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volume 2, issue 3, article 38, 2001.

Received 27 March, 2001;

accepted 11 June, 2001.

Communicated by:L. Toth

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Journal of Inequalities in Pure and Applied Mathematics

SOME PROPERTIES OF THE SERIES OF COMPOSED NUMBERS

LAURENTIU PANAITOPOL

Faculty of Mathematics, University of Bucharest, 14 Academiei St.,

RO-70109 Bucharest, Romania EMail:pan@al.math.unibuc.ro

c

2000Victoria University ISSN (electronic): 1443-5756 027-01

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Some Properties of the Series of Composed Numbers

Lauren¸tiu Panaitopol

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J. Ineq. Pure and Appl. Math. 2(3) Art. 38, 2001

http://jipam.vu.edu.au

Abstract

If c_n denotes the n-th composed number, one proves inequalities involving c_n, p_c_n, c_p_n, and one shows that the sequences(p_n)n≥1and(c_p_n)n≥1are neither convex nor concave.

2000 Mathematics Subject Classification:11A25, 11N05.

Key words: Prime Numbers, Composed Numbers, Asymptotic Behavior, Inequali- ties, Sums and Series.

The author gratefully acknowledges for partial support from Grant No. 7D/2000 awarded by the Consiliul Na¸tional al Cercet˘arii ¸Stiin¸tifice din Înv˘a¸t˘amântul Superior, România.

1. Introduction

We are going to use the following notation

π(x)the number of prime numbers ≤x, C(x)the number of composed numbers ≤x,

p_nthen-th prime number,

c_nthen-th composed number; c₁ = 4, c₂ = 6, . . . , log₂n= log(logn).

Forx≥1we have the relation

(1.1) π(x) +C(x) + 1 = [x].

Bojarincev proved (see [1], [4]) that (1.2)

c_n=n

1 + 1

logn + 2

log²n + 4

log³n +19 2 · 1

log⁴n+181 6 · 1

log⁵n+o 1

log⁵n

.

Let us remark that

(1.3) c_k+1−c_k=

1 ifc_k+ 1is composed, 2 ifc_k+ 1is prime.

In the proofs from the present paper, we shall need the following facts related toπ(x)andp_n:

(1.4) forx≥67, π(x)> x

logx−0.5

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(see [7]);

(1.5) forx≥3299, π(x)> x

logx−²⁸₂₉

(see [6]);

(1.6) forx≥4, π(x)< x

logx−1.12

(see [6]);

(1.7) forn≥1, π(x) = x logx

n

X

k=0

k!

log^kx +O

x logⁿ⁺¹x

,

(1.8) forn≥2, p_n> n(logn+ log₂n−1) (see [2] and [3]);

(1.9) forn ≥6, p_n< n(logn+ log₂n) (see [7]).

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2. Inequalities Involving c

_n

Property 2.1. We have

(2.1) n

1 + 1

logn + 3 log²n

> c_n > n

1 + 1

logn + 1 log²n

whenevern ≥4.

Proof. If we takex=c_nin (1.1), then we get

(2.2) π(c_n) +n+ 1 =c_n.

Now (1.4) implies that forn ≥48we have

cn> n+π(cn)> n+ n logn and then

c_n> n+π(c_n)> n+π

n

1 + 1 logn

> n+

n

1 + _log¹_n

logn+ log

1 + _log¹_n

−0.5

> n+ n

1 + _log¹_n logn

=n

1 + 1

logn + 1 log²n

.

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By (1.6) and (2.2) it follows that

c_n·logc_n−2.12

logc_n−1.12 < n+ 1.

Sincecn > n, it follows that ^log_log^c_cⁿ^−2.12

n−1.12 > ^log_log^n−2.12_n−1.12 hence (2.3) n+ 1> c_n· logn−2.12

logn−1.12. Assume that there would existn≥1747such that

c_n ≥n

1 + 1

logn + 3 log²n

. Then a direct computation shows that(12)implies

1

n ≥ 0.88 logn−6.36 log²n(logn−1.12).

