General Discussion => Other Discussions => Math and Science => Topic started by: Munchor on April 08, 2011, 12:21:06 pm
Title: Funny sequence
Post by: Munchor on April 08, 2011, 12:21:06 pm
Even if most of you have seen this sequence before, here it is, good luck deciphring it and getting the next element (it's infinite)
1 11 21 1211 111221 312211 12112221 1112213211
:D
Title: Re: Funny sequence
Post by: turiqwalrus on April 08, 2011, 03:15:47 pm
Spoiler For Spoiler:
one one '1' 2 1's 1 '2', 1 '1' ...
Title: Re: Funny sequence
Post by: Munchor on April 08, 2011, 03:19:41 pm
Spoil it now. So others can't see.
Title: Re: Funny sequence
Post by: Binder News on April 08, 2011, 03:23:54 pm
Oh yeah, I think I've seen this before. does it have a name?
Title: Re: Funny sequence
Post by: phenomist on April 08, 2011, 06:57:35 pm
Some exploratory questions to ask (sorted by difficulty?):
1) Prove that every digit in numbers of this series are either "1", "2", or "3". 2) Prove that for every term, the last digit is always 1. 3) Prove that you cannot have 3 3's in a row. 4) Does the length of the terms ever decrease? Find the first instance if it does. 5) Asymptotically speaking, what is the length of each term of the sequence? 6) As the terms grow large, what will the distributions for each digit be?
The first three are definitely doable.
Title: Re: Funny sequence
Post by: PeonHero on April 25, 2011, 09:32:15 pm
Let me try this:
1) Prove that every digit in numbers of this series are either "1", "2", or "3".
It's impossible that there will be anything above a 3. Why? The numbers are always grouped together in a way that makes it only possibly to get up to 3 of a number.
2) Prove that for every term, the last digit is always 1.
It will always be 1 because the initial number was 1. If the initial number was 2, it will always be two, same with any other number. The first two rows are the most important, since N is defined in the first row, the second row will always be 1N. And after that, you can't get rid of it.
3) Prove that you cannot have 3 3's in a row.
Because the maximum you can have is 2 3's in a row. 33111 will be 2331, 22233111 will be 322331, you need 3 3's to make 333 possible in the first place, but since you will never encounter 3 3's, you won't encounter 3 3's.
4) Does the length of the terms ever decrease? Find the first instance if it does.
No. I've been doing this for 15 minutes on a calculator, it's not possible.
5) Asymptotically speaking, what is the length of each term of the sequence?
I don't know what this question is asking for. It looks like an exponential increase in number of terms in the sequence per row.
6) As the terms grow large, what will the distributions for each digit be?
The terms won't really "Grow Large," since it's all just 1s,2s and 3s. The last digit will always be a 1. And the first digit alternates between 1 and 3 in the later stages. The digits in the center are mostly made up of 1's, with a few 2's and 3's in between. There are always more 1's, more 2's, more 3's, in the next row than the previous one. Etc.