Aug 2, 2017 - The first few terms of the integer sequence of square-free numbers are. 1,2,3,5,6,7,10,11 ... The arithmetical function u(n) is well-known in the literature, it is called kernel ... different primes are greater than or equal to h. That is ...
International Mathematical Forum, Vol. 12, 2017, no. 15, 721 - 730 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/imf.2017.7759
The Kernel of Powerful Numbers Rafael Jakimczuk Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina c 2017 Rafael Jakimczuk. This article is distributed under the Creative Copyright Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract We generalize a theorem of the author to h-ful numbers and, for example, we obtain an asymptotic formula for the sum of the kernels of h-ful numbers. The methods used are very elementary.
Mathematics Subject Classification: 11A99, 11B99 Keywords: Kernel function, h-ful numbers
1
Notation and Preliminary Results
A square-free number is a number without square factors, a product of different primes. The first few terms of the integer sequence of square-free numbers are 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, . . . Let us consider the prime factorization of a positive integer n ≥ 2 n = q1s1 q2s2 · · · qtst where q1 , q2 , . . . qt are the different primes in the prime factorization. We have the following two arithmetical functions u(n) = q1 q2 · · · qt The arithmetical function u(n) is well-known in the literature, it is called kernel of n, radical of n, etc. There are many papers dedicated to this arithmetical function. n = q1s1 −1 q2s2 −1 · · · qtst −1 v(n) = u(n)
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Rafael Jakimczuk
We call v(n) the remainder of n. Note that v(n) = 1 if and only if n is a square-free. We also define the arithmetical function w(n) = (q1 + 1)(q2 + 1) · · · (qt + 1) In a previous article [8] the author proves using very elementary methods the following theorem on the kernel function. Theorem 1.1 Let k be an arbitrary but fixed positive integer. The following asymptotic formula holds X
u(n)k =
n≤x
6 Ck k+1 k+1 x + o x π2 k + 1
(1)
where ∞ X
Y 1 1 1 Ck = = 1+ k+1 (p + 1)(pk+1 − 1 p n=1 w(n) n
!
(2)
The case k = 1 was studied in [1](see also [2]) and a better error term is obtained. We have
u(n) = Cx2 + O x3/2 log x
X
n≤x
where 1 1Y 1− C= 2 p p(p + 1)
!
This value of C can be obtained from our formulae (see (1) and (2) with k = 1), since 1 6 1Y C = 1 2 π2 2 p
1 1− 2 p
!
1 1+ (p + 1)(p2 − 1)
!!
1Y 1 1− = 2 p p(p + 1)
!
Note that Theorem 1.1 is also true if k = 0, in this case we obtain the trivial equation X
1 = x + o(x)
n≤x
since ∞ X
1 1 Y 1 C0 = = 1+ (p + 1)(p − 1 p n=1 w(n) n
!
=
Y p
1 π2 = 6 1 − p12
(3)
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Kernel of powerful numbers
Equation (3) was proved in a former article of the author [7]. In this article we generalize Theorem 1.1 to h-ful numbers. A h-ful number is a positive integer such that in its prime factorization all multiplicities of the different primes are greater than or equal to h. That is, the prime factorization of a h-ful number is of the form nh = q1s1 q2s2 · · · qtst
(4)
where q1 , q2 , . . . qt are the different primes in the prime factorization and si ≥ h (i = 1, 2, . . . , t). We denote a h-ful number nh . If h = 2 then the 2-ful numbers n2 are also called either squareful or powerful numbers. If h = 1 then the 1-ful numbers n1 are the positive integers. Note that h is a positive integer. Let us consider the h-ful number (4). The kernel of nh is u(nh ) = q1 q2 · · · qt . We define the h-kernel of nh as u(nh )h = q1h q2h · · · qth and the h-remainder of nh as vh (nh ) = u(nnhh )h = q1s1 −h q2s2 −h · · · qtst −h . We shall need the following well-known lemma (see ([3], chapter XXII) ) Lemma 1.2 Let cn (n ≥ 1) a sequence of real numbers. Let us consider the function X A(x) = cn n≤x
Suppose that f (x) has a continuous derivative f 0 (x) on the interval [1, ∞], then the following formula holds X
cn f (n) = A(x)f (x) −
n≤x
Z x
A(t)f 0 (t) dt
1
The following general theorem is well-known (see [8]). Theorem 1.3 Let us consider a strictly increasing sequence of positive integers, we denote b a positive integer in this sequence. Let A(x) be the number P of positive integers in this sequence not exceeding x. That is A(x) = b≤x 1. Suppose that A(x) = ρx + o(x), where ρ is a positive real number, that is, ρ is the positive density of these integers. Then X b≤x
bk =
ρ xk+1 + o xk+1 k+1
where k is an arbitrary but fixed positive integer. We shall need the following theorems on the distribution of square-free numbers. In this note a square-free number will be denoted q1 .
