If n is even divide n by 2. If n is odd, multiply n by three then add one. Take that answer and repeat the process such that you have a sequence. For any positive even integer n, does the sequence ever not terminate at 1?

My thoughts:

- Any sequence which reaches 1 terminates as it enters the loop 1>4>2>1

[*]Any sequence which contains a number known to terminate at 1, also terminates at one[*]All odd numbers do not need to be checked

[*][*]Take a and b to both be positive odd integers.

[*]a can be written as 2c+1 by the definition of an odd number

[*]b can be written as 2d+1 by the definition of an odd number

[*]a*b = (2c+1)(2d+1) by subsitution

[*](2c+1)(2d+1) = 4cd+2c+2d+1 by distribution

[*]4cd+2c+2d+1 = (4cd+2c+2d)+1 by association

[*](4cd+2c+2d)+1 = 2(2cd+c+d)+1 by distribution

[*]2(2cd+c+d)+1 is an odd number by definition. any integer multiplied by 2 and added to 1 is odd.

[*]Any odd number plus one is even by definition

[*]Therefore, after one iteration any odd number will become even.[*]For any integer a where n is equal to 2^a the sequence will terminate at 1

[*]Since 2^a is even, and can be divided by 2 exactly a times, the product of each iteration becomes 2^a-k where k is the current iteration

[*]When a=k then the product of the iteration will be 2^a-a which is 2^0 which is equal to 1

For any positive integer n; If n is even divide n by 2. If n is odd, multiply n by three then add one. Take that answer and repeat the process such that you have a sequence. For any positive integer n, does the sequence ever not contain a value 2^n?