# Binary Sequence

1. Apr 4, 2005

### Moo Of Doom

Not sure if this has been done... I sort of discovered this sequence myself, but who knows...

01000101010001000100010101000101...

What comes next?

2. Apr 4, 2005

### dextercioby

I hope this isn't a joke.Anyone could take a # as 197847389748838393384848484949393822636464785885984 and pass it into base 2

Daniel.

On normal basis,it should be "01"

3. Apr 4, 2005

### BicycleTree

I don't know, but here's my answer: Since every 1 is followed by a 0 in the pattern so far, I'm going to guess that the next item in the sequence is a 0

4. Apr 4, 2005

### Moo Of Doom

You're both correct as to the terms, but have not found out the pattern...

I'll post some more terms, though.

0100010101000100010001010100010101000101010001000100010101000100

That should be a big help.

P.S. This is in no way a joke. (And it has only to do with "binary" in the sense of needing two symbols)

EDIT: This includes the original terms.

EDIT 2: It seems to be showing a space in the terms... there shouldn't be one...

Last edited: Apr 4, 2005
5. Apr 4, 2005

### BicycleTree

Is that really 00 at the end and not 01?

6. Apr 4, 2005

### Moo Of Doom

Yes. That should be a clue.

7. Apr 4, 2005

### BicycleTree

Well, if you chop off the last 0 it's a palindrome.

8. Apr 4, 2005

### Moo Of Doom

Here's a hint: Consider the number of terms I revealed the first post and the second post.[In White]

And Bicycle Tree: I totally didn't even notice that. Cool.

9. Apr 4, 2005

### BicycleTree

The last 0 should totally be a 1.

10. Apr 4, 2005

### Moo Of Doom

Hint: That's pretty much the idea behind the sequence.

That and my previous one should probably be enough, but I can keep trying. Tell me if you really want the answer.

11. Apr 4, 2005

### BicycleTree

I don't know, what's the answer? There are some repetitive patterns and they all predict a final 1.

Last edited: Apr 4, 2005
12. Apr 4, 2005

### Moo Of Doom

EDIT: Answer removed in order to give other people a chance. Hints still apply.

Last edited: Apr 5, 2005
13. Apr 4, 2005

### BicycleTree

That's pretty good.

14. Apr 4, 2005

### Moo Of Doom

15. Apr 4, 2005

### Moo Of Doom

01101001100101101001011001101001...

16. Apr 4, 2005

### BicycleTree

Answer:Associate the block 0110 with 0 and 1001 with 1. The sequence starts with 0110 and the n'th block of 4 thereafter is determined by the n'th entry in the pattern. For example, the fourth block of 4 is determined by the 4th entry, namely 0, so it is 0110.

That one was much easier, took only a minute or two.

17. Apr 4, 2005

### BicycleTree

Well, I don't know why that one was so easy and the other one wasn't because I just tried the same idea on the first one and got this: 01 associates with 0, 00 associates with 1, start with 01 and proceed as in the post above, and that generates the original pattern you posted.

Last edited: Apr 4, 2005
18. Apr 4, 2005

### Moo Of Doom

Wow. That works indeed, but is not my original thinking. Good job, BicycleTree!

Instead of changing the last letter of the previous sequence, just write the opposite of the previous sequence by changing all the 0s to 1s and the 1s to 0s:

0->1
01->10
0110->1001
01101001->10010110
etc.

Can you prove your version is equivalent to my version? :)

EDIT: By the way, for both of those sequences (or any like it), associating any group of 2^n terms with the digits of the sequence will result in the same sequence :)

Last edited: Apr 4, 2005
19. Apr 4, 2005

### BicycleTree

Yeah, I figured that. Might be able to use that to prove it, though you'd have to prove that property first. Not trying it though right now.