A simple iterative sequence, beginning with 1:

1

11

21

1211

111221

...

It's easy to generate. Simply write down what numbers you saw in the previous line: the first line has one 1, so the second line is 11. That's two 1's, so we write 21. Then one 2 and one 1, so we write 1211. Repeat to fade.

As you might expect, the number of digits in the sequence increases at a very fast rate. Jongleur points out that the sequence is in fact doubly exponential; the number of digits in the nth term is approximately

(1.3035772690342963912570991121525518907307025046594...)^{n},

so the actual value of the number is double exponential.

This enormous irrational but nonetheless algebraic number in parantheses is the solution to a horendous polynomial of degree 71! The polynomial, should you wish to solve it yourself, is:

x^{71} - x^{69} - 2x^{68} - x^{67} + 2x^{66} + 2x^{65} + x^{64} - x^{63} - x^{62} - x^{61} - x^{60} - x^{59}

+ 2x^{58} + 5x^{57} + 3x^{56} - 2x^{55} - 10x^{54} - 3x^{53} - 2x^{52} + 6x^{51}

+ 6x^{50} + x^{49} + 9x^{48} - 3x^{47} - 7x^{46} - 8x^{45} - 8x^{44} + 10x^{43}

+ 6x^{42} + 8x^{41} - 5x^{40} - 12x^{39} + 7x^{38} - 7x^{37} + 7x^{36} + x^{35}

- 3x^{34} + 10x^{33} + x^{32} - 6x^{31} - 2x^{30} - 10x^{29} - 3x^{28} + 2x^{27}

+ 9x^{26} - 3x^{25} + 14x^{24} - 8x^{23} - 7x^{21} + 9x^{20} + 3x^{19} - 4x^{18}

- 10x^{17} - 7x^{16} + 12x^{15} + 7x^{14} + 2x^{13} - 12x^{12} - 4x^{11} - 2x^{10}

+ 5x^{9} + x^{7} - 7x^{6} + 7x^{5} - 4x^{4} + 12x^{3} - 6x^{2} + 3x - 6 = 0

The series is base dependent, since it's based on the digits as they appear, but I've only dealt with it in base 10. It's fairly simple to write a script to generate as many terms of this as you fancy, but the first few (and yes, I did this by hand as usual, so corrections are welcome) are:

1

11

21

1211

111221

312211

13112221

1113213211

31131211131221

13211311123113112211

11131221133112132113212221

3113112221232112111312211312113211

1321132132111213122112311311222113111221131221

11131221131211131231121113112221121321132132211331222113112211

311311222113111231131112132112311321322112111312211312111322212311322113212221

132113213222133112132113311211131222121321131211132221123113112221131112311332111213211322211
31113211

You can actually see the simple exponential pattern appearing in those columns of digits. Interesting.

The only digits present are always 1, 2 and 3. That's because in order to get a 4, you'd need to have 4 consecutive identical digits in the row above. That can't happen because, for example, 2222 never appears since it's just written as 42.

How does this sequence look in other bases? Anyone who fancies writing the script can let me know.