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Let $a_0$ be any positive integer,defining:

$$a_{n+1} = \begin{cases}\frac{a_n}{2} &, a_n \text{ even}\\ 3a_n + 3 &, a_n \text{ odd}. \end{cases}$$

then, $a_k=3$ where $k$ is some +ve integer,

For example the sequence$${97,294,147,444,222,111,336,168,84,42,21,66,33,102,51,156,78,39,120,60,30,15,48,24 ,12,6,3}$$

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Did you try $a_0=81$? ;) – Hagen von Eitzen Apr 5 '14 at 18:02
@ Hagen von Eitzen,yes, it does lead to $$3$$ – Tom Lynd Apr 5 '14 at 18:08
up vote 12 down vote accepted

After finitely many steps, whatever the starting value, you have $3\mid a_n$ for all $n \geqslant n_0$. Namely, if $a_0 = 2^{k}\cdot m$ with $m$ odd, then $a_k = m$ and $a_{k+1} = 3(m+1)$, and if $a_n \equiv 0 \pmod{3}$, then $a_{n+1}\equiv 0 \pmod{3}$ too, so $a_n \equiv 0 \pmod{3}$ for all $n > k$. Then look at the sequence $b_n = \frac{a_n}{3}$. We have $b_n \equiv a_n \pmod{2}$ then, and hence for $b_n$ even, we find $b_{n+1} = \frac{1}{3}a_{n+1} = \frac{1}{3}\frac{a_n}{2} = \frac{b_n}{2}$, while for $b_n$ odd, we find $b_{n+1} = \frac{1}{3}a_{n+1} = \frac{1}{3}(3a_n+3) = a_n+1 = 3b_n+1$, i.e.

$$b_{n+1} = \begin{cases}\frac{1}{2}b_n &, b_n \text{ even}\\ 3b_n + 1 &, b_n \text{ odd}. \end{cases}$$

So the question is equivalent to the Collatz conjecture. Nobody has yet proved or disproved it, but in principle, no reason is known why it should not be possible.

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@ Daniel Fischer, how does it become $3b_n+1$ Can you explain it please:( – Tom Lynd Apr 5 '14 at 17:32
I've added the steps. If it isn't clear, ask. – Daniel Fischer Apr 5 '14 at 17:42
@ Daniel Fischer, nicely written, I have accepted your answer.. – Tom Lynd Apr 5 '14 at 17:46

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