relations
(1.3)
Go Back
Close
pn = n(log n + log log n − 1) + o(n)
Quit
and
Page 3 of 16
(1.4)
II
I
log log n−2
pn = n log n + log log n−1 +
log n
+o
n
log n
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
that he generalizes to
Theorem 1.1. There exists a sequence (Pm )m≥1 of polynomials with integer
coefficients such that, for any integer m,
"
#
m
X
(−1)j−1 Pj (log log n)
1
(1.5) pn = n log n+log log n − 1+
+o
.
logm n
logj n
j=1
In the same paper, M. Cipolla gives recurrence formulae for Pm ; he finds
that every Pm has degree m and leading coefficient (m − 1)!.
As far as (rn )n is concerned, L. Panaitopol [2] proves the asymptotic equivalence
Gabriel Mincu
rn ∼ n2 log2 n.
(1.6)
We prove in this paper that (rn )n has an asymptotic expansion comparable
to that of Theorem 1.1 .
We will need the next results of L. Panaitopol:
√
√
(1.7)
π̃(x) − π( x) = O( 3 x),
(from [2]), and
Proposition 1.2. There exist a sequence of positive integers k1 , k2 , . . . and for
every n ≥ 1 a function αn , limx→∞ αn (x) = 0, such that:
x
(1.8)
π(x) =
.
kn (1+αn (x))
k1
k2
log x − 1 − log
−
−
·
·
·
−
n
2
log x
x
log x
Moreover, k1 , k2 , . . . are given by the recurrence relation
(1.9)
kn + 1! · kn−1 + 2! · kn−2 + · · · + (n − 1)! · k1 = n · n!,
An Asymptotic Expansion
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 4 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
n ≥ 1.
http://jipam.vu.edu.au
(from [3]).
We will also establish a result similar to Proposition 1.2 for π̃(x) and the
evaluation
√
n
rn − p n = o
logs n
for every positive integer s.
An Asymptotic Expansion
Gabriel Mincu
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 5 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
2.
On the Asymptotic Behaviour of π̃
Proposition 2.1. For every positive integer n, there exists a function βn ,
limx→∞ βn (x) = 0, such that
√
x
(2.1)
π̃(x) =
,
√
k (1+βn (x))
k1√
n−1
√ − n n√
log x − 1 − log x − . . . − logkn−1
log
x
x
(kn )n being the sequence of (1.9).
An Asymptotic Expansion
Proof. Let us set
Gabriel Mincu
√
(2.2)
π̃(x) =
log
√
x−1−
k1√
log x
x
− ... −
kn−1
√
logn−1 x
−
kn (1+βn (x))
√
logn x
.
Title Page
Contents
(1.8) and (1.7) give us:
(2.3)
JJ
J
√
√ kn [βn (x) − αn ( x)]
√
3
x·
=
O(
x),
logn+2 x
Go Back
so
(2.4)
II
I
Close
√
kn [βn (x) − αn ( x)] = O
n+2
log
x
√
6
x
,
Quit
Page 6 of 16
leading to limx→∞ βn (x) = 0.
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
3.
Initial Estimates for rn
Equation (2.1) gives:
√
2 x
π̃(x) ∼
.
log x
(3.1)
If we put x = rn , we get
√
2 rn
n∼
,
log rn
(3.2)
An Asymptotic Expansion
Gabriel Mincu
so
(3.3)
lim (log 2 + log
√
n→∞
rn − log n − log log rn ) = 0,
Title Page
Contents
whence
√
log rn
= 1,
n→∞ log n
lim
(3.4)
JJ
J
II
I
Go Back
leading to:
(3.5)
Close
lim (log log rn − log 2 − log log n) = 0.
n→∞
Page 7 of 16
(3.3) and (3.5) give:
(3.6)
Quit
log
√
rn = log n + log log n + o(1).
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
(2.1) implies
√
(3.7)
π̃(x) =
log
√
x
.
x − 1 + o(1)
For x = rn we get (in view of (3.6)):
√
rn
(3.8)
= log n + log log n − 1 + o(1).
n
By taking logarithms of both sides we get:
√
1
log log n − 1
(3.9) log rn − log n = log log n + log 1 +
+o
.
log n
log n
log log n−1
1
For big enough n we have log n + o log n < 1, which means that we
can expand the logarithm. We derive:
√
1
log log n − 1
(3.10)
log rn = log n + log log n +
+o
.
log n
log n
Gabriel Mincu
Title Page
Contents
JJ
J
II
I
Go Back
(2.1) also gives:
√
(3.11)
An Asymptotic Expansion
π̃(x) =
log
√
x−1−
Close
x
1√
log x
+o
1
log x
.
