The 3n+1 Conjecture
The conjecture is an unsolved conjecture in mathematics. It is named after Lothar Collatz, who first proposed it in 1937. It is also known as the Collatz conjecture, as the Ulam conjecture (after Stanislaw Ulam), or as the Syracuse problem.
The 3n+1 operation
Consider the following operation on positive integers n.
- If n is even, then divide it by 2.
- If n is odd, then multiply it by 3 and add 1.
For example, if we apply this transformation to 6, then we get 3 since 6 is even; and if we apply this operation to 11, then we get 34 since 11 is odd.
Exercise: Write a function that implements this operation, and compute the images of 1, 2, ..., 100.
Statement of the conjecture
If we start with n=6 and apply this operation, then we get 3. If we now apply this operation to 3, then we get 10. Applying the operation to 10 outputs 5. Continuing in this way, we get a sequence of integers. For example, starting with n=6, we get the sequence
6, 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, ....
Notice that this sequence has entered the loop . The conjecture is
3n+1 conjecture: For every n, the resulting sequence will always reach the number 1.
Exercise: Write a function that takes a positive integer and returns the sequence until it reaches 1. For example, for 6, your function will return [6, 3, 10, 5, 16, 8, 4, 2, 1]. Find the largest values in the sequences for 1, 3, 6, 9, 16, 27.
(Hint: You might find a while loop helpful here. Below is a very simple example that repeatedly adds 2 to the variable x until x is no longer less than 7.)
x = 0
while x < 7:
x = x + 2
print x
Exercise: Use the line command to plot the sequence for 27.
Exercise: Write an @interact function that takes an integer n and plots the sequence for n.
Stopping Time
The number of steps it takes for a sequence to reach 1 is the stopping time. For example, the stopping time of 1 is 0 and the stopping time of 6 is 8.
Exercise: Write a function that returns the stopping time of a poisitve integer n. Plot the stopping times for 1, 2, ..., 100 in a bar chart.
Exercise: Find the number less than 1000 with the largest stopping time. What is its stopping time? Repeat this for 2000, 3000, ..., 10000.
Extension to Complex Numbers
Exercise: If is odd, then is even. So we can instead consider the operation that maps to , if is even; and to , if is odd.
.
Construct as a symbolic function and use Sage to show that for all , where is the -operator. Afterwards, argue that is a smooth extension of to the complex plane (you have to argue that applying to a positive integer has the same effect as applying to that integer. You don't need Sage to do this, but it might offer you some insight!)
Exercise: Let be the complex function:
.
Construct as a symbolic function, and show that and are equal.
Hint: One way of doing this is to use a combination of .trig_expand(), .trig_reduce(), .trig_simplify().
Exercise: Use the complex_plot command to plot g in the domain and .
Exercise: Consider the composition (where there are copies of in this composition). Use complex_plot and graphics_array to plot , , , ..., on the domain and .
(Hint: To speed things up or control the percision of the computations, you may want to replace pi in your equation with CDF.pi(). Type CDF? and CDF.pi? for more information.)
Exercise: Generate some really nice images of that illustrate the fractal-like behaviour of . (Hint: You may want to explore the plot_points and interpolation options for the complex_plot command.)