Journal of Inequalities in Pure and
Applied Mathematics
ASYMPTOTIC FORMULAE
RAFAEL JAKIMCZUK
División Matemática
Universidad Nacional de Luján
Luján, Buenos Aires
Argentina.
EMail: jakimczu@mail.unlu.edu.ar
volume 5, issue 3, article 62,
2004.
Received 09 January, 2004;
accepted 08 June, 2004.
Communicated by: S.S. Dragomir
Abstract
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Abstract
Let ts,n be the n-th positive integer number which can be written as a power
pt , t ≥ s, of a prime p (s ≥ 1 is fixed). Let πs (x) denote the number of prime
powers pt , t ≥ s, not exceeding x. We study the asymptotic behaviour of the
sequence ts,n and of the function πs (x). We prove that the sequence ts,n has
an asymptotic expansion comparable to that of pn (the Cipolla’s expansion).
Asymptotic Formulae
2000 Mathematics Subject Classification: 11N05, 11N37.
Key words: Primes, Powers of primes, Cipolla’s expansion.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Function πs (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1
3
The Sequences (ts ,n ) s and pn . . . . . . . . . . . . . . . . . . . . . . . . . 9
4
The Asymptotic Behaviour of ts,n . . . . . . . . . . . . . . . . . . . . . . . 15
References
Rafael Jakimczuk
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1.
Introduction
Let pn be the n-th prime. M. Cipolla [1] proved the following theorem:
There exists a unique sequence Pj (X) (j ≥ 1) of polynomials with rational
coefficients such that, for every nonnegative integer m,
(1.1) pn = n log n + n log log n − n
m
X
(−1)j−1 nPj (log log n)
n
+
+o
.
logm n
logj n
j=1
The polynomials Pj (X) have degree j and leading coefficient 1j .
P1 (X) = X − 2,
P2 (X) =
X 2 − 6X + 11
,
2
....
If m = 0 equation (1.1) is:
pn = n log n + n log log n − n + o(n).
(1.2)
Let π(x) denote the number of prime numbers not exceeding x, then
!
m
X
(i − 1)!x
(m − 1)!x
(1.3)
π(x) =
+ ε(x)
(m ≥ 1),
m
i
log
x
log
x
i=1
where lim ε(x) = 0.
Asymptotic Formulae
Rafael Jakimczuk
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x→∞
Lemma 1.1. There exists a positive number M such that in the interval [2, ∞),
|ε(x)| ≤ M.
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Proof. Let us consider the closed interval [2, a]. In this interval, π(x) ≤ x, so
logm x
π(x) is bounded. The functions (i−1)!x
, i = 1, . . . , m and (m−1)!x
are continuous
logi x
on the compact [2, a], so they are also bounded.
As
!#
"
m
X
(i − 1)!x
logm x
ε(x) = π(x) −
,
i
(m
−
1)!x
log
x
i=1
ε(x) is in its turn bounded on [2, a].
Since a is arbitrary and lim ε(x) = 0, the lemma is proved.
x→∞
Let us consider the sequence of positive integer numbers which can be written as a power pt of a prime p (t ≥ 1 is fixed). The number of prime powers pt
not exceeding x will be (in view of (1.3))
!
1
m
1 (m − 1)!x 1t
X
1
(i − 1)!x t
(1.4)
+ ε xt
π(x t ) =
1
i 1t
log
x
logm x t
i=1
!
1
m
1 tm (m − 1)!x 1t
X
ti (i − 1)!x t
=
+ +ε x t
i
logm x
log
x
i=1
!
!
1
1
m
X
ti (i − 1)!x t
xt
.
=
+o
m
i
log
x
log
x
i=1
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Rafael Jakimczuk
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2.
The Function πs(x)
Let ts,n be the n-th positive integer number (in increasing order) which can be
written as a power pt , t ≥ s, of a prime p (s ≥ 1 is fixed). Let πs (x) denote the
number of prime powers pt , t ≥ s, not exceeding x.
Theorem 2.1.
(2.1)
πs (x) =
1
m
X
si (i − 1)!x s
!
logi x
i=1
1
+o
xs
logm x
!
.
Asymptotic Formulae
Rafael Jakimczuk
Proof. If x ∈ [2s+k , 2s+k+1 ) (k ≥ 1), then
πs (x) = π x
1
s
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k
1 X
+
π x s+i .