Quit
Page 8 of 16
For x = rn and in view of (3.4), we obtain:
√
rn
√
1
1
.
(3.12)
= log rn − 1 −
√ +o
log rn
log n
n
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
(3.10) and (3.12) allow us to write
√
(3.13)
rn
log log n − 1
= log n + log log n − 1 +
n
log n
1
1
h
i + o
−
.
log n−1
1
log n
+
o
log n 1 + logloglogn n + loglog
2
2
n
log n
For big enough n we have
log log n log log n − 1
1
< 1;
+o
2
2
log n +
log n
log n we can therefore use the expansion of
(3.14)
√
1
1+x
An Asymptotic Expansion
Gabriel Mincu
Title Page
in (3.13) and we get
log log n − 2
rn = n log n + log log n − 1 +
log n
+o
n
log n
Contents
.
JJ
J
II
I
Go Back
Close
Quit
Page 9 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
4.
Main Result
Theorem 4.1. For every positive integer s we have
√
rn − p n
1
(4.1)
=o
.
n
logs n
Proof. Induction with respect to s.
For s = 1 the statement is true because of (1.4) and (3.14).
Now let s ≥ 1; suppose that
√
rn − p n
1
(4.2)
=o
.
n
logs n
An Asymptotic Expansion
Gabriel Mincu
(4.2) and (1.5) lead to
"
#
s
X
√
(−1)j−1 Pj (log log n)
1
(4.3) rn =n log n+log log n−1+
+o
.
s
j
log
n
log
n
j=1
By taking logarithms of both sides in (1.5) we derive
(4.4)
log pn = log n + log log n
"
#
s
j−1
X
log log n−1
(−1) Pj (log log n)
1
+log 1+
+
+o
.
s+1
j+1
log n
log n
log n
j=1
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 10 of 16
(1.8) gives us
(4.5)
Title Page
x
π(x) =
log x − 1 −
k1
log x
−
k2
log2 x
− ... −
ks+1
logs+1 x
+o
.
1
log
s+1
x
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
For x = pn , this relation becomes (in view of (1.2)):
pn
k1
ks+1
= log pn − 1 −
− ... −
+o
n
log pn
logs+1 pn
(4.6)
1
logs+1 n
.
By taking logarithms of both sides in (4.3) we get
(4.7)
log
√
rn = log n + log log n
"
#
s
log log n−1 X (−1)j−1 Pj (log log n)
1
+log 1+
+
+o
.
s+1
j+1
log n
log
n
log
n
j=1
An Asymptotic Expansion
Gabriel Mincu
(2.1) gives
(4.8)
x
π̃(x) =
log
√
s+1
√ +o
x−1− logk1√x − logk22√x −· · ·− logks+1
x
1 √
logs+1 x
.
For x = rn , this relation becomes (in view of (3.4)):
√
rn
√
k1
1
ks+1
(4.9)
= log rn − 1 −
.
√ − ··· −
√ +o
n
log rn
logs+1 rn
logs+1 n
If x and y are ≥ 1, Lagrange’s theorem gives us the inequality
Contents
JJ
J
II
I
Go Back
Close
Quit
|log y − log x| ≤ |y − x| ;
(4.10)
Title Page
Page 11 of 16
with (4.4) and (4.7), it leads to:
(4.11)
log
√
rn − log pn = o
log
s+1
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
1
n
.
http://jipam.vu.edu.au
This last relation gives for every t ∈ {1, 2, . . . , s + 1}
1
1
1
1
(4.12)
−
=o
.
√ =o
logt pn logt rn
logs+t+2 n
logs+1 n
(4.6), (4.9), (4.11) and (4.12) give
√
rn − p n
1
(4.13)
=o
n
logs+1 n
An Asymptotic Expansion
and the proof is complete.
Gabriel Mincu
Theorem 4.2. There exists a unique sequence (Rm )m≥1 of polynomials with
integer coefficients such that, for every positive integer m,
(4.14) rn = n2 log2 n + 2(log log n − 1) log n + (log log n)2
#
m
X
(−1)j−1 Rj (log log n)
n2
.
−3+
+o
m
j
log
n
(j
+
1)!
