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# On The Height Of The qx 1 P roblem

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

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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

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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

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