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On the Inverse Problem Relative to 9 Dynamics of the w Function 0 0 2 Chaohua Jia n a J 4 Abstract. In this paper we shall study the inverse problem relative to dynamics 2 ofthew functionwhichisaspecialarithmeticfunctionandshallgetsomeresults. ] T N . h 1. Introduction t a m In 2006, Wushi Goldring [4] proposed some problems and conjectures [ on dynamics of the w function and gave some interesting results. Recently 1 Yong-Gao Chen and Ying Shi [2], [3] made further progress on these prob- v 4 lems. In this paperwe shall study the inverse problem relative to dynamics 3 8 of the w function. 3 . We begin by introducing some notations. Let P be the set of prime 1 0 numbers and P(n) denote the largest prime factor of integer n > 1. Write 9 0 v: C3 = {p1p2p3 : pi ∈ P(i = 1,2,3), pi 6= pj(i 6= j)}, i X B3 = {p1p2p3 : pi ∈ P(i = 1,2,3), p1 = p2 or p1 = p3 r a or p2 = p3, but not p1 =p2 = p3}, D = {p3 : p ∈ P}. 3 Then {p p p : p ∈ P(i = 1,2,3)} = C ∪B ∪D , 1 2 3 i 3 3 3 where no any two of C , B and D intersect. Let 3 3 3 A = C ∪B . 3 3 3 For n = p p p ∈ A , define the w function by 1 2 3 3 w(n) = P(p +p )P(p +p )P(p +p ) 1 2 1 3 2 3 1 and define w0(n)= n, wi(n)= w(wi−1(n)), i = 1, 2, ··· , which is reasonable according to Lemma 1 in section 2. Wushi Goldring [4] proved that for any n ∈ A , there exists i such 3 that wi(n) = 20. The smallest of such i is denoted by ind(n). He [4] proposed two conjectures on ind(n) (Conjectures 2.9 and 2.10) and gave the firstupperboundfor ind(n) which has been improved greatly by Yong- Gao Chen and Ying Shi [2] recently. Wushi Goldring [4] also asked the following inverse problems: 1. For n ∈A , can we find m ∈ A such that w(m) = n? 3 3 2. If so, how many such elements are there? 3. What form do they have? For n ∈ A , if there is m ∈ A such that w(m) = n, then we call m 3 3 a parent of n. If this m ∈ S ⊂ A , then we call it S-parent of n. Wushi 3 Goldring[4]provedthatthereareinfinitelymanyelementsofB whichhave 3 at least seven parents. He proposed the following conjecture (Conjecture 2.16 in [4]). Conjecture (Wushi Goldring). Every element of A (respectively 3 B ) has infinitely many C -parents (respectively B -parents). 3 3 3 Yong-Gao Chen and Ying Shi [3] proved that for any given positive integer k, there are infinitely many elements of B which have at least k 3 B -parents. On the other hand, they [3] proved that there are infinitely 3 many elements of B which have no B -parent. 3 3 In this paper we shall study parents of elements of C . It is obvious 3 that the element of C has no B -parent. We shall prove that there are 3 3 infinitely many elements of C which have enough C -parents. 3 3 Inthefollowing, p, p , p , p , q, r, r , r denoteprimenumbersandc , 1 2 3 1 2 1 c , ··· denote positive constants. The expression f ≪ g means f = O(g). 