Help!!

[no genius]

can you give the next two rows?

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
-------------------
-------------------

[Zurtex]

You may find this website intresting: http://www.research.att.com/~njas/sequences/


:biggrin: hahaha, that's a great sequence, try reading one of them out loud (digit at a time) and compare it with the previous term.

[marcus]

that's neat

here's the next after the two
that Zurtex posted
(if I didnt make a mistake, that is)

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

[marcus]

there is a simple proof that if one continues the sequence it will
never use any other digit besides 1,2, and 3

notice that the rule generating the sequence does not explicitly say
that a 4 or a 5 cannot occur

but in fact a 4 never comes up

you never get a segment that says "....1111...."
or ""....2222..."

so you never need a 4

why?

[Zurtex]

Ignoring the 1st term each number is split into a series of 2 digits, the former explaining how man terms and the latter explaining what the term was

So "1111" is One 1 and One 1, which means the previous term was 11. However 11 will in fact be written out as 21 in the next term. So this means you can have two repeating digits that will get you as high as 2. To get a three it must be composed of a double, such as 11, 22 or 33 and part of another term such as 12, 13 for 11 previous etc..

This of course assumes the original term was in base three and did not have 4 or more repeating digits together.


Sorry for my bad explanation, I'm good with numbers not with description.