Problem 1

For each integer

a_0>1

, define the sequence

a_0,a_1,a_2,\ldots

for

n\ge 0

a_{n+1}=\begin{cases}\sqrt{a_n} & \text{if }\sqrt{a_n}\text{ is an integer},\\ a_n+3 & \text{otherwise}.\end{cases}

Determine all values of

a_0

for which there exists a number

A

such that

a_n=A

for infinitely many values of

n

Step 4 of 5: If

3\mid a_0

, the sequence is bounded

Previous step Next step

a_n\ge 12,\ 3\mid a_n\implies a_{n'}<a_n

Detailed analysis

By the previous invariant every term is a multiple of $3$ . A perfect square that is a multiple of $3$ must be a multiple of $9$ (again since $3$ is prime). So while a term $a_n\ge 12$ is not yet a square, following $a_n,a_n+3,a_n+6,\ldots$ reaches its first square multiple of $9$ , namely $9t^2$ for the least $t$ with $3t\ge\sqrt{a_n}$ ; the next term is then $3t\le\sqrt{a_n}+3<a_n$ . Iterating this descent, the sequence must eventually drop below $12$ , landing on one of $3,6,9$ , and from there it cycles forever: $3\to 6\to 9\to 3\to\cdots$ . So the whole sequence is bounded.