Problems & Puzzles: Puzzles Puzzle 213. Hailstone Champion Sequences Hailstone sequences of numbers, {a_{i}} are these produced in the socalled Collatz problem for a given starting number a_{1} (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 a_{1} there is only one maximal member, M(a_{1}) in the corresponding sequence. For example if a_{1}=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(a_{1})=a_{1} only if a_{1}=2^k, for k=>2; otherwise, M(a_{1})>a_{1}. The quotient Q(a_{1})=M(a_{1})/a_{1} is a measure of the height of M(a_{1}) relative to the initial value a_{1} of the corresponding hailstone sequence. I have produced a table of "champion hailstone sequences", for a_{1 }>0. In this table we annotate in the first row the triad of values {a_{1, }M(a_{1}), Q(a_{1}) } for a_{1}=1; then we annotate in the next row, the next earliest a_{1 }value such that its Q(a_{1}) is larger than the Q(a_{1}) of the previous row; and so on... _{ }
Questions: 1. Does Q(a_{1}) grow beyond any limit? 2. Do you devise any special (nontrivial) property for the a_{1 }values in the table of champions hailstone sequences? Solution: William Lipp solved the question 1:
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