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Binary Sequence

  1. Apr 4, 2005 #1
    Not sure if this has been done... I sort of discovered this sequence myself, but who knows...


    What comes next?
  2. jcsd
  3. Apr 4, 2005 #2


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


    On normal basis,it should be "01"
  4. Apr 4, 2005 #3
    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
  5. Apr 4, 2005 #4
    You're both correct as to the terms, but have not found out the pattern...

    I'll post some more terms, though.


    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
  6. Apr 4, 2005 #5
    Is that really 00 at the end and not 01?
  7. Apr 4, 2005 #6
    Yes. That should be a clue.
  8. Apr 4, 2005 #7
    Well, if you chop off the last 0 it's a palindrome.
  9. Apr 4, 2005 #8
    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.
  10. Apr 4, 2005 #9
    The last 0 should totally be a 1.
  11. Apr 4, 2005 #10
    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.
  12. Apr 4, 2005 #11
    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
  13. Apr 4, 2005 #12
    EDIT: Answer removed in order to give other people a chance. Hints still apply.
    Last edited: Apr 5, 2005
  14. Apr 4, 2005 #13
    That's pretty good.
  15. Apr 4, 2005 #14
    Thanks. Glad it was challenging. :biggrin:
  16. Apr 4, 2005 #15
    How about this sequence, then?

  17. Apr 4, 2005 #16
    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.
  18. Apr 4, 2005 #17
    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
  19. Apr 4, 2005 #18
    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:


    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
  20. Apr 4, 2005 #19
    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.
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