Forn ≥1747, one easily shows that ^{0.88 log}_log_n−1.12^n−6.36 > ₃₁¹ , hence ¹_n > _{31 log}¹2n. But this is impossible, since forn≥1724we have ¹_n < _{31 log}¹2n.

Consequently we have c_n < n

1 + _log¹_n+ ³

log²n

. By checking the cases whenn≤1746, one completely proves the stated inequalities.

Property 2.2. Ifn ≥30,398, then the inequality pn> cnlogcn

holds.

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Proof. We use (1.8), (2.1) and the inequalities

log

1 + 1

logn + 3 log²n

< 1

logn + 3 log²n, and

n(logn+log logn−1)> n

1 + 1

logn + 3

log²n logn+ 1

logn + 3 log²n

, that islog logn >2 +_log⁴_n+ ⁴

log²n+ ⁶

log³n+ ⁹

log⁴n, which holds ifn≥61,800.

Now the proof can be completed by checking the remaining cases.

Proposition 2.1. We have

π(n)pn> c²_n whenevern ≥19,421.

Proof. In view of the inequalities (1.5), (1.8) and (2.1), forn≥3299it remains to prove that ^log_log^n+log_n−²28ⁿ⁻¹

29

>

1 + _log¹_n +_log³2n

2

, that is log logn > 59

29+ 5.069

logn − 0.758

log²n + 3.207

log³n − 8.68 log⁴n. It suffices to show that

log logn > 59

29+ 5.069 logn.

Forn = 130,000, one gets2.466· · · > 2.4649. . .. The checking of the cases whenn <130,000completes the proof.

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3. Inequalities Involving c

_p_n

and p

_c_n

Proposition 3.1. We have

(3.1) pn+n < cpn < pn+n+π(n) fornsufficiently large.

Proof. By (1.2) and (1.7) it follows that for n sufficiently large we have c_n = n+π(n) + _logⁿ2n +O

n log³n

, hence (3.2) c_p_n =p_n+n+ pn

log²p_n +O n

log²n

. Thus fornlarge enough we havecpn > pn+n.

Since the functionx7→ _log^x2x is increasing, one gets by (1.9) p_n

log²p_n < n(logn+ log₂n) (logn+ log(logn+ log₂n))²

< n(logn+ log₂n) logn(logn+ 2 log₂n)

< n· logn− ¹₂log₂n log²n

=π(n)− 1

2 ·nlog₂n log²n +O

n log²n

.

Both this inequality and (3.2) show that fornsufficiently large we have indeed cpn < pn+n+π(n).

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Proposition 3.2. Ifnis large enough, then the inequality p_c_n > c_p_n

holds.

Proof. By (2.1) it follows that

(3.3) cpn =π(cpn) +pn+ 1.

Now (3.1) and (3.3) imply that forn sufficiently large we have π(c_p_n) < n+ π(n). But by (2.1) it follows thatc_n > n+π(n), hence c_n > π(c_p_n). If we assume thatc_p_n > p_c_n, then we obtain the contradictionπ(c_p_n)≥ π(p_c_n) =c_n. Consequently we must havec_p_n < p_c_n.

It is easy to show that the sequence(c_n)_n≥1 is neither convex nor concave.

We are lead to the same conclusion by studying the sequences (c_p_n)n≥1 and (p_c_n)n≥1. Let us say that a sequence (a_n)n≥1 has the property P when the inequality

a_n+1−2a_n+an−1 >0 holds for infinitely many indices and the inequality

a_n+1−2a_n+an−1 <0

holds also for infinitely many indices. Then we can prove the following fact.

Proposition 3.3. Both sequences(cpn)n≥1and(pcn)n≥1have the propertyP. In order to prove it we need the following auxiliary result.

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Lemma 3.4. If the sequence(a_n)n≥n1 is convex, then form > n ≥n₁we have

(3.4) a_m−a_n

m−n ≥an+1−an.

