MathLabs

Problem 2

Let mm be a fixed positive integer. The infinite sequence {an}n≥1\{a_n\}_{n\ge1} is defined in the following way: a1a_1 is a positive integer, and for every integer n≥1n\ge1, an+1=an2+2ma_{n+1}=a_n^2+2^m if an<2ma_n<2^m, and an+1=an/2a_{n+1}=a_n/2 if an≥2ma_n\ge2^m. For each mm, determine all possible values of a1a_1 such that every term of the sequence is an integer.
Step 1 of 7: Bound the odd part
ai=bi2ci, bi odd;bi>2m ⟹ eventually a division of an odd number by 2a_i=b_i2^{c_i},\ b_i\ \text{odd};\qquad b_i>2^m\ \Longrightarrow\ \text{eventually a division of an odd number by }2
Detailed analysis

Write each term as ai=bi2cia_i=b_i2^{c_i} with bib_i odd. If some bi>2mb_i>2^m, then as long as ci>0c_i>0 we have ai≥bi>2ma_i\ge b_i>2^m, so the halving rule applies and cc decreases by 11 while bib_i is unchanged; once cic_i reaches 00, the term equals the odd number bi>2mb_i>2^m, which still triggers halving, but bib_i odd makes bi/2b_i/2 non-integer. So if every term is an integer, bi≤2mb_i\le2^m for all ii.