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Analyze a_n as a power of 2?
First Sequence: 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, ...
Second Sequence: 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, ...
Conclusion: See solution for the explanation.
Let a_1=34. Since a_1=34 is even, the second term is a_2= 12a_1= 12(34)=17. Since a_2=17 is odd, the third term is a_3=3a_2+1=3(17)+1=52. We will continue to do the same until we come across a pattern.
n | a_(n-1) | Is a_(n-1) even? | a_n |
---|---|---|---|
1 | - | - | a_1=34 |
2 | a_1=34 | Yes âś“ | a_2=1/2(34)=17 |
3 | a_2=17 | No * | a_3=3(17)+1=52 |
4 | a_3=52 | Yes âś“ | a_4=1/2(52)=26 |
5 | a_4=26 | Yes âś“ | a_5=1/2(26)=13 |
6 | a_5=13 | No * | a_6=3(13)+1=40 |
7 | a_6=40 | Yes âś“ | a_7=1/2(40)=20 |
8 | a_7=20 | Yes âś“ | a_8=1/2(20)=10 |
9 | a_8=10 | Yes âś“ | a_9=1/2(10)=5 |
10 | a_9=5 | No * | a_(10)=3(5)+1=16 |
11 | a_(10)=16 | Yes âś“ | a_(11)=1/2(16)=8 |
12 | a_(11)=8 | Yes âś“ | a_(12)=1/2(8)=4 |
13 | a_(12)=4 | Yes âś“ | a_(13)=1/2(4)=2 |
14 | a_(13)=2 | Yes âś“ | a_(14)=1/2(2)=1 |
15 | a_(14)=1 | No * | a_(15)=3(1)+1=4 |
16 | a_(15)=4 | Yes âś“ | a_(16)=1/2(4)=2 |
17 | a_(16)=2 | Yes âś“ | a_(17)=1/2(2)=1 |
We got the term 1 again. This tells us that the sequence 4, 2, 1 will repeat indefinitely.
We will continue to do the same method as the first sequence.
n | a_(n-1) | Is a_(n-1) even? | a_n |
---|---|---|---|
1 | - | - | a_1=25 |
2 | a_1=25 | No * | a_2=3(25)+1=76 |
3 | a_2=76 | Yes âś“ | a_3=1/2(76)=38 |
4 | a_3=38 | Yes âś“ | a_4=1/2(38)=19 |
5 | a_4=19 | No * | a_5=3(19)+1=58 |
6 | a_5=58 | Yes âś“ | a_6=1/2(58)=29 |
7 | a_6=29 | No * | a_7=3(29)+1=88 |
8 | a_7=88 | Yes âś“ | a_8=1/2(88)=44 |
9 | a_8=44 | Yes âś“ | a_9=1/2(44)=22 |
10 | a_9=22 | Yes âś“ | a_(10)=1/2(22)=11 |
11 | a_(10)=11 | No * | a_(11)=3(11)+1=34 |
Notice that we get a_(11)= 34. Therefore, after this term we will get the same sequence as in the first sequence, where a_1= 34.
Notice that that each time an even number is reached, it is divided until it reaches an odd number. Then, it grows again, but it appears that at some point we reach the number 4, from which the sequence has the subsequence 4, 2, 1, that repeats indefinitely.
The Collatz conjecture is a mathematical conjecture related to the sequence in this exercise. The conjecture is that no matter what value of a_1 we start with, the given sequence will always reach the subsequence 4, 2, 1, which repeats indefinitely.
Mathematicians may not be ready for such problems.
This is an extraordinarily difficult problem, completely out of reach of present day mathematics.
Even today, mathematicians do not know if the Collatz conjecture is true or false.