Next in the sequence

I saw this on another site, pretty good one...

What row of numbers comes next?
1
11
21
1211
111221
312211
13112221
..........

Originally posted by Crowley
I saw this on another site, pretty good one...

What row of numbers comes next?
1
11
21
1211
111221
312211
13112221
..........

1113213211
...but I won't say why.

Originally posted by godzillion
1113213211
...but I won't say why.

and the row after that is
31131211131221

seems this is a well known one (1?)

Originally posted by Toe
and the row after that is
31131211131221

seems this is a well known one (1?)

... followed by 13211311123113112211 🙄🙄🙄

Mick 🙂

Here'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?

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?

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?

Hmm, too easy 😉
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.

Thanks...actually, think about what happens when 3 ones show up in a sequence and you can tell why there can't be a four.

In 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...