If the sequence(a_n)n≥n₂ is concave, then forn > p≥n₂we have

(3.5) a_n−a_p

n−p ≥a_n+1−a_n

wheneverm > n ≥n₁.

Proof. In the first case, for i ≥ n we have ai+1 − ai ≥ an+1 − an, hence Pm−1

i=n (a_i+1−a_i)≥(m−n)(a_n+1−a_n), that is (3.4). The inequality (3.5) can be proved similarly.

Proof of Proposition3.3. Erdös proved in [3] that, with d_n = p_n+1 −p_n, we have

lim sup_n→∞ ^min(d_logⁿ^,d_nⁿ⁺¹⁾ =∞. In particular, the setM ={n |min(d_n, d_n+1)>

2 logn}is infinite.

For every n, at least one of the numbersn and n+ 1 is composed, that is, eithern = cm orn+ 1 = cm for somem. Consequently, there exist infinitely many indicesm such thatp_c_m₊₁−p_c_m > 2 logc_m. Sincec_m+1 ≥ c_m+ 1 and c_m > m, we get infinitely many values ofmsuch that

(3.6) pcm+1−pcm >2 logm.

LetM⁰be the set of these numbersm.

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If we assume that the sequence(p_c_n)n≥n1 is convex, then (3.4) implies that form∈M⁰ we have

p_c_2m −p_c_m

m ≥p_c_m+1−p_c_m >2 logm,

hencep_c_2m > 2mlogm+p_c_m. But this is a contradiction becausec_n ∼ n and p_n ∼nlogn, that isp_c_2m ∼2mlog 2mandp_c_m ∼mlogm.

On the other hand, if we assume that the sequence(p_c_n)n≥n₂ is concave, then (3.5) implies that forx∈M⁰ we have

p_c_m −p_c[m/2]

m−_m

2

≥pcm+1 −pcm >2 logm, that is

1> 2 m−_m

2

logm+pc[m/2]

p_c_m .

Form → ∞,m ∈ M⁰, the last inequality implies the contradiction1 ≥1 + ¹₂. Consequently the sequence(p_c_n)n≥1 has the propertyP.

Now let us assume that the sequence(c_p_n)n≥n₁ is convex. Then forn ∈M, n ≥n₁, we get by (3.4)

cp2n−cpn

n ≥c_p_n+1−c_p_n ≥p_n+1−p_n>2 logn.

If we taken → ∞,n ∈M, in the inequality1>(2nlogn+c_p_n)/c_p_2n, then we obtain the contradiction1≥ ³₂.

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Finally, if we assume that the sequence(c_p_n)n≥n2 is concave, then (3.5) implies that forn∈M,n≥n2, we have

c_p_n−c_p_[n/2]

n−_n

2

≥c_p_n+1−c_p_n ≥p_n+1−p_n>2 logn, which is again a contradiction.

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References

[1] A.E. BOJARINCEV, Asymptotic expressions for the nthcomposite num- ber, Ural. Gos. Univ. Mat. Zap., 6 (1967), 21–43 (in Russian).

[2] P. DUSART, Thekth prime is greater thank(lnk+ ln lnk−1)fork ≥2, Math. Comp., 68 (1999), no. 225, 411–415.

[3] P. ERDÖS, Problems and results on the differences of consecutive primes, Publ. Math. Debrecen, 1 (1949), 33–37.

[4] J.-P. MASSIASANDG. ROBIN, Bornes effectives pour certaines fonctions concernant les nombres premiers, J. Théor. Nombres Bordeaux, 8 (1996), 215–242.

[5] D.S. MITRINOVI ´C, J. SÁNDORANDB. CRSTICI, Handbook of Number Theory, Kluwer Academic Publishers, Dordrecht - Boston - London, 1996.

[6] L. PANAITOPOL, Several approximations of π(x), Math. Ineq. & Appl., 2(3) (1999), 317–324.

[7] J.B. ROSSER AND L. SCHOENFIELD, Approximate formulas for some functions of prime numbers, Illinois J. Math., 6 (1962), 64–94.