DOC

On The Height Of The qx 1 P roblem

By Rhonda Simpson,2014-03-20 18:06
25 views 0
On The Height Of The qx 1 P roblem

     中国科技论文在线 http://www.paper.edu.cn

    On The Height Of The qx + 1 P roblem

    JiangYu, Wu Dongliang, Wei Lili, Zhao Dongfang

    ( I nstitute of M athematics and Statistics, C entral C hina

    N ormal U niversity, W uhan, P RC 430079)

     yu-jiang0406@163.com E-mail:

    Abstract: The C ollatz problem, also known as 3x + 1 problem has been studying by many scholars since last century. So far, we already have a lot of

    good results on it. During this paper, we will mainly study the qx + 1 problem. And give the general explicit expressions of the numbers with height one and wo. Finally, applying these conclusions on 3 1 problem and 7 1 problem. tx+x+

    Keywords: qx+r P roblem, qx+1 P roblem, 3x+1 P roblem, 7x+1 P roblem.

AMS sub ject classification: 11A99.

    1.Introduction.

    The open hard 3x + 1 problem, also known as the C ollatz problem, the Syracuse

    0problem, U lams problem and so on. It has been attracting many mathematicians since

    be posed last century and already have a lot of conclusions on it. But the results on

    + general qx + 1 problem (q ? D) are rare.

    + First, we need some preliminaries. In this paper, N denote the positive integers,

    + + + N N ? 0 ,and Dis the odd elements of N . = {}

    Define the function qm+r C(m) = ,qr e(m)qr 2

    + e(m) qr where q, r, m ? D, 2is the highest power of two dividing qm + r. So for

    + + any m ? D, C(m) is still in D. The open hard qx r problem can be stated + qr

    + k thus: given (q, r), for every m ? Dis there a k such that C (m) = 1? qr

    Next, we will give wo definitions about qx r problem and a lemma. t+

    + Definition1.1. For m ? D, the trajectory of m is the sequence

     中国科技论文在线 http://www.paper.edu.cn

     2 3 T= {C(m), C (m), C (m), ...},m qr qr qr

    k + which terminates upon the first occurrence of C (m) = 1, k ? Z . If there is no such qr

    [1] k, then is an infinite sequence T.m

     +Definition1.2. Suppose m ? D, the height of m, denoted by h(m), is the

    [1] length of sequence T. h(m) = ? if Tis an infinite sequence . m m

    + + Lemma1.3. If a, q ? N , (a, q) 1, q > 1, then there exists x ? N satisfies =

    x [4] congruence equation a? 1 (mod q) .

    From [1] and [2], we know there are always numbers of any given height on 3x + 1 problem. It’s natural to ask whether has the same conclusion on other q1 problem. x+At present,this question cannot has an affirmative answer on 7x + 1 problem. During

    this essay, we will mainly discuss the height of qx + 1 problem. Our cardinal results

    can be described as following two theorems.

     ??Theorem1. If (x, x) is the smallest positive solution that satisfy the equation1 2

    x1 2? qx1. Then we have following conclusion on qx 1 problem: h(m) 1 if and = += 2

    only if

     ? x k 11 2?1+ m kN = ,? .1 q

    Theorem2. On qx + 1 problem, h(m) = 2 if and only if

     ? k+xkk1 2 2 1 ?q2?2.m = 2 q

    ?In which, k? N, k? N , k? r(mod ), ksatisfy the congruence equation x1 1 2 2 2 1

     r? ??2 2? x? k? 1 ? 0 (mod q), (x, x) is the smallest positive solution of the equation1 2 1 2

    x1 2? qx1. = 2

    2.The proof of The main Theorems.

    x+ 1 Above all, we need to proof 2? qx= 1 have solutions for any q ? D. By 2

    + x1 lemma1.3, because (2, q) = 1, so exists x? N satisfies 2? 1 (mod q). It’s 1

    x+ x1 1 equivalents to 2= qx+ 1 for some x? N . Therefore, 2? qx= 1 always have 2 2 2 solutions.

    The proof of Theorem 1:

    of By definition, we have Pro

     ? ? x kxk1 1 1 1 2?1 2 C ( = 1,) =? q1 x kq 1 1e((2 ?1)/q) 2

Report this document

For any questions or suggestions please email
cust-service@docsford.com