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Put up a sequence of numbers, letters, words...whatever. Responders must fill in the next item in the sequence WITH AN EXPLANATION. For example, if I post:

1, 2, 3, 4, 5, __

You could post "6" and say the rule is +1.

However, you could also post "126" and say the rule is:

(n-1)(n-2)(n-3)(n-4)(n-5) + n

[it works -- try it]

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8 - alternate between odd numbers and powers of 2.

CC1, CCLP2, CCLP3, ___

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Nice.

CCLP1 of course!!! Rule is when official levelsets were released chronologically.

[There are other answers. I expect CCLP2DX to be released before CCLP1.]

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1, 2, 4, 8, 16, __

3, 1, 4, __, __

a, e, i, u, __

Μ, A, Γ, N, Υ, __

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1, 2, 4, 8, 16, __

3, 1, 4, __, __

a, e, i, u, __

Μ, A, Γ, N, Υ, __

32 - Powers of 2

1, 5 - Digits of pi

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Good, although both of those have other answers I was looking for.....

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I'm not sure if the 3, 1, 4 one has enough digits to be a real "sequence." You could say that the next number is two (subtract two, add three, repeat), or you could say that it's 1.3333 (divide by three, multiply by four, repeat). Etc.

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The point is that there is more than one answer.

My personal favorite is 3,1,4,2,8

Also 1,2,4,8,16, 31

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How do you get 31 from that sequence?

A, B, C, D, __

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One of my favorite explanations of why the next number can always have more than one answer.

Take a circle. Put a point on it. Now draw all the lines between all the points. (In this case, 0 lines.) Now count how many regions the circle is broken up into (1). Now add a second point, draw all the lines (1) and count the sections (2). Now add a third point, draw all lines between all points that don't already exist (for a total of three lines) and count the sections (4). So you have 1,2,4....

If you repeat this process up to 6, and MAXIMIZE the number of sections (no intersections are of more than two lines), the sequence is 1,2,4,8,16,31,.... I don't know who discovered this, but it's mostly used just to prove this point.

[in this discussion, but "line" I mean "chord or diameter" line segment.]

Also 3,1,4,2,8 are the first five digits of 22/7, a common approximation for pi (derived, I believe from the continued fraction.)

This is where my math-geekiness kills a thread.

A, B, C, D, A -- All Big Cats Dine on Antelope.

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...does it still work for a random string of numbers? I'm guessing there has to be SOME pair of numbers that divides in such a way (rational or otherwise) as to produce the string of numbers, but...

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...does it still work for a random string of numbers? I'm guessing there has to be SOME pair of numbers that divides in such a way (rational or otherwise) as to produce the string of numbers, but...

Yes. I don't want to totally geek out here (too late!), but Klein (I think it was him) basically proved that there is a mathematical formula for ANY finite sequence of numbers. If you think about it, that kind of makes sense. Anyway, Gödel used it in his incompleteness proof, which is one of the greatest things ever that no one knows about.

You can always use a polynomial approximation to get a valid "next number" in a sequence. You just have to make the power high enough to give you enough unknowns and enough formulas. (Or you can let a computer solve the matrix for you.)

OR, and this is incredibly math-geeky, you can use repeated differences to come up with a next number:

1,3,7,24,46,111,.....

2,4,17,18,65

2,13,1,43

11,-12,42

-23,54

77

77,77

-23,54,131

11,-12,42,173

2,13,1,43,216

2,4,17,18,65,281

1,3,7,24,46,111,392

Which, in theory, is a six-degree polynomial. I'm not going to bore you with that derivation.

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I'd be interested in that derivation. This discussion is strangely engrossing. I'm starting to hate myself for it.

C, H _ _

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^C, H, I, P?

(I dunno. I fail.)

Back to one of the previous examples...

Say the sequence was:

1, 2, 3, 4, 5, 126, 127.

By repeated differences (correct me if I'm getting this wrong), we'd get:

1,1,1,1,121,1

0,0,120

0,120

120

120, 120

...ok, I'm lost. There's a reason why I did all I could to avoid math classes in college.

