The hailstone number of each number in the form 2^n-1 when n>1 is, as stated on the OEIS by Alexandre Herrera, "a(2n) = 1 + a(2n-1) for n >= 2" where a(n) is the hailstone number of 2^n-1 OEIS A193688. Simply Put the hailstone numbers of integers one less than a power of 2 come in groups of 2 or ...
My "proof" :) uses the graph-tree approach.
It seems to me that proving these 3 statements are sufficient:
1) the graph is connected.
2) the graph contains all numbers (requires connectivity).
3) the graph is acyclic.
I am realistic, but my logic seems valid to me for all 3 statements.
...
My "proof" :) uses the graph-tree approach.
It seems to me that proving these 3 statements are sufficient:
1) the graph is connected.
2) the graph contains all numbers (requires connectivity).
3) the graph is acyclic.
I am realistic, but my logic seems valid to me for all 3 statements.
...