2 We suppose that x is sufficiently large throughout. Theorem 1. There exists an element r r q of C which satisfies 1 2 3 x12 logx < ri ≤ 2x21 logx(i = 1,2), q ≤ 4xandhasatleastc1logx4x different C -parents p p p with x < p ≤ 2x(i = 1,2,3). 3 1 2 3 i 2 We shall also prove that there are infinitely many elements of B which 3 have enough C -parents. 3 Theorem 2. There exists an element qr2 of B which satisfies q ≤ 3 4x, x21 logx < r ≤ 2x12 logx and has at least c2logx4x different C3-parents p p p with x < p ≤ 2x(i = 1,2,3). 1 2 3 i Moreover, we shall prove that there are infinitely many elements of B 3 that have enough B -parents, which is a quantitative improvement on the 3 result of Yong-Gao Chen and Ying Shi [3]. Theorem 3. There exists an element qr2 of B which satisfies x < q ≤ 3 2x, x12 logx < r ≤ 2x12 logx and has at least c3loxg212x different B3-parents pq2 with x < p ≤ 2x. 2. Lemmas Lemma 1. If n ∈A , then w(n) ∈ A . 3 3 This is Lemma 2.1 in [4]. Lemma 2. Let n (1 ≤ j ≤ Z) be distinct positive integers not exceed- j ing N and Z(N; r, a) denote the number of those n which are congruent j to a(modr). If X ≥ 2, then we have r Z 2 r Z(N; r, a)− ≪ (N +X2)Z. X X(cid:16) r(cid:17) r≤X a=1 This is Theorem 1 in [1], which is obtained by the large sieve method. Lemma 3. We have 1 2 x2 1− 1 ≪ . X X (cid:16) X r X (cid:17) logx x12 logx<r≤2x21 logxx<p1≤2x x<p≤2x x<p≤2x p≡−p1(modr) Proof. We see 1 2 1− 1 X X (cid:16) X r X (cid:17) x12 logx<r≤2x21 logxx<p1≤2x x<p≤2x x<p≤2x p≡−p1(modr) 1 2 ≤ 1− 1 X X (cid:16) X r X (cid:17) x21 logx<r≤2x21logxx<n1≤x+([xr]+1)r x<p≤2x x<p≤2x p≡−n1(modr) 3 r x 1 2 ≪ 1− 1 X r X(cid:16) X r X (cid:17) 1 1 a=1 x<p≤2x x<p≤2x x2 logx<r≤2x2 logx p≡a(modr) r 1 1 2 ≪ r 1− 1 . log2x X X(cid:16) X r X (cid:17) 1 a=1 x<p≤2x x<p≤2x r≤2x2 logx p≡a(modr) Then Lemma 2 and the prime number theorem yield r 1 1 2 r 1− 1 log2x X X(cid:16) X r X (cid:17) 1 a=1 x<p≤2x x<p≤2x r≤2x2 logx p≡a(modr) 1 x ≪ ·xlog2x· log2x logx x2 = . logx Hence, Lemma 3 holds true. 3. The proof of Theorem 1 1 We note that if p > n2, then p|n ⇐⇒ P(n)= p. Then we have 1 X X X x<p1≤2x x<p2≤2x x<p3≤2x 1 1 1 1 x2 logx<P(p1+p2)≤2x2 logxx2 logx<P(p1+p3)≤2x2 logx 2 = 1 X (cid:16) X (cid:17) x<p1≤2x x<p≤2x 1 1 x2 logx<P(p+p1)≤2x2 logx 2 = 1 X (cid:16) X X (cid:17) x<p1≤2x x12 logx<r≤2x21logx x<p≤2x P(p+p1)=r 2 = 1 X (cid:16) X X (cid:17) x<p1≤2x x21 logx<r≤2x21logx x<p≤2x p≡−p1(modr) 1 = 1 X (cid:16) X r X x<p1≤2x x21 logx<r≤2x21logx x<p≤2x 1 2 + 1− 1 (1) X (cid:16) X r X (cid:17)(cid:17) 1 1 x<p≤2x x<p≤2x x2 logx<r≤2x2 logx p≡−p1(modr) 4 1 2 = 1 X (cid:16) X r X (cid:17) x<p1≤2x x21 logx<r≤2x21 logx x<p≤2x 1 +2 1 X (cid:16) X r X (cid:17) x<p1≤2x x12 logx<r≤2x21logx x<p≤2x 1 · 1− 1 X (cid:16) X r X (cid:17) 1 1 x<p≤2x x<p≤2x x2 logx<r≤2x2 logx p≡−p1(modr) 1 2 + 1− 1 . X (cid:16) X (cid:16) X r X (cid:17)(cid:17) x<p1≤2x x21 logx<r≤2x21logx x<p≤2x x<p≤2x p≡−p1(modr) The prime number theorem yields 1 2 1 X (cid:16) X