Here are the first few numbers of a sequence. Can you figure out what the next term in the sequence is?

1, 11, 21, 1211, 111221, ?

Most people who see this sequence for the first time aren't able to figure out the next number. However, if you say each digit in each number, you have:

one, one one, two one, one two one one, one one one two one one, ...

The first term consists of one "one," which gives the second term, 11. The second term consists of two "one"s, which gives the third term, 21. The second term consists of one "two" and one "one"; which gives the fourth term, 1211. In general, you can generate the next term in this manner. So, the fifth term concists of three ones, two twos, and one one; put those digits together and you have 312211, the next term in the sequence. This sequence is called the *likeness sequence*, also known by its German name *die Gleichniszahlen-Reihe* or as the *look and say sequence*.

This sequence gets large quite quickly; here are the first 15 terms in the sequence:

1,
11,
21,
1211,
111221,
312211,
13112221,
1113213211,
31131211131221,
13211311123113112211,
11131221133112132113212221,
3113112221232112111312211312113211,
1321132132111213122112311311222113111221131221,
11131221131211131231121113112221121321132132211331222113112211,
311311222113111231131112132112311321322112111312211312111322212311322113212221

Looking through the first 15 terms of the sequence, you might notice that:

- Only numbers 1–3 appear; there are never any fours, for example.
- The sequence "333" never appears.
- There are never more than three "1"s or "2"s in a row.

It's not too difficult to prove that these hold for all terms of the sequence. I'll prove below that "333" can never appear.

The proof can be done by *reductio ad absurdum*. Start by assuming that there is a smallest term that contains "333". There are two possibilities:

- the first "3" appears in an even-numbered position.
- The first "3" appears in an odd-numbered position.

Consider possibility 1. If the first "3" appears in an even-numbered position, some digit, call it `x`, will immediately precede it. So we have ...`x`333... . Working backwards, the previous number in the likeness sequence had `x` copies of "3", followed by 3 copies of "3." However, the rules of the likeness sequence would mean that the number should have been constructed as ...(`x` + 3)3..., so no number with "333" in an even-numbered position cannot be a valid member of the likeness sequence.

Now, consider possibility 2. Working backwards, the previous number in the likeness sequence had 3 copies of 3, followed by 3 copies of something else. However, that means that the previous term contained "333," which contradicts our initial assumption that the term in question is smallest term containing "333". Therefore, there can be no smallest number in the likeness sequence containing "333."

Sources used (see bibliography page for titles corresponding to numbers): 57.