Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 62, 2004
ASYMPTOTIC FORMULAE
RAFAEL JAKIMCZUK
D IVISIÓN M ATEMÁTICA
U NIVERSIDAD NACIONAL DE L UJÁN
L UJÁN , B UENOS A IRES
A RGENTINA .
jakimczu@mail.unlu.edu.ar
Received 09 January, 2004; accepted 08 June, 2004
Communicated by S.S. Dragomir
A BSTRACT. 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).
Key words and phrases: Primes, Powers of primes, Cipolla’s expansion.
2000 Mathematics Subject Classification. 11N05, 11N37.
1. I NTRODUCTION
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,
m
X
(−1)j−1 nPj (log log n)
n
(1.1)
pn = n log n + n 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:
(1.2)
pn = n log n + n log log n − n + o(n).
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
009-04
....
2
R AFAEL JAKIMCZUK
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.
x→∞
Lemma 1.1. There exists a positive number M such that in the interval [2, ∞), |ε(x)| ≤ M.
Proof. Let us consider the closed interval [2, a]. In this interval, π(x) ≤ x, so π(x) is bounded.
logm x
The functions (i−1)!x
, i = 1, . . . , m and (m−1)!x
are continuous on the compact [2, a], so they
logi x
are also bounded.
As
"
!#
m
X
logm x
(i − 1)!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
π(x t ) =
(1.4)
+ ε xt
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
2. T HE F UNCTION 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
!
.
Proof. If x ∈ [2s+k , 2s+k+1 ) (k ≥ 1), then
πs (x) = π x
1
s
k
1 X
+
π x s+i .
i=1
J. Inequal. Pure and Appl. Math., 5(3) Art. 62, 2004
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A SYMPTOTIC F ORMULAE
3
Using (1.4), we obtain
(2.2) πs (x) =
1
m
X
si (i − 1)!x s
k
X
1
s
sm (m − 1)!x 1s
logm x
!
1
m
X
(s + j)i (i − 1)!x s+j
j=1
=
+ε x
logi x
i=1
+
!
+ε x
logi x
i=1
1
m
X
si (i − 1)!x s
!
+ε x
1
s
1
s+j
1
(s + j)m (m − 1)!x s+j
!
logm x
sm (m − 1)!x 1s
logm x
!
1
1
m
k
k
1 (s + j)m (m − 1)!x s+j
X
X
X
(s + j)i (i − 1)!x s+j
s+j
+
+
ε x
.
logm x
logi x
i=1
j=1
j=1
i=1
logi x
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
j=1
=
1
sm (m−1)!x s
logm x
≤
1
(s+j)i (i−1)! s+j
· (m−1)! x
sm
logm−i x
1
xs
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
X
(s + k)i (s + k + 1)m−i
≤
1 k
j=1
2 s(s+1)
k (s + k)i (s + k + 1)m−i
=
1 k
2 s(s+1)
(i = 1, . . . , m).
Now, since
k (s + k)i (s + k + 1)m−i
lim
=0
1 k
k→∞
s(s+1)
2
(i = 1, . . . , m),
we find that
k
P
(2.3)
lim
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
m
X
1 (s + j)m (m − 1)!x s+j
X
s+j
(s
+
j)
(m
−
1)!x
ε x s+j
.
≤M
m
m
log
x
log
x
j=1
j=1
This inequality and (2.3) with i = m give
1 1
k
m
P
(m−1)!x s+j
ε x s+j (s+j) log
m
x
j=1
(2.4)
lim
= 0.
1
x→∞
J. Inequal. Pure and Appl. Math., 5(3) Art. 62, 2004
sm (m−1)!x s
logm x
http://jipam.vu.edu.au/
4
R AFAEL JAKIMCZUK
Finally, from (2.2), (2.3) and (2.4) we find that
πs (x) =
1
m
X
si (i − 1)!x s
!
logi x
i=1
1
sm (m − 1)!x s
+ ε (x)
,
logm x
0
where lim ε0 (x) = 0. The theorem is proved.
x→∞
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).
1
3. T HE S EQUENCES (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)
πs (x)
1
x→∞
= 1.
sx s
log x
If we put x = ts,n , we get
1
(ts,n ) s
lim
(3.3)
n→∞
1
= 1.
n log (ts,n ) s
From (3.3) we find that
(3.4)
lim
n→∞
1
s
log s + log (ts,n ) − log n − log log ts,n = 0.
Now, since
1
log (ts,n ) s
lim
= 1,
n→∞
log n
(3.5)
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
From (3.5) we find that
(3.6)
lim (− log s + log log ts,n − log log n) = 0.
n→∞
(3.4) and (3.6) give
(3.7)
1
log (ts,n ) s = log n + log log n + o(1).
J. Inequal. Pure and Appl. Math., 5(3) Art. 62, 2004
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A SYMPTOTIC F ORMULAE
5
Equation (2.1) gives (m = 2)
πs (x) =
x
1
s
log x
1
s