r X (cid:17) x<p1≤2x x21 logx<r≤2x21logx x<p≤2x x3 1 2 x3 ≫ ≫ . log3x(cid:16) X r(cid:17) log5x 1 1 x2 logx<r≤2x2 logx By the Cauchy inequality and Lemma 3, we have 1 2 1− 1 X (cid:16) X (cid:16) X r X (cid:17)(cid:17) x<p1≤2x x12 logx<r≤2x21logx x<p≤2x x<p≤2x p≡−p1(modr) ≤ 1· 1 X (cid:16) X X (cid:16) X x<p1≤2x x21 logx<r≤2x21logx x21 logx<r≤2x21 logx x<p≤2x p≡−p1(modr) 1 2 − 1 r X (cid:17) (cid:17) x<p≤2x 1 1 2 ≪ x2 1− 1 X X (cid:16) X r X (cid:17) x12 logx<r≤2x12 logxx<p1≤2x x<p≤2x x<p≤2x p≡−p1(modr) 5 ≪ x2. We also have 1 1 X (cid:16) X r X (cid:17) x<p1≤2x x21 logx<r≤2x21logx x<p≤2x 5 1 · 1− 1 X (cid:16) X r X (cid:17) 1 1 x<p≤2x x<p≤2x x2 logx<r≤2x2 logx p≡−p1(modr) 1 2 1 2 ≪ 1 (cid:16) X (cid:16) X r X (cid:17) (cid:17) x<p1≤2x x21 logx<r≤2x21 logx x<p≤2x 1 2 1 2 · 1− 1 (cid:16) X (cid:16) X (cid:16) X r X (cid:17)(cid:17) (cid:17) x<p1≤2x x21 logx<r≤2x21logx x<p≤2x x<p≤2x p≡−p1(modr) ≪ x3 1 2 21(x52)21 (cid:16) (cid:16) X r(cid:17) (cid:17) 1 1 x2 logx<r≤2x2 logx 3 5 11 ≪ x2 ·x4 = x 4 . Combining the above estimates, we get 1 1 |{p1p2p3 : x < pi ≤ 2x(i = 1, 2, 3), x2 logx < P(p1+p2) ≤ 2x2 logx, 1 1 x2 logx < P(p1+p3)≤ 2x2 logx}| x3 ≫ . log5x Now we shall confine w(p p p ) to C . We have 1 2 3 3 1 X X X x<p1≤2x x<p2≤2x x<p3≤2x 1 1 1 1 x2 logx<P(p1+p2)≤2x2 logxx2 logx<P(p1+p3)≤2x2 logx P(p1+p3)=P(p2+p3) ≤ 1 X X X X x21 logx<r≤2x21logxx<p3≤2x x<p1≤2x x<p2≤2x P(p1+p3)=rP(p2+p3)=r = 1 X X X X x12 logx<r≤2x21logxx<p3≤2x x<p1≤2x x<p2≤2x p1≡−p3(modr)p2≡−p3(modr) ≤ 1 X X X X x12 logx<r≤2x21logxx<n3≤2x x<n1≤2x x<n2≤2x n1≡−n3(modr)n2≡−n3(modr) x2 ≪ X X r2 x<n3≤2xx21 logx<r≤2x21 logx 1 5 ≪ x3 ≪ x2. X n2 1 x2<n 6 Similarly, 5 1 = O(x2). X X X x<p1≤2x x<p2≤2x x<p3≤2x 1 1 1 1 x2 logx<P(p1+p2)≤2x2 logxx2 logx<P(p1+p3)≤2x2 logx w(p1p2p3)6∈C3 Therefore 1 1 |{p1p2p3 ∈ C3 : x < pi ≤ 2x(i = 1, 2, 3), x2 logx < P(p1+p2)≤ 2x2 logx, 1 1 x2 logx < P(p1+p3) ≤ 2x2 logx, w(p1p2p3) ∈ C3}| (2) x3 ≫ . log5x On the other hand, the number of triples (r , r , q) is O( x2 ), where 1 2 logx 1 1 x2 logx < ri ≤ 2x2 logx(i = 1, 2), q ≤ 4x. Therefore in the set in (2), thereareatleastc x differenttriples(p , p , p )satisfyingP(p +p )= 4log4x 1 2 3 1 2 r , P(p +p ) = r , P(p +p ) = q for some (r , r , q). In other words, 1 1 3 2 2 3 1 2 there are at least 1c x different numbers n = p p p ∈ C such that 3! 4log4x 1 2 3 3 w(n) = r r q. 1 2 So far the proof of Theorem 1 is finished. 4. The proof of Theorem 2 We have 1 X X X x<p1≤2xx<p2≤2x x<p3≤2x 1 1 x2 logx<P(p1+p2)=P(p1+p3)≤2x2 logx = 1 X X X X x21 logx<r≤2x21 logxx<p1≤2x x<p2≤2x x<p3≤2x P(p2+p1)=rP(p3+p1)=r 2 = 1 X X (cid:16) X (cid:17) x12 logx<r≤2x21 logxx<p1≤2x x<p≤2x P(p+p1)=r 2 = 1 . X X (cid:16) X (cid:17) x12 logx<r≤2x21 logxx<p1≤2x x<p≤2x p≡−p1(modr) By the inequality 1 a2+b2 ≥ (a−b)2, 2 7 we can get 2 1 X X (cid:16) X (cid:17) x12 logx<r≤2x21 logxx<p1≤2x x<p≤2x p≡−p1(modr) 1 1 2 ≥ 1 X X 2(cid:16)r X (cid:17) x21 