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SOME RESULTS ON THE UNITARY ANALOGUE OF THE LEHMER PROBLEM MARTA SKONIECZNA INSTITUTE OFMATHEMATICS CASIMIR THEGREATUNIVERSITY PL

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Volume 9 (2008), Issue 2, Article 55, 4 pp.

SOME RESULTS ON THE UNITARY ANALOGUE OF THE LEHMER PROBLEM

MARTA SKONIECZNA INSTITUTE OFMATHEMATICS

CASIMIR THEGREATUNIVERSITY

PL. WEYSSENHOFFA11, 85- 072 BYDGOSZCZ, POLAND [email protected]

Received 10 April, 2008; accepted 29 May, 2008 Communicated by J. Sándor

ABSTRACT. LetM be a positive integer withM 4, and letϕ denote the unitary analogue of Euler’s totient functionϕ. Using Grytczuk-Wójtowicz’s techniques from the paper [2] we strengthen considerably the lower estimations of the solutionsnof the equation M ϕ(n) = n1. Moreover, we show that the set of positive integers, which do not fulfil this equation for anyM 2, contains an interesting subset generated by Ramsey’s theorem.

Key words and phrases: Lehmer problem, Unitary analogue of Lehmer problem.

2000 Mathematics Subject Classification. 11A25.

1. INTRODUCTION

Throughout this paperNdenotes the set of positive integers, and the numbers M, n∈ Nare fixed with M ≥ 2. Let ϕ be the Euler’s totient function, and let ϕ(n)be the number of all natural numbersk ≤nsuch that(k, n) = 1, where(k, n) is the greatest divisordofk, which is also a unitary divisor ofn(i.e., such that(d, n/d) = 1).

A classical (and still unsolved) problem proposed by Lehmer concerns the existence of a composite numbernwhich fulfils the equation

(1.1) M ϕ(n) = n−1

(see e.g. [3, p. 212-215]). Subbarao, Siva Rama Prasad and Dixit studied in [4, 5] an analogous equation for the functionϕ:

(1.2) M ϕ(n) = n−1.

Let

(1.3) n =pα11 ·pα22 · · · · ·pαrr

be the prime factorization ofn, wherep1 < p2 <· · · < prandα1, . . . , αr ∈ N. Putω(n) =r.

It is known (and easy to verify), that every solutionn of the equation (1.1), must be odd and squarefree. Moreover, since fornof the form (1.3) we have

ϕ(n) = (pα11 −1)·(pα22 −1)· · · · ·(pαrr −1)

108-08

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2 MARTASKONIECZNA

(see [4]), no solutionnof the equation (1.2) can be the power of a prime number.

Put SM := {n ∈ N : M ϕ(n) = n−1}, and S := S

M≥2SM . In the papers [4, 5] the authors obtained the following estimations ofn ∈ S:

(1.4) n <(r−2.3)2r−1, wherer=ω(n),

(1.5) if 3-n, thenω(n)≥11 if 5|n, andω(n)≥17 if 5-n,

(1.6) ω(n)≥1850 when 3|n,

(1.7) ω(n)≥17 when the number 455 is not a unitary divisor of n,

(1.8) ω(n)≥33 forM = 3,4 or 5.

In this paper, we show that the techniques of [2] allow us to obtain lower estimations for the elements ofSM , whereM ≥4, which are considerably stronger than cited in (1.5) – (1.8) and unconditional.

Our main result reads as follows.

Theorem 1.1. LetM ≥4and letn ∈ SM be of the form(1.3).

(a) Ifp1 = 3,thenω(n)≥3049M/4−1509.

(b) Ifp1 >3,thenω(n)≥143M/4−1.

Thus, forn∈ SM , whereM ≥4, we have (in general): ω(n)≥1540when3|n(forM = 4 this result is slightly weaker than (1.6)), andω(n)≥ 142 when3 - n (forM = 4 this result is stronger than (1.8)). Moreover,

• ω(n)≥21147when3|n, andω(n)≥493when3-n— forM = 5;

• ω(n)≥166849when3|n, andω(n)≥1709when3-n— forM = 6; and

• ω(n)>1249543when3|n, andω(n)≥5912when3-n— forM ≥7.

Further, by an argument similar to that of [2, Proof of corollary], we obtain Corollary 1.2. LetM ≥4, and letn∈ SM be of the form(1.3).

(a) Ifp1 = 3,thenn >(cM6M)6M, wherec= 0.597...= log 63 . (b) Ifp1 >3,thenn >(dM3M)3M, whered= 0.366...= log 33 .

Using estimation (1.4) we obtain the following analogue of [2, Theorem 2].

Theorem 1.3. LetP ={P1, P2, . . .}, wherePi < Pi+1for alli≥1, denote the set of all prime numbers. For every integerk≥2there exists an infinite subsetP(k)of the setP such that

(a) for every pairwise distinct primes p1, p2, . . . , pk ∈ P(k) and α1, α2, . . . , αk ∈ N the numbern=pα11pα22 pα33 · · ·pαkk does not fulfil equation(1.2);

(b) P(k)is maximal with respect to inclusion.

