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volume 5, issue 3, article 62, 2004.

Received 09 January, 2004;

accepted 08 June, 2004.

Communicated by:S.S. Dragomir

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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

2000c Victoria University ISSN (electronic): 1443-5756 009-04

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Asymptotic Formulae Rafael Jakimczuk

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J. Ineq. Pure and Appl. Math. 5(3) Art. 62, 2004

Abstract

Letts,nbe then-th positive integer number which can be written as a power p^t,t≥s, of a primep(s≥1is fixed). Letπ_s(x)denote the number of prime powersp^t,t≥s, not exceedingx. We study the asymptotic behaviour of the sequencets,nand of the functionπs(x). We prove that the sequencets,nhas an asymptotic expansion comparable to that ofpn(the Cipolla’s expansion).

2000 Mathematics Subject Classification:11N05, 11N37.

Key words: Primes, Powers of primes, Cipolla’s expansion.

1. Introduction

Letp_nbe then-th prime. M. Cipolla [1] proved the following theorem:

There exists a unique sequencePj(X) (j ≥1)of polynomials with rational coefficients such that, for every nonnegative integerm,

(1.1) p_n=nlogn+nlog logn−n +

m

X

j=1

(−1)^j−1nPj(log logn) log^jn +o

n log^mn

.

The polynomialsP_j(X)have degreej and leading coefficient ¹_j. P₁(X) =X−2, P₂(X) = X²−6X+ 11

2 , . . . . Ifm= 0equation (1.1) is:

(1.2) pn =nlogn+nlog logn−n+o(n).

Letπ(x)denote the number of prime numbers not exceedingx, then (1.3) π(x) =

m

X

i=1

(i−1)!x logⁱx

!

+ε(x)(m−1)!x

log^mx (m≥1), where lim

x→∞ε(x) = 0.

Lemma 1.1. There exists a positive numberM 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 π(x)is bounded. The functions^(i−1)!x_logix , i= 1, . . . , mand_(m−1)!x^log^m^x are continuous on the compact[2, a], so they are also bounded.

As

ε(x) =

"

π(x)−

m

X

i=1

(i−1)!x logⁱx

!# log^mx (m−1)!x, ε(x)is in its turn bounded on[2, a].

Sinceais arbitrary and lim

x→∞ε(x) = 0, the lemma is proved.

Let us consider the sequence of positive integer numbers which can be written as a powerp^tof a primep(t≥ 1is fixed). The number of prime powersp^t not exceedingxwill be (in view of (1.3))

π(x¹^t) =

m

X

i=1

(i−1)!x¹^t logⁱx¹^t

! +ε

x¹^t(m−1)!x¹^t log^mx¹^t (1.4)

=

m

X

i=1

tⁱ(i−1)!x¹^t logⁱx

!

+ +ε

x¹^tt^m(m−1)!x¹^t log^mx

=

m

X

i=1

tⁱ(i−1)!x¹^t logⁱx

!

+o x¹^t log^mx

! .

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2. The Function π

_s

(x)

Letts,n be then-th positive integer number (in increasing order) which can be written as a powerp^t,t≥ s, of a primep(s≥1is fixed). Letπ_s(x)denote the number of prime powersp^t,t ≥s, not exceedingx.

Theorem 2.1.

(2.1) πs(x) =

m

X

i=1

sⁱ(i−1)!x¹^s logⁱx

!

+o x¹^s log^mx

! .

Proof. Ifx∈[2^s+k,2^s+k+1) (k ≥1), then π_s(x) = π

x¹^s +

k

X

i=1

π x^s+i¹

. Using (1.4), we obtain

πs(x) =

m

X

i=1

! +ε

x¹^s

s^m(m−1)!x¹^s log^mx (2.2)

+

k

X

j=1

_m X

i=1

(s+j)ⁱ(i−1)!x^s+j¹ logⁱx

!

+ε

x^s+j¹ (s+j)^m(m−1)!x^s+j¹ log^mx

!

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=

m

X

i=1

! +ε

x¹^ss^m(m−1)!x¹^s log^mx +

m

X

i=1 k

X

j=1

(s+j)ⁱ(i−1)!x^s+j¹ logⁱx

!

+

k

X

j=1

ε

x^s+j¹ (s+j)^m(m−1)!x^s+j¹

log^mx .

In the given conditions, the following inequalities hold forx:

k

P

j=1

(s+j)ⁱ(i−1)!x

1 s+j

logⁱx s^m(m−1)!x¹s

log^mx

=

k

P

j=1 (s+j)ⁱ

s^m ·_(m−1)!^(i−1)!x^s+j¹ log^m−ix x¹^s

≤

k

X

j=1 (s+j)ⁱ

s^m · _(m−1)!^(i−1)! log^m−i 2^s+k+1 2

(s+k)j−s s(s+j)

≤

k

X

j=1

(s+k)ⁱ(s+k+ 1)^m−i

2^s(s+1)¹ k

= k(s+k)ⁱ(s+k+ 1)^m−i

2^s(s+1)¹ k (i= 1, . . . , m).

