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

DivisiónMatemática,Universidad Nacional de Luján 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

  Y 1 γ  1  = 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→∞ 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→∞ 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   B(x) = X  X q . (18) √   N+1≤n≤ x q≤ x n2 Now, we have (see (14), (15) and (18))       N X X X X  X  X  X  a(k) =  q =  q =  q + B(x) √ 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))   $x% X  X  X X 0 ≤ B(x) =  q ≤ q ≤ x 1 √ 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. Then (see [3]) the number of squarefree relatively prime to n2 not exceeding x, that is the number of squarefree relatively primes to n0 not exceeding x is 6 n0 Q (x) = x + o(x) (22) 2 π2 n00 and consequently their sum is 6 n0 Q (x) = x2 + o(x2) (23) 3 2π2 n00 We have the following theorem Theorem 2.2 The following asymptotic formula holds.

X 0 1 6 2 2 a (k) = 2 Ax + o(x ), k≤x 2 π where

∞ 0 X n 1 A = 00 4 n=1 n n Proof. The proof is the same as the proof of Theorem 2.1. In this case we use (22) and (23). The theorem is proved.

Let Q0(x) be the number of positive integers not exceeding x with relatively prime squarefree part. We have the following theorem. Theorem 2.3 The following formula holds 6 Q0(x) = Cx + o(x), (24) π2 where

∞ 0 X n 1 C = 00 2 . (25) n=1 n n Proof. The proof is the same as the proof of Theorem 2.2. In this case we use only (22). The theorem is proved.

Let s ≥ 2 a positive integer. A positive integer such that all prime in its prime factorization has exponent less than or equal to s − 1 is called s-th free number and will be denoted qs−1. For example, if s = 2 we obtain the squarefree numbers (q1 = q). The number of s-th free numbers not exceeding x will be denoted Qs(x), for example, if s = 2 we have Q2(x) = Q(x) (see 338 Rafael Jakimczuk above, equation (13)). It is well-known the following generalization of equation (13)(see, for example [4], for a simple proof) 1 Qs(x) = x + o(x). ζ(s)

s Clearly, every positive integer k can be written in the form k = qs−1.n where s qs−1 is a s-th free number and n is the greatest s-th power that divides k. The positive integer qs−1 is called the s-th free part of k and will be denoted as−1(k). For example, if s = 2 then a1(k) = a(k) (see above). We have the following generalization of Theorem 2.1. Theorem 2.4 The following formula holds

X 1 ζ(2s) 2 2 as−1(k) = x + o(x ) 1≤k≤x 2 ζ(s) Proof. The proof is the same as the proof of Theorem 2.1. The theorem is proved.

Acknowledgements. The author is very grateful to Universidad Nacional de Luj´an.

References

[1] T. M. Apostol, Introduction to Analytic Number Theory, Springer, 1976. https://doi.org/10.1007/978-3-662-28579-4

[2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford, 1960.

[3] R. Jakimczuk, On the distribution of certain subsets of quadratfrei numbers, International Mathematical Forum, 12 (2017), no 4, 185 - 194. https://doi.org/10.12988/imf.2017.612176

[4] R. Jakimczuk, A simple proof that the square-free numbers have density 6/π2, Gulf Journal of Mathematics, 2 (2013), 189 - 192.

[5] W. J. LeVeque, Topics in Number Theory, Volume I, Addison-Wesley, 1958.

[6] N. J. A. Sloane, Sequence A007913, The On-Line Encyclopedia of Integer Sequences.

Received: February 8, 2017; February 22, 2017