**This sentence has two ones, three twos, two threes and one four.**

I thought of that one, then I thought of a smaller one:

**This sentence has two twos.**

Someone is sure to have done this before. Let’s call these *A*-sentences. Here is how I generated the first *A*-sentence:

1

11

21

1112

3112

211213

312213

212223

11421314

41121324

31221324

21322314

21322314

And the process has stabilized! (Get it, each line just describes the one above it. Keep going until you get something repeating.

**The blue sentence has four ones, two twos, two threes, two fours, one five and one six.**

**The red sentence has three ones, four twos, one three, two fours, one five and one six.**

How about this one:

**This sentence has Three ones, two twos, three threes, one four, and one five.**

Of course you can break this into two sentences:

**This sentence has Three ones, three threes, one four, and one five.**

and

**This sentence has two twos.**

The “two twos” part is independent of the rest. This brings up some interesting questions:

**Q1: Is there an infinite set of A-sentences that are pairwise independent of each other? Meaning you could combine any two to get an A-sentence.**

Or even stronger:

**Q2: Is there an infinite set of sentences that are independent as a set? Meaning you could combine any number of them into one big A-sentence.**

Question three below would imply question two:

**Q3: For any whole number N, is there an A-sentence all of whose numbers are bigger than N?**

I don’t even have guess as to the answers to questions one, two or three.

**Q4: Is there an A-sentence containing exactly the numbers one through ten?**

I don’t know the answer to question four either, but at least it should be possible to answer.

This is a linguistic comment, not a mathematical one. In you discussion, clearly “twos” counts as an occurrence of “two”. Now there are people named Twomey. Really. If you stuck “Twomey” in a sentence would you count it as an occurrence of “two”? What about “twofold”?. What about “at work”?

I’ve purposely left this open. Go ahead and create whatever sentences you want, making up your own rules as you go along. On another note, someone told me of the existence of sentences that enumerate the letters used in them “This sentence has so-many a’s, so many b’s, etc.” I haven’t seen any of these, but would love to! The way I’ve done my sentences here there is no dependence on the English, things are kind of “universal,” but go ahead and use whatever rules you wish.

I killed some time in a boring class today generating sentences. I noticed that 1, 2, 3, 4, 11, 12, 13, 14, 33, and 123 all generate the same sentence, 21 32 23 14. 22, of course, is its own sentence. 44 and 124 generate 31 12 33 14. 1n with n greater than 4 yields 31 22 33 14 1n. 125 is different from all of the above. 126 never stabilizes. It goes between 41 22 23 24 15 16 and 41 22 23 24 15 16. I assume 12n does this for all n greater than five since 6 is never a “coefficient”. I’ve only been working on the single-sentence problem, but maybe the two semi-stable 126 sentences can be used in a multi-sentence problem.

I may revisit the process in future boring classes.

Cool. Thanks, Evelyn. I have been meaning to mention that I think it is not hard to convince yourself that the answers to questions 1, 2 and 3 are no. The answer to question 4 is yes.

I worked on 3 today in CMS. Some nomenclature I came up with: in the phrase “two ones”, two is the coefficient and one is the base. My argument for no is that if N is the smallest number that appears, it must appear at least N times. So there are at least N-2 bases greater than N with coefficient N. Let M be the smallest base (other than N) that has coefficient M. So M appears N times. That means that there need to be N-1 bases greater than N with coefficient M, and none of these have coefficient N. So now we’re up to needing 2N-2 bases. I haven’t made it rigorous, and I could definitely be wrong, but I think that if you keep going, you’ll be able to prove that you won’t be able to do this with a finite sentence. If my argument is correct, I think it proves that you can’t even have such a sentence for N=3. The reason N=2 is OK is that N-2=0, so in the second sentence, you’ve got a zero there, and you can stop.

Have you found any A-sentences besides two twos that don’t contain one? I have not even been able to do that.

Oh yeah, I’m still working on 4. My original attempt yielded two sentences:

A: Sentence B has eight ones, two twos, one three, two fours, one five, one six, two sevens, one eight, one nine, and one ten.

B: Sentence A has seven ones, four twos, one three, one four, one five, one six, one seven, two eights, one nine, and one ten.

I haven’t been able to whittle it down to one sentence yet.

No, I haven’t been able to produce any others without ones. I think similar reasoning can also show that that is impossible.

Also, A-sentences can be thought of as having a cycle length of one: they describe themselves (call them A1-sentences). The red and blue sentences above have a cycle length of two (they describe each other, call them A2-sentences). In https://toomai.wordpress.com/2008/10/09/a-couple-more-sentences/ sentences D, C and E together have a cycle length of three (A3-sentences). Based on computer experiments I’m conjecturing that sentences D, C and E are the only A3-sentences (up to swapping the 6s and 7s out for arbitrary integers x and y), and that there are no An-sentences for n greater then 3. The only basis for this conjecture is a random (and really rather small) computer search, so it could very well be wrong.

That’s interesting. I thought I had an A4 set, but I misremembered. If your conjecture is true, I wonder why.

http://en.wikipedia.org/wiki/Look-and-say_sequence

Awesome! Thank you!