logx<r≤2x21logxx<p1≤2x x<p≤2x 1 2 − 1− 1 . X X (cid:16) X r X (cid:17) x21 logx<r≤2x21logxx<p1≤2x x<p≤2x x<p≤2x p≡−p1(modr) The prime number theorem yields 1 1 2 1 X X 2(cid:16)r X (cid:17) x12 logx<r≤2x21 logxx<p1≤2x x<p≤2x x3 1 ≫ log3x X r2 1 1 x2 logx<r≤2x2 logx 5 x2 ≫ log5x and Lemma 3 yields 1 2 1− 1 = O(x2). X X (cid:16) X r X (cid:17) x21 logx<r≤2x21logxx<p1≤2x x<p≤2x x<p≤2x p≡−p1(modr) Therefore 1 |{p1p2p3 ∈ C3 : x < pi ≤ 2x(i = 1, 2, 3), x2 logx < P(p1+p2)= 1 P(p1+p3) ≤ 2x2 logx}| (3) 5 x2 ≫ , log5x since the contribution from terms with p p p 6∈C is O(x2). 1 2 3 3 1 On the other hand, the number of (q, r) with q ≤ 4x, x2 logx < r ≤ 2x12 logx is O(loxg23x). Therefore in the set in (3), there are at least c5 logx4x different triples (p , p , p ) satisfying P(p +p ) = P(p +p ) = r, P(p + 1 2 3 1 2 1 3 2 p ) = q forsome(q, r). Inotherwords,thereareatleast 1 c x different 3 3! 5 log4x numbers n = p p p ∈ C such that w(n) = qr2. By Lemma 1, we know 1 2 3 3 q 6= r. 8 So far the proof of Theorem 2 is finished. 5. The proof of Theorem 3 By Lemma 3, we have 1 1− 1 X X (cid:16) X r X (cid:17) 1 1 x<q≤2x x<p≤2x x<p≤2x x2 logx<r≤2x2 logx p≡−q(modr) 1 2 ≪ 1 (cid:16) X X (cid:17) (cid:16) X X 1 1 x<q≤2x 1 1 x<q≤2x x2 logx<r≤2x2 logx x2 logx<r≤2x2 logx 1 2 1 2 · 1− 1 (cid:16) X r X (cid:17) (cid:17) x<p≤2x x<p≤2x p≡−q(modr) 3 1 1 7 ≪ (x2)2(x2)2 = x4. Hence, 1 X X X 1 1 x<q≤2x x<p≤2x x2 logx<r≤2x2 logx p6=q P(p+q)=r = 1+O(x) X X X 1 1 x<q≤2x x<p≤2x x2 logx<r≤2x2 logx P(p+q)=r = 1+O(x) (4) X X X 1 1 x<q≤2x x<p≤2x x2 logx<r≤2x2 logx p≡−q(modr) 1 7 = 1+O(x4) X X r X 1 1 x<q≤2x x<p≤2x x2 logx<r≤2x2 logx 1 x2 1 7 ≥ · +O(x4) 2 log2x X r 1 1 x2 logx<r≤2x2 logx 1 x2 ≥ · . 3 log3x 1 Thereforetheremustbeonepair(q, r)withx < q ≤ 2x, x2 logx < r ≤ 2x12 logx such that there are at least 110 · loxg221x different p(=6 q) satisfying 9 x <p ≤ 2x, P(p+q)= r. Otherwise we should have 1 x2 · ≤ 1 3 log3x X X X 1 1 x<q≤2x x<p≤2x x2 logx<r≤2x2 logx p6=q P(p+q)=r 1 1 x2 ≤ · 1 10 log2x X X 1 1 x<q≤2x x2 logx<r≤2x2 logx 1 x2 ≤ · , 4 log3x which is a contradiction. 1 For this pair (q, r), there are at least 1 · x2 different p(=6 q) such 10 log2x that w(pq2) = qr2. By Lemma 1, we know q 6= r. Hence, Theorem 3 holds true. Acknowledgements InJune2007,ProfessorYong-GaoChenvisitedMorningsideMathemat- ical Center of Academia Sinica in Beijing and gave two talks to introduce his joint work with Ying Shi on dynamics of the w function. I would like to thank Professor Yong-Gao Chen for his wonderful talks which attract my interest to this topic. I also thank all my colleagues and friends in the “ergodicprimenumbertheorem”seminarinMorningsideCenterforhelpful discussion. Results in this paper were reported at the “combinatorial and analytic number theory” seminar in Nanjing Normal University in September, 2007 and were reported at the fifth Japan-China seminar on number theory in Osaka in August, 2008. 10