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Some Observations on the Greatest Prime Factor of an Integer

Some Observations on the Greatest Prime Factor of an Integer Annales Mathematicae Silesianae 37 (2023), no. 1, 67–81 DOI: 10.2478/amsil-2022-0018 SOME OBSERVATIONS ON THE GREATEST PRIME FACTOR OF AN INTEGER Rafael Jakimczuk Abstract. We examine the multiplicity of the greatest prime factor in k-full numbers and k-free numbers. We generalize a well-known result on greatest prime factors and obtain formulas related with the Riemann zeta function. 1. Introduction We examine the multiplicity of the greatest prime factor in k-full numbers and k-free numbers (Theorem 2.1 and Theorem 2.2). We define some new arithmetical functions related with the greatest prime factor and obtain in Theorem 2.3 asymptotic formulas related with the Riemann zeta function. Finally in Theorem 2.4 we obtain an asymptotic formula for the sum of m-th powers of greatest prime factors obtaining a better result than the previous one of the author. Let us consider the prime factorization of a positive integral number n, 1 s namely n = q  q , where the q (i = 1; : : : ; r) are its distinct prime 1 r factors and the s (i = 1; : : : ; r) are their respective multiplicities. We shall need the following well-known arithmetical functions, (n) is the sum of the positive divisors of n, k(n) = q  q is the kernel of n and hence (k(n)) = 1 r (q + 1) (q + 1). 1 r Received: 03.03.2022. Accepted: 02.11.2022. Published online: 23.11.2022. (2020) Mathematics Subject Classification: 11A99, 11B99. Key words and phrases: greatest prime factor, zeta function. 2022 The Author(s). This is an Open Access article distributed under the terms of the Creative Commons Attribution License CC BY (http://creativecommons.org/licenses/by/4.0/). 68 Rafael Jakimczuk A positive integer n is square-free if and only if its prime factorization has not factors with an exponent larger than one, that is, n = q  q , where 1 r the q (i = 1; : : : ; r) are distinct. Let Q (x) denote the cardinality of the set i 2 of square-free numbers not exceeding x. It is well-known that these numbers 1 6 have density = , where (s) denotes the Riemann zeta function. More (2) exactly, p p 1 6 (1.1) Q (x) = x + O( x) = x + O( x); (2) where the error term can be improved. Lemma 1.1. Let Q (x) denote the number of square-free not exceeding q q 1 r x relatively prime to the square-free q  q . The following formula holds 1 r 6 q  q 1 r Q (x) = x + O(2 x): q q 1 r (q + 1) (q + 1) 1 r Proof. See [4]. Let s  2 be an arbitrary fixed integer. A positive integer is said to be s-full if all the factors in its prime factorization have exponent greater than 1 s or equal to s. That is, the number q  q is s-full if s  s (i = 1; : : : ; r). Let n denote a general s-full number. If s = 1 then we obtain the positive integers. If s = 2 these numbers are called square-full or powerful. Lemma 1.2. Let s  1 be an arbitrary fixed integer. Let A (x) denote the cardinality of the set of s-full numbers not exceeding x. We have 6 1 1 s s A (x) = C x + o x ; s s where X Y 1 1 1 C = = 1 + : 1 1 (k(n)) s s n (p + 1)(p 1) n=1 p Proof. See [5]. Lemma 1.3. Let s  1 be an arbitrary fixed integer. The following series converges 1 s where Q runs over the (s + 1)-full numbers Q = q  q . 1 r Some observations on the greatest prime factor of an integer 69 Proof. Let a denote the n-th (s + 1)-full number. We have (see s+1 Lemma 1.2) A (x)  c x, where c is a constant. Therefore if x = a s+1 1 1 n s+1 s+1 then we obtain n = A (a )  c a , that is, a  . The lemma s+1 n 1 n n s+1 follows by the comparison criterion, since the series converges. The s+1 lemma is proved. Let us consider the prime factorization of a positive integral number n, 1 r namely n = q  q , where the q (i = 1; : : : ; r) are the distinct prime 1 r factors and the s (i = 1; : : : ; r) are their respective multiplicities. Let k  2 be an arbitrary fixed positive integer. We shall say that n is k-free if s  k1 (i = 1; : : : ; r). In particular, if k = 2 we obtain the square-free numbers. Let s denote a general k-free number. Let Q (x) be the cardinality of the set k k of k-free numbers not exceeding x. It is well-known that these numbers have density , that is, (k) 1=k (1.2) Q (x) = x + O x ; (k  2): (k) In particular, if k = 2 then equation (1.2) becomes equation (1.1). We shall need the following well-known lemmas. Lemma 1.4. We have = log log x + M + O ( (x)) ; px where M  0; 26149 : : : is the Mertens’s constant and  (x) is the usual num- 3=5 1=5 c(log x) (log log x) ber theoretic function  (x) = e for some c > 0. Note that for all positive integer N we have  (x) = O . c N log x Lemma 1.5. Let (x) denote the prime counting function. We have (x) = 1 = Li(x) + O (x (x)) ; px dt where Li(x) = is the logarithmic integral. 