·
log
n
j=1
Proof. (4.9) allows us to write
"
(4.15) rn = n
2
log n + log log n − 1 +
Title Page
Contents
JJ
J
II
I
Go Back
Close
m+1
X
j=1
(−1)j+1 P j(log log n)
j! · logj n
2
1
+o
.
logm+1 n
Quit
Page 12 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
If we set
(4.16)
R1 := 4(X − 1)P1 − 2P2
and
(4.17) Rj := −2Pj+1 +2(j +1)(X −1)Pj −
j−1
X
(j +1)
i=1
j
i
Pi Pj−i
,j ≥ 2
(4.15) gives for every m ≥ 1:
2
rn = n log2 n + 2(log log n − 1) log n + (log log n)2
− 3+
m
X
(−1)j−1 Rj (log log n)
j=1
(j + 1)! · logj n
An Asymptotic Expansion
Gabriel Mincu
#
+o
n2
logm n
Title Page
,
so the existence is proved.
Suppose now the existence of two different sequences (Rm )m≥1 and (Sm )m≥1
satisfying the conditions of the theorem. For the least j such as Sj 6= Rj we can
write
1
Rj (log log n) − Sj (log log n)
=o
,
(j + 1)! · logj n
logj n
so Rj (log log n) − Sj (log log n) = o(1), a contradiction.
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 13 of 16
Corollary 4.3. We have
rn = n2 log2 n + 2n2 (log log n − 1) log n + n2 (log log n)2 − 3n2 + o(n2 ).
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
5.
Computing the Coefficients of the
Polynomial Rm
Proposition 5.1. For every m ≥ 1, the degree of Rm is m + 1 and its leading
coefficient is 2(m − 1)!.
Proof. If we recall from the introduction that every Pn has degree n and leading
coefficient (n − 1)!, the statement follows from (4.16) and (4.17).
An Asymptotic Expansion
(1.4) gives
P1 (X) = X − 2.
Gabriel Mincu
We can easily derive from M. Cipolla’s paper [1] the relations
0
Pk0 = k(k − 1)Pk−1 + k · Pk−1
,k ≥ 2
and
Title Page
Contents
JJ
J
Pk+1 (0)
( k−1 )
X k−1 0
= −k
Pj (0)[Pk−j (0) + Pk−j
(0)] + [Pk (0) + Pk0 (0)]
j
j=1
0
(0).
− (k + 1)Pk (0) − Pk+1
Computations gave him
P2 (X) = X 2 − 6X + 11;
P3 (X) = 2X 3 − 21X 2 + 84X − 131;
II
I
Go Back
Close
Quit
Page 14 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
P4 (X) = 6X 4 − 92X 3 + 588X 2 − 1908X + 2666;
P5 (X) = 24X 5 − 490X 4 + 4380X 3 − 22020X 2 + 62860X − 81534;
P6 (X) = 120X 6 − 3084X 5 + 35790X 4 − 246480X 3 + 1075020X 2
− 2823180X + 3478014;
P7 (X) = 720X 7 − 22428X 6 + 322224X 5 − 2838570X 4 + 16775640X 3
− 66811920X 2 + 165838848X − 196993194.
In view of (4.16) and (4.17), we get in turn:
R1 (X) = 2X 2 − 14;
R2 (X) = 2X 3 − 6X 2 − 42X + 172;
R3 (X) = 4X 4 − 24X 3 − 144X 2 + 1544X − 3756;
R4 (X) = 12X 5 − 110X 4 − 600X 3 + 12300X 2 − 64060X + 122298;
R5 (X) = 48X 6 − 600X 5 − 2940X 4 + 102000X 3 − 842520X 2
+ 3319512X − 5484780;
R6 (X) = 240X 7 − 3836X 6 − 16380X 5 + 913080X 4 − 10543400X 3
+ 63989100X 2 − 215203884X + 323035480.
An Asymptotic Expansion
Gabriel Mincu
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 15 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au
References
[1] M. CIPOLLA, La determinazione assintotica dell nimo numero primo,
Rend. Acad. Sci. Fis. Mat. Napoli, Ser. 3,.8 (1902), 132–166.
[2] L. PANAITOPOL, On some properties of the π ∗ (x) − π(x) function, Notes
Number Theory Discrete Math., 6(1) (2000), 23–27.
[3] L. PANAITOPOL, A formula for π(x) applied to a result of Koninck-Ivić,
Nieuw Arch. Wiskunde, 5(1) (2000), 55–56.
An Asymptotic Expansion
Gabriel Mincu
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 16 of 16
J. Ineq. Pure and Appl. Math. 4(2) Art. 30, 2003
http://jipam.vu.edu.au