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Rafael Jakimczuk
Theorem 1.4 Let Q1 (x) be the number of squarefree not exceeding x, we have X 6 Q1 (x) = 1 = 2 x + o(x) π q1 ≤x That is, the squarefree have positive density
6 . π2
Proof. See either ([3], chapter XVIII) or (for an alternative simple proof) [5]. In this note a square-free multiple of the different and fixed primes q1 , q2 , . . . , qt , that is multiple of the square-free q1 q2 · · · qt , will be denoted qq1 q2 ···qt . Theorem 1.5 Let Qq1 q2 ···qt (x) be the number of squarefree multiple of the different and fixed primes q1 , q2 , . . . , qt not exceeding x, we have X
Qq1 q2 ···qt (x) =
qq1 q2 ···qt ≤x
1=
1 6 x + o(x) 2 π (q1 + 1)(q2 + 1) · · · (qt + 1)
That is, these squarefree have positive density
6 1 . π 2 (q1 +1)(q2 +1)···(qt +1)
Proof. See [6]. Theorem 1.6 If α > 0 the following two series of positive terms are convergent ∞ X
1 , α n=1 w(n)n
∞ X
1 α n=1 u(n)n
and besides the following two equations hold ∞ X
Y 1 1 = 1 + α (p + 1)(pα − 1 p n=1 w(n)n ∞ X
Y 1 1 = 1 + α p(pα − 1 p n=1 u(n)n
where the notation
Q
p
!
!
mean that the product runs on all positive primes p.
Proof. We have ∞ X
Y 1 1 1 1 = 1 + + + + ··· α (p + 1)pα (p + 1)(pα )2 (p + 1)(pα )3 p n=1 w(n)n
=
Y p
1 1 Y 1 1 + = 1+ 1 α (p + 1)p (p + 1)(pα − 1 1 − pα p
!
!
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Kernel of powerful numbers
Now, the product 1 1+ (p + 1)(pα − 1
Y p
!
converges to a positive number, since the series of positive terms X p
1 (p + 1)(pα − 1)
clearly converges. The theorem is proved.
2
Main Results
Now, we establish our main theorem. Theorem 1.1 is a particular case of this theorem when h = 1. Theorem 2.1 Let h be an arbitrary but fixed positive integer and let k be an arbitrary but fixed nonnegative integer. The following asymptotic formula holds u(nh )k =
X nh ≤x
k+1 6 Ck,h k+1 h + o x h x π2 k + 1
(5)
where Ck,h
∞ X
Y 1 1 1 = 1+ k+1 = k+1 (p + 1)(p h − 1 p n=1 w(n) n h
!
(6)
Proof. Let us consider the prime factorization of a positive integer a ≥ 2 a = q1s1 q2s2 · · · qtst where q1 , q2 , . . . qt are the different primes in the prime factorization of a. We put a0 = q 1 q 2 · · · q t and a00 = (q1 + 1)(q2 + 1) · · · (qt + 1) If a = 1 then we put a0 = a00 = 1. Therefore we have (see Theorem 1.3, Theorem 1.4 and Theorem 1.5) X qa0 ≤x
qak0 =
6 1 xk+1 k+1 + o x π 2 a00 k + 1
(7)
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Rafael Jakimczuk
Let us consider the set H of all h-full numbers nh not exceeding x. Now, let us consider the set Ta of all h-full numbers nh not exceeding x with the same h-remainder a, that is, Ta = {nh : nh ≤ x, vh (nh ) = a}. Note that if a1 6= a2 we have Ta1 ∩ Ta2 = φ, that is, the sets Ta1 and Ta2 are disjoint. Suppose that Ax (depending of x) is the greatest h-remainder among the numbers in the set H. Then Ax [
Ta = H
a=1
Therefore, the sets Ta are a partition of the set H. Note that some Ta can be empty. The set of the h-kernel of the numbers in the set Ta will be denoted Sa . Hence,
Sa = The series
P∞
1 a=1 a00
qah0
1 a
k+1 h
:
x ≤ a
qah0
(
qah0
=
x(1/h) 0 : qa ≤ (1/h) a
)
(8)
converges (see Theorem 1.6). Hence ∞ X
1 1 = Ck,h 00 k+1 a=1 a a h
(9)
We choose B such that (see Theorem 1.6) ∞ X
1 1
