What Number Comes Next?

[GlassPilot]

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ????

[EquisDeXD]

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ????

 

 

 

I thought at first glance that it was Pascal's triangle or a Fibonacci sequence, but it's actually some kind of recursive sequence, so the next number is 31131211131221

 

 

row one, contains "one one", which contain's "two ones", which contains "one two, and one one", and ect.

 

 

1113213211 is three ones, one three, one two, one one, one three, one two, and two ones, which is 31131211131221

 

 

1

11

21

1211

111221

312211

13112221

1113213211

31131211131221

 

Edited October 21, 2012 by EquisDeXD

[GlassPilot]

Nice work! Just don't start with "22".

Edited October 21, 2012 by GlassPilot

[EquisDeXD]

I wonder of there's some mathematical theorem for it, it's not actually a recursive sequence, but it's a system of values created from a single pattern, there must be some kind of equation for it.

[md65536]

I wonder of there's some mathematical theorem for it, it's not actually a recursive sequence, but it's a system of values created from a single pattern, there must be some kind of equation for it.

Here's a puzzle: When would you expect to see a 4?

[GlassPilot]

1350782239[/url]' post='709377']

Here's a puzzle: When would you expect to see a 4?

 

 

 

I believe the answer is: 'never'.

 

Edited October 21, 2012 by GlassPilot

[EquisDeXD]

Here's a puzzle: When would you expect to see a 4?

 

Don't know, I would have to keep testing numbers until I found one, but I don't know if you even could given the circumstance, I don't see when you'd ever ask about 4 when the limit is 3 things, one three, two three, one three, there's no combination that can yield a 4.

Edited October 21, 2012 by EquisDeXD

[ewmon]

A very novel sequence indeed.

 

I ran through a few iterations, and upon right justifying these numbers, I noticed that apparently —

• the last digit is always a 1
  
• the penultimate digit alternates between a 1 and 2
  
• the antepenultimate digit is always a 2
  
• the one previous to that has a repeat sequence of: 1, 2, 2, 3
  
• the one previous to that has a repeat sequence of: 1, 1, 1, 3
  

(After that, it gets to be quite the eye exercise.)

 

Beginning with 2 results in a steady state of 22.

 

Beginning with 3, apparently results in:

• the last digit is always a 3
  
• the penultimate digit is always a 1
  
• the antepenultimate digit is always a 1
  
• the one previous to that is always a 2
  
• the one previous to that is always a 2
  
• so, the numbers always end in ...22113
  
• and, the numbers end alternately as ...123222113 and ... 213322113.
  

Edited October 21, 2012 by ewmon

[md65536]

 

 

I believe the answer is: 'never'.

 

 

Yes, because the first 4 would have to be of the form 4n, with a previous number including "nnnn", but there's no way to get that without splitting up a count of the digit n into two counts.

 

 

Some patterns:

 

1) not including the first, all strings will end in alternating "11" and "21".

 

Rough inductive proof: Suppose a number ends in "21". Then the next string will end in "?211", and the next will end in "?221". (Repeat induction step ad infinitum.)

 

2) Once the string starts with 13n (where n is not 3), it will repeat...

1113...

31m3... (for some possibly varying m)

13p1... for some p not 3.

 

So the N'th number's first 2 and last 2 digits are a simple function of N. There seem to be bigger patterns too, at the starts and ends of the strings. I wonder if it's possible to figure out a function for the entire N'th string of numbers.

 

 

Are there any possible numbers to start a similar sequence that form a "closed loop" and return to the same number? (I see that "22" is already mentioned.)

Edited October 21, 2012 by md65536

[GlassPilot]

1

11

21

1211

111221

312211

13112221

1113213211

31131211131221

13211311123112112211

11131221133112132112212221

31131122212321121113122122113211

132113213211121312211231131122112221131221

1113122113121113123112111311222112132113212221322113112211

31131122211311123113111213211231132132211211131221131211321113222113212221

132113213221133112132113311211131221121321131211132221123113112221131112211312311332211312113211

111312211312111322212321121113122123211231131122211211131221131112311332211213211321322113312221131112132123222113111221131221

311311222113111231133211121312211231131122111213122112132113213221123113112221133112132123222112111312211312111322212311322113311211131211121332211331222113112211

 

[HuMoDz]

"4x" (x = a random number) comes from "xxxx".

"xxxx" means there was a x number of x's, and then a x number of x's again, which means there was a 2x number of x's. (xxxx = x.x + x.x = 2x.x)

Therefore, the next step shoud've been written as (2x)x.

Higher numbers won't show up neither, because they'll be made by "xxxx" + "nx". Since "xxxx" can't happen, higher numbers can't appear neither