(Notice that, by the general inequalityω(n)≥11 (see(1.4)), we haveP(k) =Pfork ≤10.)

J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 55, 4 pp. http://jipam.vu.edu.au/

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SOMERESULTS ON THEUNITARYANALOGUE OF THELEHMERPROBLEM 3

2. PROOFS

Proof of Theorem 1.1. We give here only an outline of the proof of Theorem 1.1, in which we essentially use the technique used in the proof of [2, Theorem 1].

Letnbe of the form (1.3), and letn0 be the squarefree kernel ofn, i.e.,n0 =p1·p2· · · · ·pr. Notice first that

(2.1) ϕ(n)

n = ϕ(n0) n0 .

The first step of the proof of [2, Theorem 1] is the inequality4≤ M < n/ϕ(n)forn odd and squarefree (n =n0). An exact analysis of this proof shows that, by equality (2.1) the following result is true:

Lemma 2.1. LetM ≥4be an integer, letnbe of the form(1.3)withp1 ≥3, and suppose that

(2.2) M < n

ϕ(n). Then

(a) ω(n)≥3049M/4−1509ifp1 = 3andpj ≡5(mod 6)for2≤j ≤ω(n), (b) ω(n)≥143M/4−1ifp1 >3.

Sincen ∈ SM andM ≥4, by equation (1.2) and the forms ofϕandϕ, we obtain:

M < n ϕ(n) =

r

Y

i=1

pαii pαii−1

=

r

Y

i=1

1 + 1 pαii −1

r

Y

i=1

1 + 1 pi−1

=

r

Y

i=1

pi

pi −1 = n0

ϕ(n0) = n ϕ(n). Therefore every elementn∈ SM fulfils inequality (2.2).

Further, if3|n(i.e. p1 = 3), then from (1.2) and the form ofϕ,we obtain that3-(pαjj−1), whence3-(pj−1)forj ≥2; thuspj ≡5(mod 6). Now we can apply condition (a) of Lemma 2.1, which finishes the proof of case (a) of our theorem.

Case (b) of our theorem follows from case (b) of Lemma 2.1.

Proof of Theorem 1.3. We will use here the idea and symbols used in the proof of [2, Theorem 2]. Let[N]kbe the set ofk-element increasing sequences ofN, wherek ≥2.

Consider the function f : [N]k → {0,1} of the form f(i1, i2, . . . , ik) = 0 iff the number Piα11Piα22· · ·Piαkk fulfils equation (1.2) for someα1, . . . , αk∈N.

By the Ramsey Theorem [1], there is an infinite subsetN(k)of the setNsuch that f([N(k)]k) = {0} or f([N(k)]k) = {1}.

Respectively, there is an infinite subsetP(k)ofP such that

(*) Piα1

1 Piα2

2 . . . Piαk

k ∈ S for some α1, . . . , αk ∈N, or

(**) Piα1

1 Piα2

2 . . . Piαk

k ∈ S/ for all α1, . . . , αn ∈N,

J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 55, 4 pp. http://jipam.vu.edu.au/

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4 MARTASKONIECZNA

for all pairwise distinct elementsPi1, . . . , Pik ∈ P(k). From inequality (1.4) we obtain that, for everyk ≥2the number#{n ∈N:ω(n)≤k}is finite, and thus case(∗)is impossible. Hence case(∗∗)takes place, which implies that the setP(k)fulfils condition (a) of Theorem 1.3.

The existence of a maximal (with respect to inclusion) set P(k) follows from Kuratowski-

Zorn’s Lemma.

REFERENCES

[1] R.L. GRAHAM, B.L. ROTHSCHILDAND J.H. SPENCER, Ramsey Theory, John Wiley & Sons, Inc., New York, 1980.

[2] A. GRYTCZUKANDM. WÓJTOWICZ, On a Lehmer problem concerning Euler’s totient function, Proc. Japan Acad. Ser.A, 79 (2003), 136–138.

[3] J. SÁNDOR AND B. CRISTICI, Handbook of Number Theory II, Kluwer Academic Publishers, Dortrecht/Boston/London, 2004.

[4] V. SIVA RAMA PRASADAND U. DIXIT, Inequalities related to the unitary analogue of Lehmer problem, J. Inequal. Pure and Appl. Math., 7(4) (2006), Art. 142. [ONLINE: http://jipam.

vu.edu.au/article.php?sid=761].

[5] M.V. SUBBARAOAND V. SIVA RAMA PRASAD, Some analogues of a Lehmer problem on the totient function, Rocky Mountain J. Math., 15 (1985), 609–619.

J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 55, 4 pp. http://jipam.vu.edu.au/

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