2 log t We also shall need the following fundamental lemma. 70 Rafael Jakimczuk Lemma 1.6. Let j, ` nonnegative integers and a positive integer. Then for all positive integer N `+1 `+1 j N1 X X x x a x j;h;`; p = + O ; h N+1 p log x (log x) (log x) h=0 px where 1 j fug (log u) a = du j;h;`; `+1 1+ 1 u and (as usual) fxg = xbxc is the fractional part of x. The error term depends at most on j; `; ; N: Proof. See [3]. Let us consider the prime factorization of a positive integer n, namely 1 s n = q  q , where q > q >  > q are the distinct prime factors and 1 2 r the s (i = 1; : : : ; r) are the multiplicities. Note that q is the greatest prime i 1 factor of n and note that n can be considered as a product of prime powers q (i = 1; : : : ; r). Let G(n) = q denote the greatest prime factor in the prime factorization of n and let A(n) denote the sum of all prime factors in the prime factorization of n, that is, A(n) = s q + s q + + s q . Alladi and Erdős ([1]) proved 1 1 2 2 r r the following formulas 2 2 2 x x G(n) = + O ; 12 log x log x nx 2 2 2 x x A(n) = + O : 12 log x log x nx Therefore 2 2 X X (1.3) A(n)  G(n)  : 12 log x nx nx Below, in Remark 3.1, we shall prove (1.3) and more precise formulas, as a consequence of our results in Theorem 2.3. We shall establish the following definitions. Let a (n) be the sum of prime s s i i powers q such that s = k. If all prime powers q are such that s 6= k we i i i i put a (n) = 0. Let A (n) be the sum of prime powers q such that s  k. k k i i Some observations on the greatest prime factor of an integer 71 If all prime powers q are such that s < k we put A (n) = 0. We put i k s s 1 1 g (n) = q if s = k and g (n) = 0 if s 6= k. We put G (n) = q if k 1 k 1 k 1 1 s  k and G (n) = 0 if s < k. We recall that q is the greatest prime 1 k 1 1 4 5 5 4 4 5 5 5 5 factor of n. For example, a (19 11 5 ) = 19 , a (19 11 5 ) = 11 + 5 , 4 5 4 5 5 4 5 5 4 5 5 4 5 5 4 a (19 11 5 ) = 0, A (19 11 5 ) = 19 + 11 + 5 , A (19 11 5 ) = 19 + 3 4 3 5 5 4 5 5 4 5 5 4 4 5 5 11 + 5 , A (19 11 5 ) = 0, g (19 11 5 ) = 19 , g (19 11 5 ) = 0 (h 6= 4), 8 4 h 4 5 5 4 4 5 5 G (19 11 5 ) = 19 , G (19 11 5 ) = 0. 2 7 Alladi and Erdős ([1]) proved that 2 2 (2) x x (1.4) G(n) = + O : 2 log x log x nx Jakimczuk ([6]) proved the following generalization m+1 (m + 1) x G(n)  ; m + 1 log x nx where m is an arbitrary fixed positive integer. In Theorem 2.4 we obtain a more precise result. 2. Main Results Theorem 2.1. Let s  1 be an arbitrary fixed integer. Let G (x) denote the cardinality of the set of s-full numbers n not exceeding x such that the greatest prime factor of n has multiplicity s in the prime factorization of n . s s Then G (x)  A (x). That is, in almost all s-full numbers the greatest prime s s factor has multiplicity s. If s = 1 then in almost all integers the greatest prime factor has multiplicity 1. Theorem 2.2. Let k  3 be an arbitrary fixed integer. Let H (x) denote the cardinality of the set of k-free numbers s not exceeding x such that the greatest prime factor of s has multiplicity 1 in the prime factorization of s . k k Then H (x)  Q (x). That is, in almost all k-free numbers s the greatest k k k prime factor has multiplicity 1. 72 Rafael Jakimczuk Theorem 2.3. Let k and N be arbitrary fixed positive integers. We have the following asymptotic formulas 1 1 N1 1+ 1+ X X k k x b x h;k (2.1) a (n) = + O ; h N+1 log x (log x) (log x) nx h=0 1 N1 1 1+ 1+ X X k k x b x h;k (2.2) A (n) = + O ; h N+1 log x (log x) (log x) nx h=0 1 1 N1 1+ 1+ X X k k x b x h;k (2.3) g (n) = + O ; h N+1 log x (log x) (log x) nx h=0 1 1 N1 1+ 1+ X X k k x b x h;k (2.4) G (n) = + O ; h N+1 log x (log x) (log x) nx h=0 where the error terms depend at most on k and N and where b depends on h;k (0) (i) the zeta function (s) =  (s) and its successive derivatives  (s) in the point s = 1 + , as it is showed by the following formula i (i) (1)  1 + (2.5) b = h! : h;k h+1i i! 1 + i=0 The first coefficient is 1 1 b =  1 + : 0;k 1 + k Therefore we have 1+ X X X X 1 1 x A (n)  a (n)  g (n)  G (n)   1 + : k k k k 1 + k log x nx nx nx nx Theorem 2.4. Let m be an arbitrary fixed positive integer. Then m+1 m+1 (m + 1) x x (2.6) G(n) = + O : m + 1 log x log x nx If m = 1 then equation (2.6) becomes equation (1.4). Some observations on the greatest prime factor of an integer 73 3. Proofs Proof of Theorem 2.1. If the s-full number n is of the form q , where q is a square-free, then