When π(n) does not divide n

Transcription

1 Whe π() does ot divide Germá Adrés Paz Jauary 26, 2015 Abstract Let π() deote the prime-coutig fuctio ad let / log 1 log 1 f() = log log 0.1. I this paper we prove that if is a iteger ad f() = 0, the π() does ot divide. We also show that if ad π() divides, the f() = 1. I additio, we prove that if ad /π() is a iteger, the is a multiple of log 1 located i the iterval [e log 1 +1, e log ]. This allows us to show that if c is ay fixed iteger 12, the i the iterval [e c, e c+0.1 ] there is always a iteger such that π() divides. Let S deote the sequece of itegers geerated by the fuctio d() = /π() (where Z ad > 1) ad let S k deote the kth term of sequece S. Here we ask the questio whether there are ifiitely may positive itegers k such that S k = S k+1. Keywords: bouds o the prime-coutig fuctio, explicit formulas for the prime-coutig fuctio, itervals, prime umbers, sequeces 2010 Mathematics Subject Classificatio: 00-XX 00A05 11-XX 11A41 11Bxx (2000) Rosario, Sata Fe, Argetia; germapaz ar@hotmail.com 1

2 0 Notatio Throughout this paper the umber is always a positive iteger. Moreover, we use the followig otatio: (absolute value) (ceilig fuctio) (divides) (does ot divide) (floor fuctio) frac( ) (fractioal part) log (atural logarithm) 1 Itroductio Determiig how prime umbers are distributed amog atural umbers is oe of the most difficult mathematical problems. This explais why the prime-coutig fuctio π() (which couts the umber of primes less tha or equal to a give umber ) has bee oe of the mai objects of study i Mathematics for ceturies. I [2] Gaitaas obtais a explicit formula for π() that holds ifiitely ofte. His proof is based o the fact that the fuctio d() = /π() takes o every iteger value greater tha 1 (as proved by Golomb [3]) ad o the fact that x/(log x 0.5) < π(x) < x/(log x 1.5) for x 67 (as show by Rosser ad Schoefeld [4]). I this paper we fid alterative expressios that are valid for ifiitely may positive itegers, ad we also prove, amog other results, that if ad / log 1 log 1 log log 0.1 equals 0, the π() does ot divide. We will place emphasis o the followig three theorems, which were proved by Golomb, Dusart, ad Gaitaas respectively: 2

3 Theorem 1.1 [3]. The fuctio d() = /π() takes o every iteger value greater tha 1. Theorem 1.2 [1]. If is a iteger 60184, the log 1 < π() < log 1.1. Remark 1.3. Dusart s paper states that for x we have x/(log x 1) π(x) x/(log x 1.1), but sice log is always irratioal whe is a iteger > 1, we ca state his theorem the way we did. Theorem 1.4 [2]. The formula π() = log 0.5 is valid for ifiitely may positive itegers. 2 Mai results We are ow ready to prove our mai results: Theorem 2.1. The formula π() = log 1 holds for ifiitely may positive itegers. Proof. Accordig to Theorem 1.2, for we have log 1 < π() < If we multiply by, we get log 1.1 log 1.1 < log 1.1 π() < 1 π() < log 1. < log 1. (1) Sice log 1.1 ad log 1 are both irratioal (for > 1), iequality (1) implies that whe /π() is a iteger we must have π() = log 1 = log = log 1.1 = log 1 1. (2) 3

4 Takig Theorem 1.2 ad equality (2) ito accout, we ca say that for every whe /π() is a iteger we must have π() = log 1 π() = log 1. Sice Theorem 1.1 implies that /π() is a iteger ifiitely ofte, it follows that there are ifiitely may positive itegers such that π() = / log 1. I fact, the followig theorem follows from Theorems 1.1, from Gaitaa s proof of Theorem 1.4, ad from the proof of Theorem 2.1: Theorem 2.2. For every whe /π() is a iteger we must have = log 1.5 = log 0.5 = log 1 = π() = log = log 1.1 = log 1 1. I other words, for whe /π() is a iteger we must have π() = log 1.5 = = log 0.5 = log 1.1 = log 1 = log 1 1. (3) log = Theorem 2.3. Let be a iteger If frac(log ) = log log > 0.1, the π() (that is to say, /π() is ot a iteger). Proof. Accordig to Theorem 2.2, if ad /π() is a iteger, the π() = log 1 = log 1.1. I other words, for whe /π() is a iteger we have log 1 = log 1.1 log 1 = log frac(log 1) 0.1 log 1 log

5 log log frac(log ) 0.1 log log 0.1. Suppose that P is the statemet /π() is a iteger ad Q is the statemet log log 0.1. Accordig to propositioal logic, the fact that P Q implies that Q P. Similar theorems ca be proved by usig Theorem 2.2 ad equality (3). Remark 2.4. We ca also say that if ad the π(). > e 0.1+ log, Remark 2.5. Because log is irratioal for > 1, aother way of statig Theorem 2.3 is by sayig that if ad the first digit to the right of the decimal poit of log is 1, 2, 3, 4, 5, 6, 7, 8, or 9, the π(). Example: log = The first digit after the decimal poit of log (i red) is 3. This implies that π(10 31 ) does ot divide We ca also say that if ad π() divides, the the first digit after the decimal poit of log ca oly be 0. Now, if y is a positive oiteger, the the first digit after the decimal poit of y is equal to 10 frac(y) = 10y 10 y. So, we ca say that if ad 10 log 10 log = 0, the π(). O the other had, if ad π() divides, the 10 log 10 log = 0. The followig theorem follows from Theorem 2.3: Theorem 2.6. Let e be the base of the atural logarithm. If a is ay iteger 11 ad is ay iteger cotaied i the iterval [e a+0.1, e a+1 ], the π(). (The umber e r is irratioal whe r is a ratioal umber 0.) Example 2.7. Take a = 18. If is ay iteger i the iterval [e 18.1, e 19 ], the π(). Corollary 2.8. If a is ay positive iteger > 1, the π( e a ) e a. Proof. For a 12 the proof follows from Theorem 2.6. O the other had, e a /π( e a ) is ot a iteger wheever 2 a 11, as show i the followig table: 5

