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Sum of Divisors of the Primorial and Sum of Squarefree Parts

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Sum of Divisors of the Primorial and Sum of Squarefree Parts International Mathematical Forum, Vol. 12, 2017, no. 7, 331 - 338 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/imf.2017.7113 Two Topics in Number Theory: Sum of Divisors of the Primorial and Sum of Squarefree Parts Rafael Jakimczuk Divisi´onMatem´atica,Universidad Nacional de Luj´an Buenos Aires, Argentina Copyright c 2017 Rafael Jakimczuk. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduc- tion in any medium, provided the original work is properly cited. Abstract Let pn be the n-th prime and σ(n) the sum of the positive divisors of n. Let us consider the primorial Pn = p1:p2 : : : pn, the sum of its Qn positive divisors is σ(Pn) = i=1(1 + pi). In the first section we prove the following asymptotic formula n 6 σ(P ) = Y(1 + p ) = eγ P log p + O (P ) : n i π2 n n n i=1 Let a(k) be the squarefree part of k, in the second section we prove the formula π2 X a(k) = x2 + o(x2): 30 1≤k≤x We also study integers with restricted squarefree parts and generalize these results to s-th free parts. Mathematics Subject Classification: 11A99, 11B99 Keywords: Primorial, divisors, squarefree parts, s-th free parts, average of arithmetical functions 332 Rafael Jakimczuk 1 Sum of Divisors of the Primorial In this section p denotes a positive prime and pn denotes the n-th prime. The following Mertens's formulae are well-known (see [5, Chapter VI]) 1 1 ! X = log log x + M + O ; (1) p≤x p log x where M is called Mertens's constant. 1! 1 ! X log 1 − = − log log x − γ + O ; (2) p≤x p log x where γ is the Euler's constant. ! −γ ! Y 1 e 1 1 − = + O 2 ; (3) p≤x p log x log x 0 1 Y 1 γ @ 1 A = e log x + O(1): (4) p≤x 1 − p Note that equation (4) is an immediate consequence of (3) if we use the formula 1 = 1 − x(1 + o(1)) (x ! 0); 1 + x since then 1 1 ! = 1 + O : 1 log x 1 + O log x P 1 Q 1 In this section we examine the sum p≤x log 1 + p and the product p≤x 1 + p . Then, we apply the results to the sum of the divisors of the primorial. Our main theorem is the following. Theorem 1.1 The following asymptotic formulae hold. ! ! X 1 6 1 log 1 + = log log x + γ + log 2 + O ; (5) p≤x p π log x ! Y 1 6 γ 1 + = 2 e log x + O(1): (6) p≤x p π Two topics in number theory: The primorial and squarefree parts 333 Proof. If we consider the formula ex = 1 + x(1 + o(1)) (x ! 0) then equation (6) is an immediate consequence of equation (5). Since we have 1 !! 1 ! exp O = 1 + O : log x log x Therefore we have that to prove equation (5). We have the equation 1 log(1 + x) = x − x2(1 + o(1)) (x ! 0) 2 Consequently ! X 1 X 1 1 X 1 + o(1) log 1 + = − 2 : (7) p≤x p p≤x p 2 p≤x p Now, we have X 1 + o(1) X 1 + o(1) X 1 + o(1) X 1 + o(1) 1 2 = 2 − 2 = 2 + O (8) p≤x p p p p>x p p p x Note that we have used the well-known formula (see [1, Chapter 3]) X 1 1 2 = O ; n>x n x where n denotes a positive integer. Substituting (1) and (8) into (7) we obtain almost (5) 1! 1 ! X log 1 + = log log x + C + O ; p≤x p log x where C is a constant. Therefore we have almost (6) 1! Y 1 + = eC log x + O(1): p≤x p It is well-known the limit (Euler's product formula)(see [1, Chapter 11]) n 1 ! 6 lim Y 1 − = : (9) n!1 2 2 i=1 pi π π2 Note that ζ(2) = 6 , where ζ(s) denotes the Riemann zeta function. Equations C 6 γ (3) and (9) give e = π2 e . The theorem is proved. 