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Using (1.4), we obtain
(2.2)
πs (x) =
1
m
X
si (i − 1)!x s
!
logi x
i=1
+
+ε x
1
s
sm (m − 1)!x 1s
logm x
1
m
X
(s + j)i (i − 1)!x s+j
k
X
j=1
i=1
1
+ε x s+j
logm x
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logi x
1
(s + j)m (m − 1)!x s+j
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!
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1
m
X
si (i − 1)!x s
=
i=1
!
+ε x
1
s
sm (m − 1)!x 1s
logm x
!
1
m
k
X
X
(s + j)i (i − 1)!x s+j
+
logi x
i=1
j=1
logi x
1
k
1 (s + j)m (m − 1)!x s+j
X
+
ε x s+j
.
logm x
j=1
Asymptotic Formulae
In the given conditions, the following inequalities hold for x:
k
P
j=1
1
k
P
(s+j)i (i−1)!x s+j
logi x
1
sm (m−1)!x s
m
log
=
j=1
1
(s+j)i (i−1)! s+j
· (m−1)! x
sm
x
logm−i x
Rafael Jakimczuk
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1
s
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x
≤
k
X
(s+j)i
sm
·
(i−1)!
(m−1)!
2
j=1
logm−i 2s+k+1
(s+k)j−s
s(s+j)
=
k (s + k) (s + k + 1)
1 k
2 s(s+1)
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k
X
(s + k)i (s + k + 1)m−i
≤
1 k
j=1
2 s(s+1)
i
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m−i
(i = 1, . . . , m).
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Now, since
k (s + k)i (s + k + 1)m−i
=0
1 k
k→∞
s(s+1)
2
lim
(i = 1, . . . , m),
we find that
k
P
lim
(2.3)
j=1
x→∞
1
(s+j)i (i−1)!x s+j
logi x
1
sm (m−1)!x s
logm x
=0
(i = 1, . . . , m).
On the other hand, from the lemma we have the following inequality
1
1
k
k
X
1 (s + j)m (m − 1)!x s+j
X
(s + j)m (m − 1)!x s+j
s+j
.
ε x
≤M
logm x
logm x
j=1
j=1
This inequality and (2.3) with i = m give
1 1
k
P
(s+j)m (m−1)!x s+j
s+j
ε x
logm x
j=1
(2.4)
lim
= 0.
1
x→∞
sm (m−1)!x s
logm x
Finally, from (2.2), (2.3) and (2.4) we find that
!
1
1
m
X
si (i − 1)!x s
sm (m − 1)!x s
0
πs (x) =
+ ε (x)
,
logm x
logi x
i=1
where lim ε0 (x) = 0. The theorem is proved.
x→∞
Asymptotic Formulae
Rafael Jakimczuk
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From (1.4) and (2.1) we obtain the following corollary
1
Corollary 2.2. The functions πs (x) and π x s have the same asymptotic behaviour (s ≥ 1).
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Rafael Jakimczuk
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3.
1
The Sequences (ts,n ) s and pn
Theorem 3.1.
1
s
(ts ,n ) = pn + o
(3.1)
n
logr n
(r ≥ 0).
Proof. We proceed by mathematical induction on r.
Equation (2.1) gives (m = 1)
lim
(3.2)
x→∞
πs (x)
1
sx s
= 1.
Asymptotic Formulae
Rafael Jakimczuk
log x
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If we put x = ts,n , we get
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lim
(3.3)
n→∞
(ts,n )
1
s
n log (ts,n )
1
s
= 1.
From (3.3) we find that
1
(3.4)
lim log s + log (ts,n ) s − log n − log log ts,n = 0.
n→∞
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Now, since
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1
(3.5)
log (ts,n ) s
lim
= 1,
n→∞
log n
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we obtain
1
(ts,n ) s
lim
n→∞
1
=1
n log (ts,n ) s
if and only if
1
(ts,n ) s
lim
= 1.
n→∞ n log n
We also derive
lim
n→∞ ns
ts,n
= 1.
logs n
Asymptotic Formulae
Rafael Jakimczuk
From (3.5) we find that
(3.6)
lim (− log s + log log ts,n − log log n) = 0.
n→∞
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(3.4) and (3.6) give
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1
s
log (ts,n ) = log n + log log n + o(1).