clearly the greatest prime factor of n has multiplicity s. The cardinality of the set of the numbers n = q not exceeding x, that is, q  x, is (see (1.1)) 1=s 1=s (3.1) x + o x : Let us consider the prime factorization of a (s + 1)-full number Q, that is, r r 1 t s Q = q  q . Let us consider the s-full numbers n of the form q Q, where 1 t q is a square-free, gcd(q; Q) = 1 and Q is fixed. Except by a finite number of cases, the greatest prime factor in q Q is in the prime factorization of q and consequently it has multiplicity s. Therefore the number of these numbers q Q not exceeding x (q Q  x) such that the greatest prime factor has multiplicity s will be (see Lemma 1.1) 1=s 6 q  q x 1 t 1=s (3.2) + o x : 2 1=s (q + 1) (q + 1) Q 1 t Let  > 0. We choose B such that (see Lemma 1.3) (3.3) < : 1=s Q>B Therefore we have (see (3.1), (3.2), (3.3) and Lemma 1.2) 0 1 6 q  q 1 1 t 1=s 1=s @ A G (x) = 1 + x + o x + F (x) 2 1=s (q + 1) (q + 1) 1 t QB 0 1 6 q  q 1 1 t 1=s 1=s @ A (3.4) = C x + o x + F (x); 1=s (q + 1) (q + 1) 1 t Q>B where (see (3.3)) 1=s 1=s X X x x 1=s (3.5) 0  F (x)    x : 1=s 1=s Q Q xQ>B Q>B 74 Rafael Jakimczuk Note that X Y q  q 1 p 1 1 1 t 1 + = 1 + + + 1 s+1 s+2 (q + 1) (q + 1) s p + 1 1 t Q s s p p Q p 0 1 1 1 1 @ A = 1 + p + 1 s 1 p 1 1 1 = 1 + = C : p + 1 p 1 Consequently (see (3.4) and (3.5)) G (x) 6 C   +  +  = 3 (x  x ): 1=s 2 That is, since  > 0 can be arbitrarily small, 1=s G (x)  C x  A (x): s s s The theorem is proved. Proof of Theorem 2.2. The proof is identical to the proof of Theo- rem 2.1. If the k-free number is a square-free q then clearly the greatest prime factor of q has multiplicity 1. The number of these numbers not exceeding x, that is, q  x, is (see (1.1)) x + o (x) : Let us consider the prime factorization of a square-full number Q, that is, r r 1 t Q = q  q , where 2  r  k 1 (i = 1; : : : ; t). Let us consider the k- free numbers s of the form qQ, where q is a square-free, gcd(q; Q) = 1 and Q is fixed. Except for a finite number of cases the greatest prime factor in s = qQ is in the prime factorization of q and consequently it has multiplicity 1. Therefore the number of these numbers s = qQ not exceeding x (qQ  x) such that the greatest prime factor has multiplicity 1 will be (see Lemma 1.1) 6 q  q x 1 t + o (x) : (q + 1) (q + 1) Q 1 t Some observations on the greatest prime factor of an integer 75 Now, the proof follows as the proof of Theorem 2.1. Note that 0 1 6 q  q 1 1 t @ A 1 + (q + 1) (q + 1) Q 1 t Y Y 1 p 1 1 = 1 1 + + + 2 2 k1 p p + 1 p p p p Y Y Y p 1 1 1 1 = 1 + 1 = 1 2 2 k2 k p p 1 p p p p p = : (k) The theorem is proved. Proof of Theorem 2.3. It is well-known the asymptotic formula N1 x h! x (3.6) Li(x) = + O : h N+1 log x (log x) (log x) h=0 We have (Lemma 1.4, Lemma 1.5, Lemma 1.6 and equality (3.6)) X X x x a (n) = p k k+1 p p nx px X X X 1 1 x x k k = x x x p + p k k+1 p p p 1 1 1 k+1 k k+1 px px px 1 1 1 k k k = x Li x + O x  x k k+1 x log log x + M + O  x !! 1 1 N1 1+ 1+ k k x a x 1;h;k;k + O h N+1 log x (log x) (log x) h=0 N1 x a x 1;h;k;k+1 + + O h N+1 log x (log x) (log x) h=0 76 Rafael Jakimczuk 1 N1 1 1+ 1+ k k 1 x a x 1;h;k;k = xLi x + O h N+1 log x (log x) (log x) h=0 1 N1 1 1+ 1+ k k x b x h;k (3.7) = + O ; h N+1 log x (log x) (log x) h=0 where fug (log u) h+1 h+1 (3.8) b = k h! a = k h! du: h;k 1;h;k;k 1 2+ Now, by application of Euler-Maclaurin summation formula ([2]), we obtain h1 h 1 (1) h! fug log u (h) h (s) = + (1) h du h+1 s+1 (s 1) u 1 h fug log u h+1 + (1) s du: s+1 Therefore by mathematical induction we find that s fug (s) = s du; s+1 s 1 u (s) s fug log u (s) = + s du; 2 s+1 s (s 1) u and in general 1 h i+1 (i) fug log u h! (1)  (s) (3.9) du = + h! : s+1 h+1 h+1i u (s 1) i! s i=0 Substituting s = 1 + into (3.9) and by application of (3.8) we obtain for- mula (2.5). Therefore equality (2.1) is proved. We have X X x x A (n) = p + F (x) k k+1 p p nx px (3.10) = a (n) + F (x): nx Some observations on the greatest prime factor of an integer 77 j k h log x log x log x Note that the inequality p > x holds for h = + 1 >  and log 2 log 2 log p therefore we have x x x k+1 k+2 h 0  F (x)  p + p + + p k+1 k+2 h p p p k+1 px 1+ k+1 log x x x (3.11)  hx 1  + 1 xc = O : N+1 log 2 k+1 (log x) log x k+1 px Properties (3.10), (3.11) and (2.1) give equality (2.2). Note that g (n)  a (n) and hence formula (2.1) gives k k 1 N1 1 1+ 1+ X X k k x b x h;k (3.12) g (n)  + O : h N+1 log x (log x) (log x) nx h=0 k k+1 k+1 k Note that if x < p  x then p  x and p > x. Now, the multiples j k j k k k k k x x of p not exceeding x are p ; p 2; : : : ; p , where < p. Therefore p is k k p p the greatest prime factor of these numbers. Consequently, we have (see (3.7)) X X X x 1 x k k k k+1 g (n)  p = x x p x x k k p p 1 k nx p x k+1 x <px 1 1 N1 1+ 1+ X X k k x x b x h;k (3.13) + p = + O ; k h N+1 p log x (log x) (log x) h=0 k+1 px since X X X x 1 k k k+1 0  p  p  x 1 1 1 1 k+1 k+1 k+1 px px px 1+ x x c = O N+1 log x (log x) and 1+ 1 1 x 1+ k+1 k+1 x x  x = O : N+1 (log x) Inequalities (3.12) and (3.13) give formula (2.3). 