- 14 May '03 18:37Here's some more:

1 ;1

11 ;2

21 ;2

1211 ;4

111221 ;6

312211 ;6

13112221 ;8

1113213211 ;10

31131211131221 ;14

13211311123113112211 ;16

11131221133112132113212221 ;22

3113112221232112111312211312113211 ;30

1321132132111213122112311311222113111221131221 ;42

The sequence on the right is the amount of digits in each row. Can anyone fit a formula to that series? - 14 May '03 19:02Just a guess: each term describes the number of times each digit appears in the previous term, with the exception of the first term, which is defined as 1?

1

11 (one one)

21 (two ones)

1211 (one two, one one)

111221 (one one, one two, two ones)

...

Interestingly, I don't think it is possible to ever get a four to appear here. Does that sound right? - 14 May '03 21:18

Hmm, too easy*Originally posted by royalchicken***Just a guess: each term describes the number of times each digit appears in the previous term, with the exception of the first term, which is defined as 1?**

1

11 (one one)

21 (two ones)

1211 (one two, one one)

111221 (one one, one two, two ones)

...

Interestingly, I don't think it is possible to ever get a four to appear here. Does that sound right?

I don't know about a four coming up, but looking at the other posts, it doesn't seem that it'll appear in a row. - 15 May '03 19:00In fact, note that any three digit sequence in a term becomes a two digit one in the next term, two digits ones become two-digit ones, and one digits ones become one-digit ones. So if you know the relative frequencies of one-, two-, and three-digit blocks you can find an easy asymptotic estimate of the number of digits in the nth term...any guesses here? I'll tell you if you're right...