(3.7)
Equation (2.1) gives (m = 2)
πs (x) =
x
1
s
1
log x s
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
1
s
+
log
x
2
so
x
1
s
+ o
1
1
s
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1
s
x = πs (x) log x −
log
x
2
x
1
s
+o
xs
log x
1
s
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!
1
xs
log x
1
s
1
s
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If we put x = ts,n , we get
1
(3.8)
1
s
1
s
(ts,n ) = n log (ts,n ) −
1
(ts,n ) s
1
log (ts,n ) s
+o
!
(ts,n ) s
1
.
log (ts,n ) s
Finally, from (3.8), (3.3) and (3.7) we find that
1
(ts,n ) s = n log n + n log log n − n + o (n) .
(3.9)
Asymptotic Formulae
Therefore, for r = 0 the theorem is true because of (1.2) and (3.9).
Let r ≥ 0 be given, and assume that the theorem holds for r, we will prove
it is also true for r + 1.
From the inductive hypothesis we have (in view of (1.1))
Rafael Jakimczuk
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(3.10) pn = n log n + n log log n − n
r
X
n
(−1)j−1 nPj (log log n)
+
+o
,
logr n
logj n
j=1
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and
(3.11)
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1
s
(ts ,n ) = n log n + n log log n − n
r
X
(−1)j−1 nPj (log log n)
n
+
.
+o
r
j
log
n
log
n
j=1
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From (3.10) we find that
(3.12)
log log n − 1
log pn = log n + log log n + log 1 +
log n
#
r
X
(−1)j−1 Pj (log log n)
1
+
+o
.
r+1
j+1
log
n
log
n
j=1
Let us write (1.3) in the form
(3.13)
π(x) =
r+3
X
(i − 1)!x
i=1
logi x
!
+o
x
logr+3 x
Asymptotic Formulae
.
If we put x = pn and use the prime number theorem, we get
!
r+3
X
1
n
(i − 1)!
(3.14)
=
+o
.
r+3
i
pn
log
n
log
p
n
i=1
Similarly, from (3.11) we find that
(3.15)
log log n − 1
log (ts ,n ) = log n + log log n + log 1 +
log n
#
r
X (−1)j−1 Pj (log log n)
1
+
+o
.
r+1
j+1
log
n
log
n
j=1
1
s
Let us write (2.1) in the form




1
1
r+3
X (i − 1)!x s
xs
1  + o
1  .
(3.16)
πs (x) = 
i
r+3
s
log
x
log
xs
i=1
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If we put x = ts,n and use (3.5), we get


r+3
X
n
1
(i
−
1)!

 + o
(3.17)
.
1 =
1
r+3
i
log
n
s
(ts ,n ) s
log
(t
,
)
i=1
s n
If x ≥ 1 and y ≥ 1, Lagrange’s theorem gives us the inequality
|log y − log x| ≤ |y − x|
Asymptotic Formulae
with (3.12) and (3.15), it leads to
(3.18)
1
s
log (ts ,n ) − log pn = o
Rafael Jakimczuk
1
logr+1 n
.
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From (3.18) we find that
(3.19)
1
1
−
1 = o
k
k
log pn log (ts ,n ) s
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1
logr+k+2 n
1
=o
(k = 1, . . . , r + 3).
logr+3 n
(3.14), (3.17) and (3.19) give
n
n
−
=o
pn (ts ,n ) 1s
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n
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that is
(3.20)
1
s
(ts ,n ) − pn = (ts ,n )
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1
1
s
log
r+2
n
o (1) .
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If we write
1
(ts ,n ) s = pn + f (n)
(3.21)
substituting (3.21) into (3.20) we find that
f (n) =
log
r+2
pn
o(1),
n + o(1)
so
Asymptotic Formulae
f (n) = o
(3.22)
n
log
r+1
Rafael Jakimczuk
n
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(3.21) and (3.22) give
1
s
(ts ,n ) = pn + o
The theorem is thus proved.
n
logr+1 n
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4.
The Asymptotic Behaviour of ts,n
Theorem 4.1. There exists a unique sequence Ps,j (X) (j ≥ 1) of polynomials
with rational coefficients such that, for every nonnegative integer m
(4.1)
ts,n =
X
ci fi (n) +
m
X
(−1)j−1 ns Ps,j (log log n)
j=1
logj n
+o
ns
logm n
.