78 Rafael Jakimczuk Finally, we have X X (3.14) G (n) = g (n) + F (x); k k 1 nx nx where (see (3.11)) 1+ (3.15) 0  F (x)  F (x) = O : N+1 (log x) Properties (2.3), (3.14) and (3.15) give equality (2.4), which completes the proof. Remark 3.1. If k = 1 then equalities (2.3) and (2.2) give N1 2 2 X X x b x h;1 g (n) = + O ; h N+1 log x (log x) (log x) nx h=0 N1 2 2 X X x b x h;1 A (n) = + O ; h N+1 log x (log x) (log x) nx h=0 where A (n) is the sum of the prime powers in the prime factorization of n, s s s 1 2 s i that is, A (n) = q +q + +q . Therefore, since s q  q (i = 1; : : : ; r), 1 i i 1 2 i we have X X X X g (n)  G(n)  A(n)  A (n): 1 1 nx nx nx nx Therefore we obtain N1 2 2 2 2 X X x b x  x h;1 G(n) = + O  ; h N+1 log x (log x) (log x) 12 log x nx h=0 N1 2 2 2 2 X X x b x  x h;1 A(n) = + O  : h N+1 log x (log x) (log x) 12 log x nx h=0 P P Consequently G(n) and A(n) have the same asymptotic expan- nx nx sion. Some observations on the greatest prime factor of an integer 79 Proof of Theorem 2.4. By the prime number theorem we have x x (x) = + O : log x log x Abel’s summation gives m+1 m+1 1 x x S (x) = p = + O : m + 1 log x log x px That is m+1 m+1 1 x x (3.16) S (x) = p = + f(x) (x  2); m + 1 log x log x px where jf(x)j < M for x  2. Note that M depends on m. j k Let us consider the positive integer s = (log x) + 1 and a positive integer k such that 1  k  s 1. Now, consider the primes p such that x x < p  . The numbers multiples of p not exceeding x are p; 2p; 3p; : : : ; kp k+1 k and since p > k, if x is sufficiently large, we obtain that p is the greatest prime factor of these k numbers. Hence if F (x) is the contribution to the sum G(n) of the numbers not exceeding x such that their greatest prime nx factor is in the interval 2; then we have s1 X X X m m G(n) = kp + F (x) x x nx k=1 <p k+1 k 0 1 s1 s1 X X X x x @ A = p + F (x) = S S + F (x) m m i s x x i=1 i=1 <p s i s1 x x (3.17) = S (s 1)S + F (x); m m i s i=1 where m+1 x x (3.18) (s 1)S = O ; log x m+1 X X x x m m1 (3.19) 0  F (x)  p = x p = O ; log x x x p p s s 80 Rafael Jakimczuk and (see (3.16)) s1 s1 X X m+1 m+1 x 1 x 1 x x 1 (3.20) S = +f  : x x i m + 1 i i i log log i=1 i=1 i Now, we have the formula = 1+g(t)t, where g(t) ! 1 as t ! 0. Therefore 1t we have (see (3.20)) s1 s1 m+1 X m+1 X 1 x 1 x 1 1 log i log i = 1 + g m+1 m + 1 i m + 1 log x i log x log x log i=1 i=1 m+1 m+1 (m + 1) x x 1 m+1 m + 1 log x (m + 1) log x i is s1 m+1 x log i log i + g 2 m+1 i log x (m + 1) log x i=1 m+1 m+1 (m + 1) x x (3.21) = + O ; m + 1 log x log x and (see (3.16)) s1 s1 m+1 X X m+1 x x 1 x x 1 1 f  = f 2 2 2 x m+1 i i i i log log x log i i=1 i i=1 log x m+1 (3.22)  2M(m + 1) : log x Substituting (3.21) and (3.22) into (3.20) and then substituting (3.20), (3.19) and (3.18) into (3.17) we obtain (2.6). The theorem is proved. Acknowledgement. The author would like to thank the editor and the anonymous referee who provided extensive arguments, constructive remarks and helpful suggestions on the earlier version of this article. The author is also very grateful to Universidad Nacional de Luján. Some observations on the greatest prime factor of an integer 81 References [1] K. Alladi and P. Erdős, On an additive arithmetic function, Pacific J. Math. 71 (1977), no. 2, 275–294. [2] T.M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York- Heidelberg, 1976. [3] O. Bordellés and L. Tóth, Additive arithmetic functions meet the inclusion-exclusion principle, II, to appear in Res. Number Theory. Avaliable at arXiv: 2112.13409. [4] R. Jakimczuk and M. Lalín, Sums of !(n) and (n) on the k-free parts and k-full parts of some particular sequences, preprint 2022. [5] R. Jakimczuk and M. Lalín, The number of prime factors on average in certain integer sequences, J. Integer Seq. 25 (2022), no. 2, Art. 22.2.3, 15 pp. [6] R. Jakimczuk, A note on sums of greatest (least) prime factors, Int. J. Contemp. Math. Sci. 8 (2013), no. 9–12, 423–432. División Matemática Universidad Nacional de Luján Buenos Aires Argentina e-mail: jakimczu@mail.unlu.edu.ar http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annales Mathematicae Silesianae de Gruyter

Some Observations on the Greatest Prime Factor of an Integer