6 a e a /π( e a ) I other words, if a Z +, the π( e a ) e a oly whe a = 1. Theorem 2.9. Let be a iteger ad let / log 1 log 1 f() = log log 0.1. If f() = 0, the π(). O the other had, if π(), the f() = 1. Proof. Part 1 Suppose that where ad f() = g()h(), g() = log log 0.1 / log 1 log 1 h() =. To begi with, if 60184, the log log ca ever be equal to 0.1. Now, whe log log < 0.1 we have 1 < log log 0.1 < 0 ad hece log log 0.1 = 1. O the other had, whe log log > 0.1 we have 0 < log log 0.1 < 1 ad hece log log 0.1 = 0. This meas that if is ay iteger 60184, 6

7 the g() equals either 0 or 1. We ca also say that if ad g() = 0, the log log > 0.1, which implies that π() (accordig to Theorem 2.3). (This meas that if ad π(), the g() = 1.) Part 2 If 60184, the log 1 log 1, which meas that / log 1 log 1 / = = h() log 1 log 1 equals either 0 or 1. If h() = 0, the is ot divisible by log 1, which implies that π() (accordig to Theorem 2.2). I other words, if ad h() = 0, the π(). (This meas that if ad π(), the h() = 1.) Part 3 There are two possible outputs for g() (0 or 1) as well as two possible outputs for h() (0 or 1). This meas that for we have either or or or g()h() = 0 0 = 0, g()h() = 0 1 = 0, g()h() = 1 0 = 0, g()h() = 1 1 = 1. If f() = g()h() = 0, the at least oe of the factors g() ad h() equals 0, which implies that π() (see Part 1 ad Part 2). This meas that if ad f() = 0, the π(). Cosequetly, if ad π(), the f() = 1. Theorem If ad /π() is a iteger, the is a multiple of log 1 located i the iterval [e log 1 +1, e log ]. 7

8 Proof. Accordig to Theorems 2.2 ad 2.3, if ad /π() is a iteger, the ad π() = log 1 = π() log 1 frac(log ) = log log 0.1. The fact that frac(log ) 0.1 implies that is located i the iterval [e k, e k+0.1 ] for some positive iteger k. I other words, we have e k < < e k+0.1 k < log < k k 1 < log 1 < k 0.9, which meas that k 1 = log 1 k = log Remark Suppose that b is ay fixed iteger 12. Theorem 2.10 implies that if is a iteger i the iterval [e b, e b+0.1 ] ad at the same time is ot a multiple of b 1, the π(). This meas that if ad π() divides, the is located i the iterval [e b, e b+0.1 ] for some positive iteger b ad is a multiple of b 1. The followig theorem follows from Theorems 1.1 ad 2.10 ad from the fact that /π() < 11 for (this fact ca be checked usig software): Theorem Let c be ay fixed iteger 12. I the iterval [e c, e c+0.1 ] there is always a iteger such that π() divides. I other words, i the iterval [e c, e c+0.1 ] there is always a iteger such that π() = /(c 1). 3 Coclusio ad Further Discussio The followig are the mai theorems of this paper: Theorem 2.9. Let be a iteger ad let / log 1 log 1 f() = log log

9 If f() = 0, the π(). O the other had, if π(), the f() = 1. Theorem If ad /π() is a iteger, the is a multiple of log 1 located i the iterval [e log 1 +1, e log ]. Theorem Let c be ay fixed iteger 12. I the iterval [e c, e c+0.1 ] there is always a iteger such that π() divides. I other words, i the iterval [e c, e c+0.1 ] there is always a iteger such that π() = /(c 1). We recall that Golomb [3] proved that for every iteger > 1 there exists a positive iteger m such that m/π(m) =. Suppose ow that R is the sequece of umbers geerated by the fuctio d() = /π() ( Z ad > 1). I other words, R = (2, 1.5, 2, , 2, 1.75, 2, 2.25, 2.5,... ). Suppose also that S is the sequece of itegers geerated by the fuctio d() = /π(). I other words, S = (2, 2, 2, 2, 3, 3, 3, 4, 4,... ). Motivated by Golomb s result ad Theorem 2.12 we ask the followig questio: Questio 3.1. Are there ifiitely may positive itegers a such that i the iterval [e a, e a+0.1 ] there are at least two distict positive itegers 1 ad 2 such that π( 1 ) 1 ad π( 2 ) 2? I other words, are there ifiitely may positive itegers that ca be expressed as m/π(m) i more tha oe way? Now, let S k deote the kth term of sequece S. Clearly, Questio 3.1 is equivalet to the followig questio: Questio 3.2. Are there ifiitely may positive itegers k such that S k = S k+1? Refereces [1] Dusart, P. Estimates of Some Fuctios Over Primes without R.H. arxiv: [math.nt],

10 [2] Gaitaas, K. N. A explicit formula for the prime coutig fuctio. arxiv: [math.nt], [3] Golomb, S. W. O the Ratio of N to π(n). The America Mathematical Mothly. Vol. 69, No. 1, pp , [4] Rosser, J. B.; Schoefeld, L. Approximate formulas for some fuctios of prime umbers. Illiois Joural of Mathematics. Vol. 6, No. 1, pp ,