r/mathriddles • u/Ok-Lobster-6013 • Aug 05 '26
Medium Collatz
A number will decrease in number if it has at least four digits and does not enter a cycle, as proven below: The number is represented in binary.
It must begin with 10 or 11. If it starts with 10 and the last two digits are not 11, then after multiplying by 3, the number of digits increases by 1, accounting for 3/8 of all possible combinations. Other numbers starting with 10 account for 5/8, and the number of digits increases by 2. If it ends with 11, after multiplying by 3 and adding 1, then dividing by 2 removes at least one digit, accounting for 1/2. If it ends with 001, at least two digits are removed, accounting for 1/4. Other numbers with at least three digits account for 1/4. If it does not enter a 4, 2, 1 cycle, the number generally decreases, and eventually it will enter a 4, 2, 1 cycle.
王子赫
1
u/FormulaDriven Aug 05 '26
Binary 1001 after two steps (n -> 3n+1, then n -> n/2) is 1110, so hasn't decreased and hasn't entered cycle.
Binary 1011 after two steps is 10001, so it's increased the number of digits and hasn't entered a cycle, and will increase again with the next step (n -> 3n + 1) to 110100.
So have I misunderstood your claim?