Jakimczuk, Rafael
Annales Mathematicae Silesianae , Volume 37 (1): 15 – Mar 1, 2023
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Annales Mathematicae Silesianae 37 (2023), no. 1, 67–81 DOI: 10.2478/amsil-2022-0018 SOME OBSERVATIONS ON THE GREATEST PRIME FACTOR OF AN INTEGER Rafael Jakimczuk Abstract. We examine the multiplicity of the greatest prime factor in k-full numbers and k-free numbers. We generalize a well-known result on greatest prime factors and obtain formulas related with the Riemann zeta function. 1. Introduction We examine the multiplicity of the greatest prime factor in k-full numbers and k-free numbers (Theorem 2.1 and Theorem 2.2). We define some new arithmetical functions related with the greatest prime factor and obtain in Theorem 2.3 asymptotic formulas related with the Riemann zeta function. Finally in Theorem 2.4 we obtain an asymptotic formula for the sum of m-th powers of greatest prime factors obtaining a better result than the previous one of the author. Let us consider the prime factorization of a positive integral number n, 1 s namely n = q  q , where the q (i = 1; : : : ; r) are its distinct prime 1 r factors and the s (i = 1; : : : ; r) are their respective multiplicities. We shall need the following well-known arithmetical functions, (n) is the sum of the positive divisors of n, k(n) = q  q is the kernel of n and hence (k(n)) = 1 r (q + 1) (q + 1). 1 r Received: 03.03.2022. Accepted: 02.11.2022. Published online: 23.11.2022. (2020) Mathematics Subject Classification: 11A99, 11B99. Key words and phrases: greatest prime factor, zeta function. 2022 The Author(s). This is an Open Access article distributed under the terms of the Creative Commons Attribution License CC BY (http://creativecommons.org/licenses/by/4.0/). 68 Rafael Jakimczuk A positive integer n is square-free if and only if its prime factorization has not factors with an exponent larger than one, that is, n = q  q , where 1 r the q (i = 1; : : : ; r) are distinct. Let Q (x) denote the cardinality of the set i 2 of square-free numbers not exceeding x. It is well-known that these numbers 1 6 have density = , where (s) denotes the Riemann zeta function. More (2) exactly, p p 1 6 (1.1) Q (x) = x + O( x) = x + O( x); (2) where the error term can be improved. Lemma 1.1. Let Q (x) denote the number of square-free not exceeding q q 1 r x relatively prime to the square-free q  q . The following formula holds 1 r 6 q  q 1 r Q (x) = x + O(2 x): q q 1 r (q + 1) (q + 1) 1 r Proof. See [4]. Let s  2 be an arbitrary fixed integer. A positive integer is said to be s-full if all the factors in its prime factorization have exponent greater than 1 s or equal to s. That is, the number q  q is s-full if s  s (i = 1; : : : ; r). Let n denote a general s-full number. If s = 1 then we obtain the positive integers. If s = 2 these numbers are called square-full or powerful. Lemma 1.2. Let s  1 be an arbitrary fixed integer. Let A (x) denote the cardinality of the set of s-full numbers not exceeding x. We have 6 1 1 s s A (x) = C x + o x ; s s where X Y 1 1 1 C = = 1 + : 1 1 (k(n)) s s n (p + 1)(p 1) n=1 p Proof. See [5]. Lemma 1.3. Let s  1 be an arbitrary fixed integer. The following series converges 1 s where Q runs over the (s + 1)-full numbers Q = q  q . 1 r Some observations on the greatest prime factor of an integer 69 Proof. Let a denote the n-th (s + 1)-full number. We have (see s+1 Lemma 1.2) A (x)  c x, where c is a constant. Therefore if x = a s+1 1 1 n s+1 s+1 then we obtain n = A (a )  c a , that is, a  . The lemma s+1 n 1 n n s+1 follows by the comparison criterion, since the series converges. The s+1 lemma is proved. Let us consider the prime factorization of a positive integral number n, 1 r namely n = q  q , where the q (i = 1; : : : ; r) are the distinct prime 1 r factors and the s (i = 1; : : : ; r) are their respective multiplicities. Let k  2 be an arbitrary fixed positive integer. We shall say that n is k-free if s  k1 (i = 1; : : : ; r). In particular, if k = 2 we obtain the square-free numbers. Let s denote a general k-free number. Let Q (x) be the cardinality of the set k k of k-free numbers not exceeding x. It is well-known that these numbers have density , that is, (k) 1=k (1.2) Q (x) = x + O x ; (k  2): (k) In particular, if k = 2 then equation (1.2) becomes equation (1.1). We shall need the following well-known lemmas. Lemma 1.4. We have = log log x + M + O ( (x)) ; px where M  0; 26149 : : : is the Mertens’s constant and  (x) is the usual num- 3=5 1=5 c(log x) (log log x) ber theoretic function  (x) = e for some c > 0. Note that for all positive integer N we have  (x) = O . c N log x Lemma 1.5. Let (x) denote the prime counting function. We have (x) = 1 = Li(x) + O (x (x)) ; px dt where Li(x) = is the logarithmic integral. 