What comes next in this binary sequence?

[Moo Of Doom]

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

01000101010001000100010101000101...

What comes next?

 

[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"

 

[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[/color]

 

[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...

 

[BicycleTree]

Is that really 00 at the end and not 01?

 

[Moo Of Doom]

Is that really 00 at the end and not 01?

Yes. That should be a clue.

 

[BicycleTree]

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

 

[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.

 

[BicycleTree]

The last 0 should totally be a 1.

 

[Moo Of Doom]

The last 0 should totally be a 1.

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.

 

[BicycleTree]

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

 

[Moo Of Doom]

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

 

[BicycleTree]

That's pretty good.

 

[Moo Of Doom]

Thanks. Glad it was challenging.

 

[Moo Of Doom]

How about this sequence, then?

01101001100101101001011001101001...

 

[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.[/color]

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

 

[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.

 

[Moo Of Doom]

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

My Answer:

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 :)

 

[BicycleTree]

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 :)

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.