334 Rafael Jakimczuk Theorem 1.2 Let us consider the primorial Pn = p1:p2 : : : pn, the sum of Qn its positive divisors is σ(Pn) = i=1(1 + pi). We have the following asymptotic formula n Y 6 γ σ(Pn) = (1 + pi) = 2 e (p1:p2 : : : pn) log pn + O ((p1:p2 : : : pn)) i=1 π 6 = eγP log p + O(P ) (10) π2 n n n Note that the primorial Pn = p1:p2 : : : pn is its greatest divisor. Proof. Put x = pn in equation (6). The theorem is proved. It is well-known (see [2, Chapter XVIII]) that σ(n) lim sup = eγ: n log log n We have Theorem 1.3 The following limit holds σ(P ) 6 lim n = eγ: (11) n!1 2 Pn log log Pn π (1+o(1))pn Proof. It is well-known that Pn = e , therefore log log Pn ∼ log pn. Now, limit (11) is an immediate consequence of (10). The theorem is proved. 2 Sum of Squarefree Parts and Generalization Every positive integer k can be written in the form k = q:n2 where q is a squarefree or quadratfrei number and n2 is the greatest square that divides k. The positive integer q is called the squarefree part of k and it is denoted a(k) (see [6]). Therefore if k is a square we have a(k) = 1, in contrary case a(k) is the product of primes which have odd exponent in the prime factorization of k. For exampe, if k = 24:53:112:235 then a(k) = 5:23. If q is relatively prime with n in the decomposition k = q:n2, that is (q; n2) = 1, we shall call q the relatively prime squarefree part of k and will be denoted a0(k), besides we shall say that k has a relatively prime squarefree part. The average of the arithmetic function a(k) is studied in the following theorem. Two topics in number theory: The primorial and squarefree parts 335 Theorem 2.1 The following formula holds π2 X a(k) = x2 + o(x2): (12) 1≤k≤x 30 Proof. The number of squarefree integers q not exceeding x (denoted Q(x)) is well-known (see either [2, Chapter XVIII] or [4]). We have 1 Q(x) = X 1 = x + o(x); (13) q≤x ζ(2) where ζ(s) denotes the Riemann zeta function. From here, we can obtain easily (see [2, Chapter XXII]) that X 1 2 2 Q1(x) = q = x + o(x ): (14) q≤x 2ζ(2) Let us consider the set Sn such that x S = q : q ≤ (15) n n2 Let us consider > 0. We choose the positive integer N such that 1 1 < (16) ζ(2) (N + 1)2 and 1 X 1 4 < (17) 2ζ(2) n≥N+1 n Besides, for sake of simplicity, we put 0 1 B(x) = X B X qC : (18) p @ A N+1≤n≤ x q≤ x n2 Now, we have (see (14), (15) and (18)) 0 1 0 1 0 1 N X X X X B X C X B X C a(k) = @ qA = @ qA = @ qA + B(x) p x x 1≤k≤x n2≤x qSn n≤ x q≤ n=1 q≤ n2 n2 N N ! X x 1 2 X 1 2 = Q1 2 + B(x) = x 4 + o(x ) + B(x) n=1 n 2ζ(2) n=1 n 2 ζ(4) 2 x X 1 2 = x − 4 + o(1)x + B(x) 2ζ(2) 2ζ(2) n≥N+1 n 336 Rafael Jakimczuk That is P 1≤k≤x a(k) ζ(4) 1 X 1 B(x) 2 − = − 4 + o(1) + 2 (19) x 2ζ(2) 2ζ(2) n≥N+1 n x n x o 2 Note that SN+1 [ SN [ SN−1 [ · · · ⊆ q : q ≤ (N+1)2 . Besides, since qn ≤ x, the number of sets Si such that qSi is less than or equal to the number of j x k multiples of q not exceeding x, namely q , where b:c denotes the integer part function. Hence (see (13) and (16)) 0 1 $x% X B X C X X 0 ≤ B(x) = @ qA ≤ q ≤ x 1 p x x q x N+1≤n≤ x q≤ q≤ q≤ n2 (N+1)2 (N+1)2 x ! 1 1 ! = xQ = + o(1) x2 ≤ 2x2 (N + 1)2 ζ(2) (N + 1)2 That is, B(x) 0 ≤ ≤ 2. (20) x2 Therefore (see (19), (17) and (20)) we have P a(k) ζ(4) 1≤k≤x − ≤ 4. (21) x2 2ζ(2) Consequently equation (12) is proved, since can be arbitrarily small, ζ(2) = π2 π4 6 and ζ(4) = 90 (see [2, Chapter XVII]). The theorem is proved. In the following theorem we examine the sum X a0(k); k≤x where the sum run on the k that have a relatively prime squarefree part. Suppose that the prime factorization of n2 is 2 h1 h2 hs n = p1 p2 ··· ps ; where p1; p2; : : : ; ps are the different primes in the prime factorization and h1; h2; : : : ; hs are the even exponents. We put 0 n = p1p2 ··· ps 00 n = (p1 + 1)(p2 + 1) ··· (ps + 1) Two topics in number theory: The primorial and squarefree parts 337 If n2 = 1 then we put n0 = n00 = 1.
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