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Now, since

k→∞lim

k(s+k)ⁱ(s+k+ 1)^m−i

2^s(s+1)¹ k = 0 (i= 1, . . . , m),

we find that

(2.3) lim

x→∞

k

P

j=1

(s+j)ⁱ(i−1)!x

1 s+j

logⁱx s^m(m−1)!x¹^s

log^mx

= 0 (i= 1, . . . , m).

On the other hand, from the lemma we have the following inequality

k

X

j=1

ε

x^s+j¹ (s+j)^m(m−1)!x^s+j¹ log^mx

≤M

k

X

j=1

(s+j)^m(m−1)!x^s+j¹

log^mx .

This inequality and (2.3) withi=mgive

(2.4) lim

x→∞

k

P

j=1

ε

x^s+j¹ _(s+j)m(m−1)!x^s+j¹ log^mx s^m(m−1)!x¹s

log^mx

= 0.

Finally, from (2.2), (2.3) and (2.4) we find that πs(x) =

m

X

i=1

!

+ε⁰(x)s^m(m−1)!x¹^s log^mx , where lim

x→∞ε⁰(x) = 0. The theorem is proved.

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From (1.4) and (2.1) we obtain the following corollary Corollary 2.2. The functionsπ_s(x)andπ

x¹^s

have the same asymptotic be- haviour(s ≥1).

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3. The Sequences (t

_s

,

_n

)

¹^s

and p

_n

Theorem 3.1.

(3.1) (t_s,_n)¹^s =p_n+o n

log^rn

(r≥0).

Proof. We proceed by mathematical induction onr.

Equation (2.1) gives (m = 1)

(3.2) lim

x→∞

π_s(x)

sx¹^s logx

= 1.

If we putx=t_s,n, we get

(3.3) lim

n→∞

(t_s,n)¹^s nlog (ts,n)¹^s

= 1.

From (3.3) we find that

(3.4) lim

n→∞

logs+ log (t_s,n)¹^s −logn−log logt_s,n

= 0.

Now, since

(3.5) lim

n→∞

log (t_s,n)¹^s logn = 1,

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we obtain

n→∞lim

(t_s,n)¹^s nlog (t_s,n)¹^s

= 1 if and only if

n→∞lim

(t_s,n)¹^s nlogn = 1.

We also derive

n→∞lim t_s,n

n^slog^sn = 1.

(3.6) lim

n→∞(−logs+ log logt_s,n−log logn) = 0.

(3.4) and (3.6) give

(3.7) log (t_s,n)¹^s = logn+ log logn+o(1).

Equation (2.1) gives (m= 2) π_s(x) = x¹^s

logx¹^s + x¹^s log²

x¹^s +o



 x¹^s log²

x¹^s



, so

x¹^s =πs(x) logx¹^s − x¹^s

1 +o x¹^s

1

! .

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If we putx=t_s,n, we get

(3.8) (t_s,n)¹^s =nlog (t_s,n)¹^s − (t_s,n)¹^s log (t_s,n)¹^s

+o (t_s,n)¹^s log (t_s,n)¹^s

! . Finally, from (3.8), (3.3) and (3.7) we find that

(3.9) (ts,n)¹^s =nlogn+nlog logn−n+o(n). Therefore, forr = 0the theorem is true because of (1.2) and (3.9).

Letr ≥ 0be given, and assume that the theorem holds forr, we will prove it is also true forr+ 1.

From the inductive hypothesis we have (in view of (1.1)) (3.10) p_n =nlogn+nlog logn−n

+

r

X

j=1

(−1)^j−1nP_j(log logn) log^jn +o

n log^rn

,

and

(3.11) (ts,n)¹^s =nlogn+nlog logn−n +

r

X

j=1

n log^rn

.

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(3.12) logp_n = logn+ log logn+ log

1 + log logn−1 logn +

r

X

j=1

(−1)^j−1P_j(log logn) log^j+1n +o

1 log^r+1n

# . Let us write (1.3) in the form

(3.13) π(x) =

r+3

X

i=1

(i−1)!x logⁱx

! +o

x log^r+3x

. If we putx=p_nand use the prime number theorem, we get

(3.14) n

p_n =

r+3

X

i=1

(i−1)!

logⁱp_n

! +o

1 log^r+3n

. Similarly, from (3.11) we find that

(3.15) log (t_s,_n)¹^s = logn+ log logn+ log

1 + log logn−1 logn +

r

X

j=1

(−1)^j−1P_j(log logn) log^j+1n +o

1 log^r+1n

# . Let us write (2.1) in the form

(3.16) π_s(x) =





r+3

X(i−1)!x¹^s

i ₁



+o





x¹^s

r+3 ₁



.

