Note on some explicit formulae for twin prime counting function


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1 Notes on Number Theory and Discrete Mathematics Vol. 9, 03, No., Note on some explicit formulae for twin prime counting function Mladen VassilevMissana 5 V. Hugo Str., 4 Sofia, Bulgaria Abstract: In the paper for any integer n 5 the validity of the formula ( ) ϕ(3 ) π (n) = + 3 = (where π denotes the twin prime counting function and ϕ is Euler s totient function) is established. Also for any integer n 5 the formula ( ) ϕ(3 ln f() ) π (n) = + 3 = (where f is arbitrary artihmetic function with strictly positive values satisfying the condition < ) is proved. =4 f() Keywords: Prime number (prime), Twin primes, Twin prime counting function, Arithmetic function. AMS Classification: A5, A4. Used denotations denotes the floor function, i.e. x denotes the largest integer that is not greater than the real nonnegative number x; π  the twin prime counting function, i.e. for any integer n 3, π (n) denotes the number of couples (p, p + ) such that p n and p and p + are both primes; ϕ denotes Euler s totient function, i.e. ϕ() = and for any integer n ϕ(n) = n p n ( p ), 43
2 where p n means that the product is taen over all prime divisors p of n (it is nown that ϕ(n) coincides with the number of all integer n which are relatively prime to n). Introduction As nown (see [, p. 59]), twin primes are couples of the ind (p, p + ), where p and p + are both primes. It is clear that (with the only exception being the couple (3, 5)) all couples of twin primes have the form (, + ) with an appropriate integer. The sequence of all twin primes is: 3, 5, 7,, 3, 7, 9, 9, 3, 4, 43,.... The couples of twin primes are: (3, 5), (5, 7), (, 3), (7, 9), (9, 3), (4, 43),.... So for any integer n 5, for the twin prime counting function π, the representation is valid, where: δ = π (n) = + = {, if (, + ) is a couple of primes 0, otherwise. δ () Using the mentioned above, in year 00, the author (in []) proposed (for the first time) fourteen different formulae for the twin prime counting function π. One of them is the following: π (n) = + = ( ) ϕ(3 ). 3 Unfortunately in this formula the floor function is applied to each term of the sum which maes it not sufficiently convenient. But it is possible to remedy this disadvantage with the help of the formulae that we propose for the first time in the present paper. In these formulae the floor function is applied to the whole sum. Preliminary result Let for integer δ is introduced by: Further we need the following result. δ def = ϕ(3 ) 3. () Lemma. (, + ) is a couple of twin primes if and only if δ =. Otherwise, δ. (3) 44
3 Proof. Since for any integer and + are relatively prime, and because of the multiplicativity of the function ϕ, we have δ = ϕ(( )( + )) ( ) = ϕ( )ϕ( + ). (4) ( ) Let for some integer the numbers and + are both primes. Then: ϕ( ) = ; ϕ( + ) = and (4) implies δ =. Conversely, let δ =. Since ϕ( ) and ϕ( + ), we must have: ϕ( ) = ; ϕ( + ) =. But the last means that the numbers and + are both primes, i.e. (, + ) is a twin prime couple. Otherwise, we have three different cases: (a) and + are composite; (b) is prime and + is composite; (c) is composite and + is prime. Let (a) hold. Then ϕ( ) 3; ϕ( + ) and (4) yields δ = Therefore, for (a) (3) is valid. Let (b) hold. Then ϕ(( )( + )) ( ) ( ) =. Hence (4) yields ϕ( ) = ; ϕ( + ). δ ( )( ) ( ) =. Therefore, for (b) (3) is valid. Let (c) hold. Then ϕ( ) 3; ϕ( + ) =. Hence (4) yields Therefore, for (c) (3) is valid. The lemma is proved. δ ( 3) ( ) = 3 <. 45
4 3 Main results The first main result of the paper is the following. Theorem. Let f is arbitrary artihmetic function with strictly positive values satisfying the condition <. (5) f() =4 Then for any integer n 5 the formula π (n) = + is valid. = Proof. Let n 5 be arbitrarily chosen. We denote Then R = + R def = + = = ( ) ϕ(3 ln f() ) () 3 ( ) ϕ(3 ln f() ). (7) 3 ln f() δ = ln f() + ln f(), where is a sum over all integer satisfying n+, such that δ = and is a sum over all integer such that δ < ( satisfying n+ ). Hence R = π (n) + ln f(), (8) since ln f() = π (n), because of the Lemma. Therefore, to prove () it remains only to establish that, under the conditions of Theorem, ln f() <. (9) where Let is such that δ <. Since (3) holds, we have δ ( ) < e, (0) ( e def = lim + m = m m) and for the monotonously increasing sequence { ( m} m) m= it is fulfilled ( e = lim m. m m) 4
5 Therefore, (because of (5)). Theorem is proved. ln f() < =4 e ln f() = f() < =4 Let n 77. Then 3 n+ and (8) yields where R = π (n) + δ + δ 8 + δ 9 + δ + ln f(), () is considered for satisfying 3 n+ and Let f(x) = e x. Then: δ i def = (δ i ) i ln f(i). and δ 4 = 5 ; δ = 7 ; δ 8 = 7 8 ; δ 9 = 0 7 ; δ = 3 4 δ i = (δ i ) i. The direct computation yields δ + δ 8 + δ 9 + δ < 0.0. () Using (0), this time we have: ln f() < =3 e ln f() = =3 f() = =3 e = e e < (3) Since <, from (), () and (3) it follows R = π (n). Therefore, () and the last equality proved () for n 77 with f(x) = e x. But it is a matter of a direct chec to see that in this case () is valid also for 5 n 7. Since ln f() = for f() = e, we proved the second main result of our paper. Theorem. For any integer n 5 the representation ( ϕ(3 ) π (n) = + 3 holds. = ) 47
6 References [] Ribenboim, P. The New Boo of Prime Number Records (3rd Edition), SpringerVerlag, New Yor, 99. [] VassilevMissana, M. Some new formulae for the twin prime counting function π, Notes on Number Theory and Discrete Mathematics, Vol 7, 00, No.,
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