The polynomials Ps,j (X) have degree j + s − 1 and leading coefficient
1
.
( j+s−1
)
s
The fi (n) are sequences of the form ns logr n (log log n)u and the ci are constants. f1 (n) = ns logs n and c1 = 1, if i 6= 1 then fi (n) = o(f1 (n)).
If m = 0 equation (4.1) is
X
(4.2)
ts,n =
ci fi (n) + o(ns ).
Proof. From (1.1) and (3.1) we obtain (4.1)
"
m+s−1
X (−1)j−1 nPj (log log n)
ts,n = n log n + n log log n − n +
logj n
j=1
s
n
+o
logm+s−1 n
m
X
X
(−1)j−1 ns Ps,j (log log n)
ns
=
ci fi (n) +
+o
logm n
logj n
j=1
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if we write
(4.3) Ps,j (X)
X
X
=
(r,k) j1 +.....+jt =j+r
r−t+1
(−1)
s s − r
r
k
(X − 1)k Pj1 (X) · · · Pjt (X),
where r + k + t = s.
The first sum runs through the vectors (r, k) (r ≥ 0, k ≥ 0, r + k ∈
{0, 1, . . . , s − 1}), such that the set of vectors (j1 , j2 , . . . , jt ) whose coordinates are positive integers which satisfy j1 + j2 + · · · + jt = j + r is nonempty.
The second sum runs through the former nonempty set of vectors (j1 , j2 , . . . , jt )
(this set depends on the vector (r, k)).
If m = 0 we obtain (4.2).
Let us consider a vector (r, k). The degree of each polynomial
s − r
r−t+1 s
(−1)
(X − 1)k Pj1 (X) · Pj2 (X) · · · Pjt (X)
r
k
is j + r + k. Hence the degree of the polynomial
s − r
X
r−t+1 s
(4.4)
(−1)
r
k
j +j +···+j =j+r
1
2
t
× (X − 1)k Pj1 (X) · Pj2 (X) · · · Pjt (X)
does not exceed j + r + k. Since r + k ∈ {0, 1, . . . , s − 1}, the greatest degree
of the polynomials (4.4) does not exceed j + s − 1. On the other hand, in (4.3)
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there are s polynomials (4.4) of degree j + s − 1. Since in this case t = 1, these
s polynomials are
s − r
r s
(X − 1)k Pj+r (X)
(r + k = s − 1)
(−1)
r
k
and their sum is
(4.5)
s−1
X
s s − r (−1)
(X − 1)s−r−1 Pj+r (X)
r
s−r−1
r=0
r
=
s−1
X
Rafael Jakimczuk
r
(−1)
r=0
s
r
s−r−1
(s − r)(X − 1)
(−1)r
s s − r
r=0
r j+r
Pj+r (X).
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Since the leading coefficient of the polynomial Pj+r (X) is
coefficient of the polynomial (4.5) will be
s−1
X
Asymptotic Formulae
=
1
j+s−1
s
1
,
j+r
the leading
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Hence the degree of the polynomial (4.3) is j + s − 1 and its leading coefficient
1
is j+s−1
. The theorem is thus proved.
( s )
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Examples.
t1,n = n log n + n log log n − n +
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m
X
(−1)j−1 nPj (log log n)
j
j=1
log n
+o
n
logm n
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t2,n = n2 log2 n + 2n2 log n log log n − 2n2 log n + n2 (log log n)2
m
X
(−1)j−1 n2 P2,j (log log n)
n2
2
− 3n +
+o
.
m
j
log
n
log
n
j=1
Corollary 4.2. The sequences ts,n and psn (s ≥ 1) have the same asymptotic
expansion, namely (4.1).
Note. G. Mincu [2] proved Theorem 3.1 and Theorem 4.1 when s = 2.
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References
[1] M. CIPOLLA, La determinazione assintotica dell’ nimo numero primo,
Rend. Acad. Sci. Fis. Mat. Napoli, 8(3) (1902), 132–166.
[2] G. MINCU, An asymptotic expansion, J. Inequal. Pure and Appl.
Math., 4(2) (2003), Art. 30. [ONLINE http://jipam.vu.edu.au/
article.php?sid=268]
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