Ifcndenotes then-th composed number, one proves inequalities involvingcn, pcn, cpn, and one shows that the sequences(pn)n≥1and(cpn)n≥1are neither convex nor concave

http://jipam.vu.edu.au/

Volume 2, Issue 3, Article 38, 2001

SOME PROPERTIES OF THE SERIES OF COMPOSED NUMBERS

LAUREN ¸TIU PANAITOPOL FACULTY OFMATHEMATICS, UNIVERSITY OFBUCHAREST,

14 ACADEMIEIST., RO-70109 BUCHAREST, ROMANIA

pan@al.math.unibuc.ro

Received 27 March, 2001; accepted 11 June, 2001.

Communicated by L. Toth

ABSTRACT. Ifcndenotes then-th composed number, one proves inequalities involvingcn, pc_n, cp_n, and one shows that the sequences(pn)n≥1and(cp_n)n≥1are neither convex nor concave.

Key words and phrases: Prime Numbers, Composed Numbers, Asymptotic Behaviour, Inequalities, Sums and Series.

2000 Mathematics Subject Classification. 11A25, 11N05.

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

.

ISSN (electronic): 1443-5756 c

2001 Victoria University. All rights reserved.

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.

027-01

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

(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]).

2. INEQUALITIESINVOLVINGc_n Property 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

c_n> n+π(c_n)> 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

.

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_c^n−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).

For n ≥ 1747, one easily shows that ^{0.88 log}_logn−1.12^n−6.36 > ₃₁¹ , hence ¹_n > _{31 log}¹2n. But this is impossible, since forn ≥1724we have _n¹ < ¹

31 log²n. Consequently we havec_n < n

1 + _log¹_n +_log³2n

. By checking the cases when n ≤ 1746,

one completely proves the stated inequalities.

Property 2. Ifn ≥30,398, then the inequality

p_n > c_nlogc_n holds.

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 +_log⁴2n+_log⁶3n+_log⁹4n, 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

logn+log₂n−1 logn−²⁸₂₉ >

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.

For n = 130,000, one gets 2.466· · · > 2.4649. . .. The checking of the cases when n <

130,000completes the proof.

3. INEQUALITIESINVOLVINGc_pn ANDp_cn Proposition 3.1. We have

(3.1) p_n+n < c_pn < p_n+n+π(n) fornsufficiently large.

Proof. By (1.2) and (1.7) it follows that fornsufficiently large we havec_n=n+π(n) +_logⁿ2n+ O

n log³n

, hence

(3.2) c_pn =p_n+n+ p_n

log²p_n +O n

log²n

.

Thus fornlarge enough we havec_pn > p_n+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 indeedc_pn < p_n+n+

π(n).

Proposition 3.2. Ifnis large enough, then the inequality p_cn > c_pn

holds.

Proof. By (2.1) it follows that

(3.3) c_pn =π(c_pn) +p_n+ 1.

Now (3.1) and (3.3) imply that fornsufficiently large we haveπ(c_pn)< n+π(n). But by (2.1) it follows thatc_n > n+π(n), hencec_n > π(c_pn). If we assume thatc_pn > p_cn, then we obtain the contradictionπ(c_pn)≥π(p_cn) = c_n. Consequently we must havec_pn < p_cn. It is easy to show that the sequence(c_n)n≥1is neither convex nor concave. We are lead to the same conclusion by studying the sequences(cpn)n≥1 and(pcn)n≥1. Let us say that a sequence (an)n≥1has the propertyP when the inequality

a_n+1−2a_n+a_n−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(c_pn)_n≥1 and(p_cn)_n≥1 have the propertyP. In order to prove it we need the following auxiliary result.

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

(3.4) a_m−a_n

m−n ≥a_n+1−a_n. If the sequence(an)n≥n2 is concave, then forn > p≥n2 we have

(3.5) a_n−a_p

n−p ≥a_n+1−a_n wheneverm > n≥n₁.

Proof. In the first case, fori≥ n we haveai+1 −ai ≥ an+1−an, hencePm−1

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

Proof of Proposition 3.3. Erdös proved in [3] that, with d_n = p_n+1 − p_n, we have lim sup_n→∞^min(d_log^n,d_nⁿ⁺¹⁾ = ∞. In particular, the set M = {n | min(d_n, d_n+1) > 2 logn} is infinite.

For every n, at least one of the numbers n and n+ 1 is composed, that is, eithern = c_m or n + 1 = c_m for some m. Consequently, there exist infinitely many indices m such that p_cm+1−p_cm >2 logc_m. Sincec_m+1 ≥c_m+ 1andc_m > m, we get infinitely many values ofm such that

(3.6) p_cm+1−p_cm >2 logm.

LetM⁰ be the set of these numbersm.

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

have p_c2m−p_cm

m ≥pcm+1 −pcm >2 logm,

hencep_c2m > 2mlogm+p_cm. But this is a contradiction because c_n ∼ n and p_n ∼ nlogn, that isp_c2m ∼2mlog 2mandp_cm ∼mlogm.

On the other hand, if we assume that the sequence (p_cn)n≥n2 is concave, then (3.5) implies that forx∈M⁰we have

p_cm−p_c[m/2]

m−_m

2

≥p_cm+1−p_cm >2 logm,

that is

1> 2 m−_m

2

logm+p_c[m/2]

p_cm .

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

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

cp2n−cpn

n ≥c_pn+1−c_pn ≥p_n+1−p_n>2 logn.

If we take n → ∞, n ∈ M, in the inequality1 > (2nlogn +c_pn)/c_p2n, then we obtain the contradiction1≥ ³₂.

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

c_pn −c_p[n/2]

n−_n

2

≥cpn+1−cpn ≥pn+1−pn>2 logn,

which is again a contradiction.

REFERENCES

[1] A.E. BOJARINCEV, Asymptotic expressions for thenthcomposite number, Ural. Gos. Univ. Mat.

Zap., 6 (1967), 21–43 (in Russian).

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[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. SÁNDOR AND B. CRSTICI, Handbook of Number Theory, Kluwer Aca- demic Publishers, Dordrecht - Boston - London, 1996.

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