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You have to take the difference of every two numbers, so each list has one less number:

1, 2, 3, 4, 5, 126, 127

1,1,1,1,121,1

0,0,0,120,-120

0,0,120,-240

0,120,-360

120,-480

-600

(then add the last number again)

-600,-600

(now move back up the lists, adding a new number to the end)

120,-480,-1080

0,120,-360,-1440

0,0,120,-240,-1680

0,0,0,120,-120,-1800

1,1,1,1,121,1,-1799

Now, this can seem very counter-intuitive. However, if you use the polynomial method you'll get a similar result. This is a high-order equation which eventually dives very negative. That's probably the only way you get a curve that hits 1 all those times with 121 in the middle.

It's a lot more common to use this method for increasing sequences:

1,3,6,10,15

2,3,4,5

1,1,1

1,1,1,1 (you can go back up once all the numbers are the same)

2,3,4,5,6

1,3,6,10,15,21

I'll make a separate posting on the polynomial method.

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A high order equation? Eh? Is it possible to derive the equation, or do we just "know" that it exists.

Also, is there a theoretical limit on the number of valid answers to any given sequence?

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The polynomial process for solving equations involves creating a set of equations based on the known numbers in the sequence.

For example, if your sequence is

1,4,10,20

You can make the following set of equations:

an3 + bn2 + cn + d = f(n) for all n

a + b + c + d = 1

8a + 4b + 2c + d = 4

27a + 9b + 3c + d = 10

64a + 16b + 4c + d = 20

(You can only get as many unknowns and equations as you have numbers in the sequence.)

Now you solve by subtracting successive equations. I won't show all my work.

7a + 3b + c = 3

19a + 5b + c = 6

37a + 7b + c = 10

12a + 2b = 3

18a + 2b = 4

6a = 1

a = 1/6

b = 1/2

c = 1/3

d = 0

Therefore the next number in the sequence is:

125a + 25b + 5c + d =

125/6 + 25/2 + 5/3 + 0 =

(125 + 75 + 10) / 6 = 210/6 = 35

If you use the difference method:

1,4,10,20

3,6,10

3,4

1

1,1

3,4,5

3,6,10,15

1,4,10,20,35

You can sort of see they use the same numbers.

Now, this gets very messy very quickly, so I recommend using Excel instead.

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an3 + bn2 + cn + d = f(n) for all n

a + b + c + d = 1

8a + 4b + 2c + d = 4

27a + 9b + 3c + d = 10

64a + 16b + 4c + d = 20

...so the coefficients are always the same; only the numbers at the end change? And so I assume that if you had five numbers in the sequence, you would include a variable "e" and include exponents of 5 in the last equation?

Crazy stuff. This thread is proving to be very educational.

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You have the idea. This gets very messy when you get to about 6 unknowns (65=7776). For "nice" sequences that you can figure out in other ways, typically the results are simple. But if you just pick random numbers -- it gets ugly quickly.

A high order equation? Eh? Is it possible to derive the equation, or do we just "know" that it exists.

Also, is there a theoretical limit on the number of valid answers to any given sequence?

Technically, it doesn't "exist" -- it's an approximation. It only gives one of the possible answers.

And no, I don't believe there is any theoretical limit. What's that number for the size of all natural numbers again? Aleph?

677, 776, ___

677, 776, ___

875, 974

Rule: +99

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875, 974

Rule: +99

Dammit, I didn't think of that -- can you think of the one I thought of for this sequence?

/confusing

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667, 776, ____

As stated previous, there are lots of options, ESPECIALLY when you only have two numbers to start with -- it could be almost anything. +99 was just the simplest (from my pov). Anyway, I think I made an error, as 776 - 667 = 109

So, what I should have said was: 667, 776, 885, 994....

And then there's 667, 776, 667, 776 .... (the rule seems obvious)

More fun is 667, 776, 677, 766, 777 .... which is derived from successive pieces of this sequence: 667776677766777

Or, in a really odd way, it could be 667, 776, 946... which would translate on a phone dial pad to "mop pro win"

It's probably something else, though.

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No, you got it Dave --- the repeat of 667 and 776 Here's a long one to make sure there's only ONE rule that can be made here!

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173, _

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