Another prize quiz

Just one question to answer.

Let t_{k+1} = t_k / 2 if t_k is even, t_{k+1} = (3*t_k + 1) / 2 if t_k is odd, for positive integers t_k. Prove or disprove: For every t_0 = 1, 2, ... there exists some k such that t_k = 1.

Prize for first good answer + proof is a fine chess.com trophy :)

42?

Prove it! ;)

On average the increased factor is roughly sqrt(3/2 * 1/2) = sqrt(3)/2 < 1. However, "on average" doesn't prove there could not be one exception of t_0 such that t_k -> infinity. Also there could be other "cycles" besides the cycle 1 -> 2 -> 1 -> 2 -> 1 -> ... which most numbers end with; then the factor also doesn't help.

An interesting co-incidence is that this problem was first posed in the year that my father was born (which was also the year that Boris Spassky and John Horton Conway were born).

Can you be more clear , please ?. I can't understand what have you write and your problem . What means t_k  and all those signs , please ?

Yes, it's a very old problem :) so I can also easily promise to give the first one with the correct solution + proof $1000 ;)

@ FT-physicist: t_k is an infinite sequence t_0, t_1, t_2, ... which is recursively defined by t_(k+1) = t_k / 2 if t_k is even and t_(k+1) = (3*t_k + 1) / 2 if t_k is odd, e.g. t_1 = t_0/2 if t_0 is even, and t_1 = (3t_0 + 1)/2 if t_0 is odd. So for example t_0 = 7 would give the sequence (7, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1, 2, 1, 2, 1, 2, ...) which "ends" in the cycle (1, 2).