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 1 of 5: Determinism reduces the problem to boundedness

Previous step Next step

In plain words

The next term depends only on the current one, so a repeated value must repeat forever.

a_{n+1}=\Phi(a_n)

Detailed analysis

Since $a_{n+1}$ is completely determined by $a_n$ alone via a fixed function $\Phi$ (either $\sqrt{a_n}$ or $a_n+3$ ), if two terms ever coincide, say $a_i=a_j$ with $i<j$ , then $a_{i+k}=a_{j+k}$ for every $k\ge 0$ : the sequence becomes exactly periodic with period $j-i$ from index $i$ onward, so $a_i$ recurs infinitely often. Conversely, if some number recurs infinitely often, two of the indices where it occurs give $a_i=a_j$ with $i<j$ , which by the same reasoning forces periodicity, so the sequence takes only finitely many values, i.e. it is bounded. Hence some $A$ occurs infinitely often if and only if the sequence $(a_n)$ is bounded.