Problem 2

Let

m

be a fixed positive integer. The infinite sequence

\{a_n\}_{n\ge1}

is defined in the following way:

a_1

is a positive integer, and for every integer

n\ge1

a_{n+1}=a_n^2+2^m

a_n<2^m

, and

a_{n+1}=a_n/2

a_n\ge2^m

. For each

m

, determine all possible values of

a_1

such that every term of the sequence is an integer.

Step 4 of 7: The landing point forces m=2

Previous step Next step

\text{descending power of two lands at exponent}\ m-1;\qquad m-1=\tfrac m2\ \Longrightarrow\ m=2

Detailed analysis

Once every term is a power of two, say $a_i=2^{k_i}$ , the large rule fires while $k_i\ge m$ and decreases $k_i$ by exactly $1$ each time, so the first exponent below $m$ that is reached is always exactly $k=m-1$ ; this is precisely where the small rule next fires. By step 3, staying at odd part $1$ forever requires every such landing to be the borderline case $2c_i=m$ , i.e. $k_i=m/2$ . Since the landing exponent is always $m-1$ , consistency forces $m-1=m/2$ , i.e. $m=2$ ; for any other $m$ , this landing would produce odd part $b_{i+1}=1+2^{|m-1-m/2|}>1$ , and repeating the argument of step 3 shows the odd part would eventually exceed $2^m$ , contradicting step 1. So $m=2$ is the only possible value.