Combinatorics and Number Theory

Form a sequence of positive integers using the following rules. Start with a positive integer \(n>1\). If \(n\) is odd, then the next number is \(3n+1\). If \(n\) is even, then the next number is \(n/2\). Halt if you ever reach\(1\). For example, if we start with \(28\), the sequence is \[\begin{aligned}\end{aligned}\]

Now suppose you start with \(19\). Then the first few terms are \[\begin{aligned}\end{aligned}\] But now we note that the integer \(22\) appears in the first sequence, so the two sequences will agree from this point on. Sequences formed by this rule are called Collatz sequences.

Pick a number somewhere between \(100\) and \(200\) and write down the sequence you get. Regardless of your choice, you will eventually halt with a\(1\). However, is there some positive integer \(n\) (possibly quite large) so that if you start from \(n\), you will never reach\(1\)?

In , we show the integer partitions of\(8\).

There are\(22\) partitions altogether, and as noted, exactly\(6\) of them are partitions of\(8\) into odd parts. Also, exactly\(6\) of them are partitions of\(8\) into distinct parts.

What would be your reaction if we asked you to find the number of integer partitions of \(25892\)? Do you think that the number of partitions of \(25892\) into odd parts equals the number of partitions of \(25892\) into distinct parts? Is there a way to answer this question without actually calculating the number of partitions of each type?