This old conjecture (1930). People have never found a counterexample through a lot of checking computations, but no one can prove it.
Try to choose any integer n. The rules are as follows: [If n is odd, calculate n * 3+1; If n is an even number, calculate N/2].
After the first result is obtained, the operation is repeated according to the rules (if n is odd, the operation is N*3+ 1, if n is even, the operation is N/2).
If you keep counting like this, you will find that the last number will cycle once, which is (4→2→ 1→4).
You can try if you don't believe me. I suggest choosing a smaller number (within 100) at the beginning, because this calculation requires patience.
Kakuguchi is a famous Japanese scholar. He put forward two very simple rules, which can transform any natural number, and finally make it fall into an infinite loop of "4-2- 1".
The transformation rule proposed by Kakuya is:
1. When n is odd, the next step becomes 3n+1;
2. When n is an even number, the next step becomes n/2.
People call it the "Corner and Valley Conjecture".
Let me give you some examples to try:
When n is a number 6, according to the rule, the strain is:
6→6÷2→3→3×3+ 1→ 10→ 10÷2→5→5×3+ 1→ 16→ 16÷2→8→8÷2→4→4÷2→2→2÷2→ 1→ 1×3+ 1→4→4÷2→2→2÷2→ 1→……
Finally, it falls into an infinite loop of "4-2- 1".
When n is a two-digit number, such as 46, it should be converted into:
46→46÷2→23→23×3+ 1→70→70÷2→35→35×3+ 1→ 106→ 106÷2→53→53×3+ 1→ 160→ 160÷2→80→8o÷2→40→40
Caught in an infinite loop of "4-2- 1".
There is no need to enumerate more examples. So far, people have not encountered any exceptions. Tested numbers will eventually fall into an infinite loop:
However, there are so many natural numbers that no one can prove, let alone deny, the conjecture about angles and valleys.
Depth expansion
Given a positive integer n, if n is divisible by a, it becomes n/a, and if it is divisible, it is multiplied by b plus c (that is, bn+c). Repeat this operation, after a limited number of steps, will you definitely get D? There are only three answers to this question: 1, not necessarily 2, not necessarily 3, not necessarily all.
The following is all about a certain capital.
A = B = C = D = M。
Two a=m b= 1 c=- 1 d=0.
Three a=m b=c=d= 1
Four A = two B = two M- 1 C =- 1 D = 1
Above (m> 1)
Five a = 2 b = 2m-1c =1d =1
6 A = 2 B = C = D = 2M- 1
M above is an arbitrary natural number.
The simplest example:
a=b=c=d=2
a=2 b= 1 c= 1 d= 1
a=2 b= 1 c=- 1 d=0
There are only five original questions. When m=2, it is said that many people in China will prove that the original problem is only a very small part.
All the above data are true, and there is not a counterexample. This question is very short, but it implies very rich mathematical ideas ... there are many things to be used. Those theorems and formulas are perfect and can express very common mathematical laws. This is a mathematical problem, not a guess. This topic focuses on cultivating students' independent thinking ability and reverse thinking. ...
Actually, this question is very simple.
I don't know if this is a holistic approach.
The first step in the overall proof of the above situation:
Firstly, a 2 yuan function is constructed, which reveals a secret: all natural numbers divisible by A are converted into natural numbers f(x, y) that are not divisible by A..
Five a = 2 b = 2m-1c =1d =1
Decomposition of natural numbers by mathematical induction and division ... Prove:
(2^(mn)- 1)/(2^n- 1)=e
When m and n are natural numbers, e is odd.
m= 1 A 1=( 1)
m=2 A2=( 1，5)
m=3 A3=( 1，9， 1 1)
m=4 A4=( 1， 17， 19，23)
m=5 A5=( 1，33，35，37，39)
m=6 A6=( 1，65，67，7 1，73，79)
...
...
...
All terms in the general term formula of the combined infinite sequence A () cannot be divisible by the m power of 2-1.
This combination sequence is very simple, just the first item of countless arithmetic progression. ....