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If we putx=t_s,nand use (3.5), we get

(3.17) n

(t_s,_n)¹^s

=





r+3

X

i=1

(i−1)!

logⁱ

(t_s,_n)¹^s



+o

1 log^r+3n

. Ifx≥1andy ≥1, Lagrange’s theorem gives us the inequality

|logy−logx| ≤ |y−x|

with (3.12) and (3.15), it leads to

(3.18) log (t_s,_n)¹^s −logp_n=o

1 log^r+1n

. From (3.18) we find that

1

log^kp_n − 1 log^k(ts,n)¹^s

=o

1 log^r+k+2n

(3.19)

=o

1 log^r+3n

(k= 1, . . . , r+ 3).

(3.14), (3.17) and (3.19) give n

p_n − n (ts,n)¹^s

=o

1 log^r+3n

, that is

(3.20) (t_s,_n)¹^s −p_n = (t_s,_n)¹^s 1

log^r+2no(1).

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If we write

(3.21) (t_s,_n)¹^s =p_n+f(n) substituting (3.21) into (3.20) we find that

f(n) = p_n

log^r+2n+o(1)o(1), so

(3.22) f(n) = o

n log^r+1n

(3.21) and (3.22) give

(t_s,_n)¹^s =p_n+o

n log^r+1n

. The theorem is thus proved.

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4. The Asymptotic Behaviour of t

_s,n

Theorem 4.1. There exists a unique sequenceP_s,j(X) (j ≥1)of polynomials with rational coefficients such that, for every nonnegative integerm

(4.1) t_s,n =X

c_if_i(n) +

m

X

j=1

(−1)^j−1n^sP_s,j(log logn)

log^jn +o

n^s log^mn

.

The polynomials P_s,j(X) have degree j +s − 1 and leading coefficient

1

(^j+s−1s ).

Thef_i(n)are sequences of the formn^slog^rn(log logn)^uand thec_iare con- stants. f₁(n) = n^slog^snandc₁ = 1, ifi6= 1thenf_i(n) = o(f₁(n)).

Ifm = 0equation (4.1) is

(4.2) t_s,n =X

c_if_i(n) +o(n^s).

Proof. From (1.1) and (3.1) we obtain (4.1)

ts,n=

"

nlogn+nlog logn−n+

m+s−1

X

j=1

(−1)^j−1nP_j(log logn) log^jn

+o

n log^m+s−1n

s

=X

c_if_i(n) +

m

X

j=1

(−1)^j−1n^sP_s,j(log logn)

log^jn +o

n^s log^mn

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if we write (4.3) P_s,j(X)

=X

(r,k)

X

j1+...+jt=j+r

(−1)^r−t+1s r

s−r

k

(X−1)^kP_j₁(X)· · ·P_j_t(X),

wherer+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, . . . , j_t) whose coordi- nates are positive integers which satisfyj₁+j₂+· · ·+j_t=j+ris nonempty.

The second sum runs through the former nonempty set of vectors(j₁, j₂, . . . , j_t) (this set depends on the vector(r, k)).

Ifm = 0we obtain (4.2).

Let us consider a vector(r, k).The degree of each polynomial (−1)^r−t+1s

r

s−r k

(X−1)^kP_j₁(X)·P_j₂(X)· · ·P_j_t(X) isj+r+k. Hence the degree of the polynomial

(4.4) X

j1+j2+···+jt=j+r

(−1)^r−t+1s r

s−r

k

×(X−1)^kP_j₁(X)·P_j₂(X)· · ·P_j_t(X) does not exceedj+r+k. Sincer+k∈ {0,1, . . . , s−1}, the greatest degree

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there arespolynomials (4.4) of degreej+s−1. Since in this caset= 1, these spolynomials are

(−1)^rs r

s−r

k

(X−1)^kP_j+r(X) (r+k=s−1) and their sum is

(4.5)

s−1

X

r=0

(−1)^rs r

s−r s−r−1

(X−1)^s−r−1P_j+r(X)

=

s−1

X

r=0

(−1)^rs r

(s−r)(X−1)^s−r−1Pj+r(X).

Since the leading coefficient of the polynomial P_j+r(X) is _j+r¹ , the leading coefficient of the polynomial (4.5) will be

s−1

X

r=0

(−1)^rs r

s−r

j+r = 1

j+s−1 s

.

Hence the degree of the polynomial (4.3) isj+s−1and its leading coefficient is ¹

(^j+s−1s ). The theorem is thus proved.

Examples.

t_1,n =nlogn+nlog logn−n+

m

X

j=1

n log^mn

,

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t_2,n =n²log²n+ 2n²lognlog logn−2n²logn+n²(log logn)²

−3n²+

m

X

j=1

(−1)^j−1n²P_2,j(log logn)

log^jn +o

n² log^mn

.

Corollary 4.2. The sequences t_s,n and p^s_n (s ≥ 1)have the same asymptotic expansion, namely (4.1).

Note. G. Mincu [2] proved Theorem3.1and Theorem4.1whens= 2.

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References

[1] M. CIPOLLA, La determinazione assintotica dell’ n^imo 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]