## Count up with Zeckendorf's theorem!

For all your silly time-killing forum games.

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Sean Quixote
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### Count up with Zeckendorf's theorem!

Wikipedia wrote:Zeckendorf's theorem states that every positive integer can be represented uniquely as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers.

1 = 1
2 = 10
3 = 100
4 = 101
5 = 1000
6 = ...?

Just for reference, and those unfamiliar with the Fibonacci sequence who want to play, the first few Fibonacci numbers are:

1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711

That should do us just fine, for now...
Last edited by Sean Quixote on Wed Sep 28, 2011 8:29 pm UTC, edited 1 time in total.

curtis95112
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### Re: Count up with Zeckendorf's theorem!

Since you posted up to 5, I'll start with 6.

6 = 111
Mighty Jalapeno wrote:
Tyndmyr wrote:
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### Re: Count up with Zeckendorf's theorem!

7 = 1010, if I'm understanding this correctly.

a-wan
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### Re: Count up with Zeckendorf's theorem!

Looks like 6 should have been 1001 since the Fibonacci numbers cannot be consecutive (i.e. your number should never have 2 ones next to each other).

8 = 10000

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

9 = 10001

That's right, a-wan. Basically the way I see it, what's going to go on here is we're gonna pretend like we're translating Zeckendorf's representation into a base system that resembles binary, in that it will contain only ones and zeros. The Fibonacci numbers will be our "orders of magnitude" or place values: from the rightmost digit, just go in your head, "1, 2, 3, 5, 8, 13, 21, 34, 55, etc..."

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### Re: Count up with Zeckendorf's theorem!

10 = 10010

This doesn't look like a very efficient number system for small numbers. I'm sure it gets better for larger ones.

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

11 = 10100

Yeah, not really. Never even though of that before, but yeah, not really. I guess, every time we reach another Fibonacci number, the efficiency ratio will improve by approximately a factor of phi?

a-wan
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### Re: Count up with Zeckendorf's theorem!

12 = 10101

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

13 = 100000

a-wan
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### Re: Count up with Zeckendorf's theorem!

14 = 100001

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

15 = 100010

a-wan
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### Re: Count up with Zeckendorf's theorem!

16 = 100100

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

17 = 100101

Ah, 17... Anyone wanna take a gander as to what it has in common with 72, 305, 1292, 5473, et cetera?

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### Re: Count up with Zeckendorf's theorem!

18 = 101000

At a guess and a look at a couple of them, it looks like they're all 100...101

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

19 = 101001

Eh.. maybe. I dunno actually, because the answer I was looking for technically has little if anything to do with the Zeckendorf's representation. It has more to do with another thing that I came up with, but I never was sure what I should call it...

a-wan
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### Re: Count up with Zeckendorf's theorem!

20 = 101010

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

21 = 1000000

Sean Quixote wrote:Ah, 17... Anyone wanna take a gander as to what it has in common with 72, 305, 1292, 5473, et cetera?

If you don't want a spoiler alert: You might not want to read my thread over in the math forum.

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### Re: Count up with Zeckendorf's theorem!

22 = 1000001

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

23 = 1000010

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### Re: Count up with Zeckendorf's theorem!

24 = 1000100

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

25 = 1000101

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### Re: Count up with Zeckendorf's theorem!

26 = 1001000

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

27 = 1001001

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### Re: Count up with Zeckendorf's theorem!

28 = 1001010

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

29 = 1010000

I've never actually written this stuff out before (and now I guess I shouldn't have to ) so I just realized another thing that's going on here: if someone came along one day and said, "I want to create a base system that only has two symbols (1 and 0), but let's say that numbers can only be written in such a way that the 1s never touch eachother..." This is also what you would come up with.

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### Re: Count up with Zeckendorf's theorem!

30 = 1010001

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

31 = 1010010

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### Re: Count up with Zeckendorf's theorem!

32 = 1010100

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

33 = 1010101

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### Re: Count up with Zeckendorf's theorem!

34 = 10000000

a-wan
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### Re: Count up with Zeckendorf's theorem!

35 = 10000001

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

36 = 10000010

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### Re: Count up with Zeckendorf's theorem!

37 = 10000100

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

38 = 10000101

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### Re: Count up with Zeckendorf's theorem!

39 = 10001000

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

40 = 10001001

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### Re: Count up with Zeckendorf's theorem!

41 = 10001010

Sean Quixote
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### Re: Count up with Zeckendorf's theorem!

42 = 10010000

gaga654
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### Re: Count up with Zeckendorf's theorem!

43 = 10010001

Sean Quixote
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44 = 10010010