2 log t We also shall need the following fundamental lemma. 70 Rafael Jakimczuk Lemma 1.6. Let j, ` nonnegative integers and a positive integer. Then for all positive integer N `+1 `+1 j N1 X X x x a x j;h;`; p = + O ; h N+1 p log x (log x) (log x) h=0 px where 1 j fug (log u) a = du j;h;`; `+1 1+ 1 u and (as usual) fxg = xbxc is the fractional part of x. The error term depends at most on j; `; ; N: Proof. See [3]. Let us consider the prime factorization of a positive integer n, namely 1 s n = q  q , where q > q >  > q are the distinct prime factors and 1 2 r the s (i = 1; : : : ; r) are the multiplicities. Note that q is the greatest prime i 1 factor of n and note that n can be considered as a product of prime powers q (i = 1; : : : ; r). Let G(n) = q denote the greatest prime factor in the prime factorization of n and let A(n) denote the sum of all prime factors in the prime factorization of n, that is, A(n) = s q + s q + + s q . Alladi and Erdős ([1]) proved 1 1 2 2 r r the following formulas 2 2 2 x x G(n) = + O ; 12 log x log x nx 2 2 2 x x A(n) = + O : 12 log x log x nx Therefore 2 2 X X (1.3) A(n)  G(n)  : 12 log x nx nx Below, in Remark 3.1, we shall prove (1.3) and more precise formulas, as a consequence of our results in Theorem 2.3. We shall establish the following definitions. Let a (n) be the sum of prime s s i i powers q such that s = k. If all prime powers q are such that s 6= k we i i i i put a (n) = 0. Let A (n) be the sum of prime powers q such that s  k. k k i i Some observations on the greatest prime factor of an integer 71 If all prime powers q are such that s < k we put A (n) = 0. We put i k s s 1 1 g (n) = q if s = k and g (n) = 0 if s 6= k. We put G (n) = q if k 1 k 1 k 1 1 s  k and G (n) = 0 if s < k. We recall that q is the greatest prime 1 k 1 1 4 5 5 4 4 5 5 5 5 factor of n. For example, a (19 11 5 ) = 19 , a (19 11 5 ) = 11 + 5 , 4 5 4 5 5 4 5 5 4 5 5 4 5 5 4 a (19 11 5 ) = 0, A (19 11 5 ) = 19 + 11 + 5 , A (19 11 5 ) = 19 + 3 4 3 5 5 4 5 5 4 5 5 4 4 5 5 11 + 5 , A (19 11 5 ) = 0, g (19 11 5 ) = 19 , g (19 11 5 ) = 0 (h 6= 4), 8 4 h 4 5 5 4 4 5 5 G (19 11 5 ) = 19 , G (19 11 5 ) = 0. 2 7 Alladi and Erdős ([1]) proved that 2 2 (2) x x (1.4) G(n) = + O : 2 log x log x nx Jakimczuk ([6]) proved the following generalization m+1 (m + 1) x G(n)  ; m + 1 log x nx where m is an arbitrary fixed positive integer. In Theorem 2.4 we obtain a more precise result. 2. Main Results Theorem 2.1. Let s  1 be an arbitrary fixed integer. Let G (x) denote the cardinality of the set of s-full numbers n not exceeding x such that the greatest prime factor of n has multiplicity s in the prime factorization of n . s s Then G (x)  A (x). That is, in almost all s-full numbers the greatest prime s s factor has multiplicity s. If s = 1 then in almost all integers the greatest prime factor has multiplicity 1. Theorem 2.2. Let k  3 be an arbitrary fixed integer. Let H (x) denote the cardinality of the set of k-free numbers s not exceeding x such that the greatest prime factor of s has multiplicity 1 in the prime factorization of s . k k Then H (x)  Q (x). That is, in almost all k-free numbers s the greatest k k k prime factor has multiplicity 1. 72 Rafael Jakimczuk Theorem 2.3. Let k and N be arbitrary fixed positive integers. We have the following asymptotic formulas 1 1 N1 1+ 1+ X X k k x b x h;k (2.1) a (n) = + O ; h N+1 log x (log x) (log x) nx h=0 1 N1 1 1+ 1+ X X k k x b x h;k (2.2) A (n) = + O ; h N+1 log x (log x) (log x) nx h=0 1 1 N1 1+ 1+ X X k k x b x h;k (2.3) g (n) = + O ; h N+1 log x (log x) (log x) nx h=0 1 1 N1 1+ 1+ X X k k x b x h;k (2.4) G (n) = + O ; h N+1 log x (log x) (log x) nx h=0 where the error terms depend at most on k and N and where b depends on h;k (0) (i) the zeta function (s) =  (s) and its successive derivatives  (s) in the point s = 1 + , as it is showed by the following formula i (i) (1)  1 + (2.5) b = h! : h;k h+1i i! 1 + i=0 The first coefficient is 1 1 b =  1 + : 0;k 1 + k Therefore we have 1+ X X X X 1 1 x A (n)  a (n)  g (n)  G (n)   1 + : k k k k 1 + k log x nx nx nx nx Theorem 2.4. Let m be an arbitrary fixed positive integer. Then m+1 m+1 (m + 1) x x (2.6) G(n) = + O : m + 1 log x log x nx If m = 1 then equation (2.6) becomes equation (1.4). Some observations on the greatest prime factor of an integer 73 3. Proofs Proof of Theorem 2.1. If the s-full number n is of the form q , where q is a square-free, then clearly the greatest prime