1
s
+
log
x
2
x
1
s
+ o

1
s
log
x
2
x
1
s
 ,
so
xs
1
s
x = πs (x) log x −
!
1
1
1
s
1
log x s
xs
+o
.
1
log x s
If we put x = ts,n , we get
1
1
s
1
s
(ts,n ) = n log (ts,n ) −
(3.8)
1
(ts,n ) s
1
!
(ts,n ) s
+o
1
.
log (ts,n ) s
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)
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))
r
X
(−1)j−1 nPj (log log n)
n
(3.10)
pn = n log n + n log log n − n +
+o
,
r
j
log
n
log
n
j=1
and
(3.11)
1
s
(ts ,n ) = n log n + n log log n − n +
r
X
(−1)j−1 nPj (log log n)
logj n
j=1
+o
n
logr n
.
From (3.10) we find that
(3.12)
log pn = log n + log log n
"
#
r
j−1
X
log log n − 1
(−1) Pj (log log n)
1
+ log 1 +
.
+
+o
r+1
j+1
log n
log n
log n
j=1
Let us write (1.3) in the form
π(x) =
(3.13)
r+3
X
(i − 1)!x
i=1
logi x
!
+o
x
logr+3 x
.
If we put x = pn and use the prime number theorem, we get
!
r+3
X
n
(i − 1)!
1
(3.14)
=
+o
.
pn
logr+3 n
logi pn
i=1
Similarly, from (3.11) we find that
(3.15)
1
log (ts ,n ) s = log n + log log n
"
#
r
j−1
X
log log n − 1
(−1) Pj (log log n)
1
+ log 1 +
+
+o
.
r+1
j+1
log n
log n
log n
j=1
J. Inequal. Pure and Appl. Math., 5(3) Art. 62, 2004
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6
R AFAEL JAKIMCZUK
Let us write (2.1) in the form




1
1
r+3
X
s
s
(i − 1)!x 
x
1 + o
1  .
πs (x) = 
i
r+3
s
x
log
xs
i=1 log
(3.16)
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|
with (3.12) and (3.15), it leads to
1
s
log (ts ,n ) − log pn = o
(3.18)
1
logr+1 n
.
From (3.18) we find that
(3.19)
1
1
−
1 = o
k
k
log pn log (ts ,n ) s
1
=o
logr+k+2 n
1
logr+3 n
(k = 1, . . . , r + 3).
(3.14), (3.17) and (3.19) give
n
n
−
=o
pn (ts ,n ) 1s
1
,
logr+3 n
that is
(3.20)
1
1
1
(ts ,n ) s − pn = (ts ,n ) s
log
r+2
n
o (1) .
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
f (n) = o
(3.22)
n
logr+1 n
(3.21) and (3.22) give
1
s
(ts ,n ) = pn + o
The theorem is thus proved.
J. Inequal. Pure and Appl. Math., 5(3) Art. 62, 2004
n
logr+1 n
.
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A SYMPTOTIC F ORMULAE
7
4. T HE A SYMPTOTIC B EHAVIOUR 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
m
X
X
(−1)j−1 ns Ps,j (log log n)
ns
(4.1)
ts,n =
ci fi (n) +
+o
.
m
j
log
n
log
n
j=1
1
.
( j+s−1
)
s
u
The fi (n) are sequences of the form ns logr n (log log n) and the ci are constants. f1 (n) =
s
n 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 ).
The polynomials Ps,j (X) have degree j + s − 1 and leading coefficient
Proof. From (1.1) and (3.1) we obtain (4.1)
"
# s
m+s−1
X (−1)j−1 nPj (log log n)
n
ts,n = n log n + n log log n − n +
+o
logm+s−1 n
logj n
j=1
m
X
X
(−1)j−1 ns Ps,j (log log n)
ns
=
ci fi (n) +
+o
j
logm n
log
n
j=1
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)
(X − 1)k Pj1 (X) · Pj2 (X) · · · Pjt (X)
r
k
j +j +···+j =j+r
1
2
t
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) 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
(−1)
(X − 1)k Pj+r (X)
(r + k = s − 1)
r
k
J. Inequal. Pure and Appl. Math., 5(3) Art. 62, 2004
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8
R AFAEL JAKIMCZUK
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
(−1)r
r=0
s
r
(s − r)(X − 1)s−r−1 Pj+r (X).
1
Since the leading coefficient of the polynomial Pj+r (X) is j+r
, the leading coefficient of the
polynomial (4.5) will be
s−1
X
1
r s s−r
(−1)
= j+s−1 .
r j+r
s
r=0
Hence the degree of the polynomial (4.3) is j + s − 1 and its leading coefficient is
1
( j+s−1
)
s
. The
theorem is thus proved.
Examples.
t1,n = n log n + n log log n − n +
m
X
(−1)j−1 nPj (log log n)
j=1
logj n
+o
n
logm n
,
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
.
logm n
logj 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.
R EFERENCES
[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]
J. Inequal. Pure and Appl. Math., 5(3) Art. 62, 2004
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