An interesting apparent pattern regarding numbers of the form f(n) = 2^n - 1
Hello everyone,
I was recently running some tests on the Collatz's conjecture and noticed a certain pattern.
It works like this: take the set of numbers generated by the function f(n) = 2^n - 1, for n >= 3, and arrange them into pairs as follows: (3, 4), (5, 6), (7, 8) ... and so on. You will then have the values (7, 15), (31, 63), (127, 255), ad infinitum.
Well then, in each of these pairs, the number of steps required for the second value in the pair to reach 1 is always the number of steps for the first value plus one. Let's take the first pair as an example to visualize this:
7 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20-> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 (16 steps)
15 -> 46 -> 23 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 (17 steps)
There is also another case: if, within this sequence, you disregard the steps of dividing by 2 and proceed directly to the odd values, then—taking the same even subset—the number of steps for the first number will equal the number of steps for the second. Using the same example, we have:
7 -> 11 -> 17 -> 13 -> 5 -> 1 (5 steps)
15 -> 23 -> 35 -> 53 -> 5 -> 1 (5 steps)
I've already tested it with pairs of considerably large numbers n, and the pattern remains. I have no idea if there are any results in the literature on this, much less if it works for infinitely many values of n.
Do you know anything about this?