factor of n has multiplicity s. The cardinality of the set of the numbers n = q not exceeding x, that is, q  x, is (see (1.1)) 1=s 1=s (3.1) x + o x : Let us consider the prime factorization of a (s + 1)-full number Q, that is, r r 1 t s Q = q  q . Let us consider the s-full numbers n of the form q Q, where 1 t q is a square-free, gcd(q; Q) = 1 and Q is fixed. Except by a finite number of cases, the greatest prime factor in q Q is in the prime factorization of q and consequently it has multiplicity s. Therefore the number of these numbers q Q not exceeding x (q Q  x) such that the greatest prime factor has multiplicity s will be (see Lemma 1.1) 1=s 6 q  q x 1 t 1=s (3.2) + o x : 2 1=s (q + 1) (q + 1) Q 1 t Let  > 0. We choose B such that (see Lemma 1.3) (3.3) < : 1=s Q>B Therefore we have (see (3.1), (3.2), (3.3) and Lemma 1.2) 0 1 6 q  q 1 1 t 1=s 1=s @ A G (x) = 1 + x + o x + F (x) 2 1=s (q + 1) (q + 1) 1 t QB 0 1 6 q  q 1 1 t 1=s 1=s @ A (3.4) = C x + o x + F (x); 1=s (q + 1) (q + 1) 1 t Q>B where (see (3.3)) 1=s 1=s X X x x 1=s (3.5) 0  F (x)    x : 1=s 1=s Q Q xQ>B Q>B 74 Rafael Jakimczuk Note that X Y q  q 1 p 1 1 1 t 1 + = 1 + + + 1 s+1 s+2 (q + 1) (q + 1) s p + 1 1 t Q s s p p Q p 0 1 1 1 1 @ A = 1 + p + 1 s 1 p 1 1 1 = 1 + = C : p + 1 p 1 Consequently (see (3.4) and (3.5)) G (x) 6 C   +  +  = 3 (x  x ): 1=s 2 That is, since  > 0 can be arbitrarily small, 1=s G (x)  C x  A (x): s s s The theorem is proved. Proof of Theorem 2.2. The proof is identical to the proof of Theo- rem 2.1. If the k-free number is a square-free q then clearly the greatest prime factor of q has multiplicity 1. The number of these numbers not exceeding x, that is, q  x, is (see (1.1)) x + o (x) : Let us consider the prime factorization of a square-full number Q, that is, r r 1 t Q = q  q , where 2  r  k 1 (i = 1; : : : ; t). Let us consider the k- free numbers s of the form qQ, where q is a square-free, gcd(q; Q) = 1 and Q is fixed. Except for a finite number of cases the greatest prime factor in s = qQ is in the prime factorization of q and consequently it has multiplicity 1. Therefore the number of these numbers s = qQ not exceeding x (qQ  x) such that the greatest prime factor has multiplicity 1 will be (see Lemma 1.1) 6 q  q x 1 t + o (x) : (q + 1) (q + 1) Q 1 t Some observations on the greatest prime factor of an integer 75 Now, the proof follows as the proof of Theorem 2.1. Note that 0 1 6 q  q 1 1 t @ A 1 + (q + 1) (q + 1) Q 1 t Y Y 1 p 1 1 = 1 1 + + + 2 2 k1 p p + 1 p p p p Y Y Y p 1 1 1 1 = 1 + 1 = 1 2 2 k2 k p p 1 p p p p p = : (k) The theorem is proved. Proof of Theorem 2.3. It is well-known the asymptotic formula N1 x h! x (3.6) Li(x) = + O : h N+1 log x (log x) (log x) h=0 We have (Lemma 1.4, Lemma 1.5, Lemma 1.6 and equality (3.6)) X X x x a (n) = p k k+1 p p nx px X X X 1 1 x x k k = x x x p + p k k+1 p p p 1 1 1 k+1 k k+1 px px px 1 1 1 k k k = x Li x + O x  x k k+1 x log log x + M + O  x !! 1 1 N1 1+ 1+ k k x a x 1;h;k;k + O h N+1 log x (log x) (log x) h=0 N1 x a x 1;h;k;k+1 + + O h N+1 log x (log x) (log x) h=0 76 Rafael Jakimczuk 1 N1 1 1+ 1+ k k 1 x a x 1;h;k;k = xLi x + O h N+1 log x (log x) (log x) h=0 1 N1 1 1+ 1+ k k x b x h;k (3.7) = + O ; h N+1 log x (log x) (log x) h=0 where fug (log u) h+1 h+1 (3.8) b = k h! a = k h! du: h;k 1;h;k;k 1 2+ Now, by application of Euler-Maclaurin summation formula ([2]), we obtain h1 h 1 (1) h! fug log u (h) h (s) = + (1) h du h+1 s+1 (s 1) u 1 h fug log u h+1 + (1) s du: s+1 Therefore by mathematical induction we find that s fug (s) = s du; s+1 s 1 u (s) s fug log u (s) = + s du; 2 s+1 s (s 1) u and in general 1 h i+1 (i) fug log u h! (1)  (s) (3.9) du = + h! : s+1 h+1 h+1i u (s 1) i! s i=0 Substituting s = 1 + into (3.9) and by application of (3.8) we obtain for- mula (2.5). Therefore equality (2.1) is proved. We have X X x x A (n) = p + F (x) k k+1 p p nx px (3.10) = a (n) + F (x): nx Some observations on the greatest prime factor of an integer 77 j k h log x log x log x Note that the inequality p > x holds for h = + 1 >  and log 2 log 2 log p therefore we have x x x k+1 k+2 h 0  F (x)  p + p + + p k+1 k+2 h p p p k+1 px 1+ k+1 log x x x (3.11)  hx 1  + 1 xc = O : N+1 log 2 k+1 (log x) log x k+1 px Properties (3.10), (3.11) and (2.1) give equality (2.2). Note that g (n)  a (n) and hence formula (2.1) gives k k 1 N1 1 1+ 1+ X X k k x b x h;k (3.12) g (n)  + O : h N+1 log x (log x) (log x) nx h=0 k k+1 k+1 k Note that if x < p  x then p  x and p > x. Now, the multiples j k j k k k k k x x of p not exceeding x are p ; p 2; : : : ; p , where < p. Therefore p is k k p p the greatest prime factor of these numbers. Consequently, we have (see (3.7)) X X X x 1 x k k k k+1 g (n)  p = x x p x x k k p p 1 k nx p x k+1 x <px 1 1 N1 1+ 1+ X X k k x x b x h;k (3.13) + p = + O ; k h N+1 p log x (log x) (log x) h=0 k+1 px since X X X x 1 k k k+1 0  p  p  x 1 1 1 1 k+1 k+1 k+1 px px px 1+ x x c = O N+1 log x (log x) and 1+ 1 1 x 1+ k+1 k+1 x x  x = O : N+1 (log x) Inequalities (3.12) and (3.13) give formula (2.3). 