Problems & Puzzles: Puzzles

Puzzle 213.  Hailstone Champion Sequences

Hailstone sequences of numbers, {a_i} are these produced in the so-called Collatz problem for a given starting number a₁ (see 1  & 2), applying recursively the following rule:

a_i+1 = 3*a_i +1 if a_i is odd; otherwise a_i+1 = a_i / 2

For each particular initial value a₁ there is only one maximal member, M(a₁) in the corresponding sequence.

For example if a₁=7 , the hailstone sequences is:

7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ....

and M(7)=52

According to the rules of the Collatz problem for generating the hailstone sequences, M(a₁)=a₁ only if a₁=2^k, for k=>2; otherwise, M(a₁)>a₁.

The quotient Q(a₁)=M(a₁)/a₁ is a measure of the height of M(a₁) relative to the initial value a₁ of the corresponding hailstone sequence.

I have produced a table of "champion hailstone sequences", for a₁ >0. In this table we annotate in the first row the triad of values {a_1, M(a₁), Q(a₁) } for a₁=1; then we annotate in the next row, the next earliest a₁ value such that its Q(a₁) is larger than the Q(a₁) of the previous row; and so on...  

Questions:

1. Does Q(a₁) grow beyond any limit?

2. Do you devise any special (non-trivial) property for the a₁ values in the table of champions hailstone sequences?

Solution:

William Lipp solved the question 1: