User:Ralf Stephan

Can the Collatz conjecture be formulated as a covering problem?

The consecutive halving steps reduce any number to its odd part. Let $S_0$ = powers of two (their odd part is 1). Then find all numbers $n$ such that $3n+1 \in S_0$. This is A002450 (a(n) = (4^n - 1)/3). Construct $S_1=A002450 \cup \{all numbers with odd part in A002450\}, this is A181666. Then find all numbers $n$ such that $3n+1 \in S_1$. Rinse and repeat...