78 Rafael Jakimczuk Finally, we have X X (3.14) G (n) = g (n) + F (x); k k 1 nx nx where (see (3.11)) 1+ (3.15) 0  F (x)  F (x) = O : N+1 (log x) Properties (2.3), (3.14) and (3.15) give equality (2.4), which completes the proof. Remark 3.1. If k = 1 then equalities (2.3) and (2.2) give N1 2 2 X X x b x h;1 g (n) = + O ; h N+1 log x (log x) (log x) nx h=0 N1 2 2 X X x b x h;1 A (n) = + O ; h N+1 log x (log x) (log x) nx h=0 where A (n) is the sum of the prime powers in the prime factorization of n, s s s 1 2 s i that is, A (n) = q +q + +q . Therefore, since s q  q (i = 1; : : : ; r), 1 i i 1 2 i we have X X X X g (n)  G(n)  A(n)  A (n): 1 1 nx nx nx nx Therefore we obtain N1 2 2 2 2 X X x b x  x h;1 G(n) = + O  ; h N+1 log x (log x) (log x) 12 log x nx h=0 N1 2 2 2 2 X X x b x  x h;1 A(n) = + O  : h N+1 log x (log x) (log x) 12 log x nx h=0 P P Consequently G(n) and A(n) have the same asymptotic expan- nx nx sion. Some observations on the greatest prime factor of an integer 79 Proof of Theorem 2.4. By the prime number theorem we have x x (x) = + O : log x log x Abel’s summation gives m+1 m+1 1 x x S (x) = p = + O : m + 1 log x log x px That is m+1 m+1 1 x x (3.16) S (x) = p = + f(x) (x  2); m + 1 log x log x px where jf(x)j < M for x  2. Note that M depends on m. j k Let us consider the positive integer s = (log x) + 1 and a positive integer k such that 1  k  s 1. Now, consider the primes p such that x x < p  . The numbers multiples of p not exceeding x are p; 2p; 3p; : : : ; kp k+1 k and since p > k, if x is sufficiently large, we obtain that p is the greatest prime factor of these k numbers. Hence if F (x) is the contribution to the sum G(n) of the numbers not exceeding x such that their greatest prime nx factor is in the interval 2; then we have s1 X X X m m G(n) = kp + F (x) x x nx k=1 <p k+1 k 0 1 s1 s1 X X X x x @ A = p + F (x) = S S + F (x) m m i s x x i=1 i=1 <p s i s1 x x (3.17) = S (s 1)S + F (x); m m i s i=1 where m+1 x x (3.18) (s 1)S = O ; log x m+1 X X x x m m1 (3.19) 0  F (x)  p = x p = O ; log x x x p p s s 80 Rafael Jakimczuk and (see (3.16)) s1 s1 X X m+1 m+1 x 1 x 1 x x 1 (3.20) S = +f  : x x i m + 1 i i i log log i=1 i=1 i Now, we have the formula = 1+g(t)t, where g(t) ! 1 as t ! 0. Therefore 1t we have (see (3.20)) s1 s1 m+1 X m+1 X 1 x 1 x 1 1 log i log i = 1 + g m+1 m + 1 i m + 1 log x i log x log x log i=1 i=1 m+1 m+1 (m + 1) x x 1 m+1 m + 1 log x (m + 1) log x i is s1 m+1 x log i log i + g 2 m+1 i log x (m + 1) log x i=1 m+1 m+1 (m + 1) x x (3.21) = + O ; m + 1 log x log x and (see (3.16)) s1 s1 m+1 X X m+1 x x 1 x x 1 1 f  = f 2 2 2 x m+1 i i i i log log x log i i=1 i i=1 log x m+1 (3.22)  2M(m + 1) : log x Substituting (3.21) and (3.22) into (3.20) and then substituting (3.20), (3.19) and (3.18) into (3.17) we obtain (2.6). The theorem is proved. Acknowledgement. The author would like to thank the editor and the anonymous referee who provided extensive arguments, constructive remarks and helpful suggestions on the earlier version of this article. The author is also very grateful to Universidad Nacional de Luján. Some observations on the greatest prime factor of an integer 81 References [1] K. Alladi and P. Erdős, On an additive arithmetic function, Pacific J. Math. 71 (1977), no. 2, 275–294. [2] T.M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York- Heidelberg, 1976. [3] O. Bordellés and L. Tóth, Additive arithmetic functions meet the inclusion-exclusion principle, II, to appear in Res. Number Theory. Avaliable at arXiv: 2112.13409. [4] R. Jakimczuk and M. Lalín, Sums of !(n) and (n) on the k-free parts and k-full parts of some particular sequences, preprint 2022. [5] R. Jakimczuk and M. Lalín, The number of prime factors on average in certain integer sequences, J. Integer Seq. 25 (2022), no. 2, Art. 22.2.3, 15 pp. [6] R. Jakimczuk, A note on sums of greatest (least) prime factors, Int. J. Contemp. Math. Sci. 8 (2013), no. 9–12, 423–432. División Matemática Universidad Nacional de Luján Buenos Aires Argentina e-mail: jakimczu@mail.unlu.edu.ar

Journal

Annales Mathematicae Silesianae – de Gruyter

Published: Mar 1, 2023

Keywords: greatest prime factor; zeta function; 11A99; 11B99

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