and I think the actual next line in that song was "chain of fools", (rock me, Aretha) and yet I continue....

If you take a number, square each of its digits and find the sum you will get a new number(usually, is there more than one number that produces itself as its own iterate under this process). Do the same to that number, and the sequence continues. But eventually you have to come back where you started. Numbers with more than three digits will always produce a smaller number and three digit numbers will always be less than 9

^{2}+9

^{2}+9

^{2}= 243, so eventually, wherever you start, you end up with a number less than 243 (and 243 --> 4+16+9 = 29 so it gets even smaller).. what would be the last number that produces a number larger than itself?

Ok, simple stuff, if you start with one, you get one and that's dull so let's go on. If you begin with two you produce the sequence 2--> 4--> 16--> 37--> 58--> 89--> 145-->42-->20-->4 and then repeats the cycle of eight numbers forever. Now the big conjecture... start with ANY integer (oooohhhh, he said you could pick ANY integer, how bold) and eventually, either it gets to one, or it jumps into this chain. Now I wonder; is there a way to prove (short of hacking them all out, which I have done) that there is no "other" chain that some numbers might drop into?

I also have a feeling that as the numbers go off to infinity, the proportion of numbers that go off to one has some non-zero limit; in fact, I suspect it might be around 1/7 or just a tiny bit more, but don't have a clue how to prove that. Any takers? All conjectural rationale will be considered.

I think the same kind of questions could be asked about forming the sum of the cube of the digits... would all roads lead back to one or two then?

A quick answer trying only a few numbers... NO, Follow the orbits of 2 or 3 or 7 and they all go to separate self-replicating numbers or one-cycles.. 4 goes into a triplet of 55 250,133,55; so I guess my new question about cubics is... How many of these finite cycles are there?