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geodave

Sequences

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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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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, __

 

32 - Powers of 2

 

1, 5 - Digits of pi

 

Not sure about the others.

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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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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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^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?

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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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Actually, mine was an alternate fibonnaci sequence starting with 0 and 2, but it amounts to the same thing.

 

10, 10, 10, 15, 10, 10, 15, 10, 40, __

 

That's tricky. Looks sort of like a tennis score.

 

Let's try the difference method

 

0,0,5,-5,0,5,-5,30

0,5,-10,5,5,-10,35

5,-15,15,0,-15,45

-20,30,-15,15,60

50,-45,30,45

-95,75,15

180,-60

-240

-240,-240

180,-60,-300

-95,75,15, -285

50,-45,30,45,-240

-20,30,-15,15,60,-180

5,-15,15,0,-15,45,-135

0,5,-10,5,5,-10,35,-100

0,0,5,-5,0,5,-5,30,-70

 

which would make the next number -30. I'm guess that's NOT what you were looking for.

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Didn't we just do fibonacci?

 

I cannot spell fibbonaci.

 

you'll never get this one (unless you google it):

 

3,7,15,1,292,1,1,1,2,1,3......

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10, 10, 10, 15, 10, 10, 15, 10, 40, __

 

That's tricky. Looks sort of like a tennis score.

 

Here are the next few numbers in the sequence:-

... 8, 30, 40, 0, 25, 25, 0, 10, __

 

By 0 actually I mean infinity (or to be more precise, 10% of infinity).

The sequence itself is finite, though.

 

- Madhav.

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10% of infinity? It's been a long time since Number Theory class, but doesn't 10% of infinity = infinity?

 

I still have no idea. Any ideas on mine?

 

Let's try this one:

 

0,2,16,54,128,250,432,686,1024,....

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(x^3)*2

 

Woah, did I actually figure one of these out? :blink:

 

Yes you did!

 

Let's have a little more fun.....

 

70,60,45,25,0...

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you'll never get this one...

3,7,15,1,292,1,1,1,2,1,3......

I took your word for it and didn't even try. :P The 292 just looks evil.

 

10, 10, 10, 15, 10, 10, 15, 10, 40, __

... 8, 30, 40, 0, 25, 25, 0, 10, __

It's been a long time since Number Theory class...

If you've learnt your Lessons well, the first eight should be a cakewalk. ;)

 

- Madhav.

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I'm still lost.

 

The 292 is pretty wild. The sequence is the continued fraction expansion of pi (I guess pieguy didn't see it.)

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Here's one that has nothing to do with math in a way.

 

82, 137, ___

 

But I bet SOMEONE will get this.

 

146 (the three levels referenced in CC1 #34)

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146 (the three levels referenced in CC1 #34)

 

That was too easy...

 

50, 9, 9, 9, ___

 

And this isn't random...

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What I posted in the "The person below me" thread:

False. I don't have any.

 

The person below me knows what should replace the question mark in "fadhatpbmkwsrt?mi"

 

EDIT: fidhatpbmkwsrt?mi

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here's one for you, since I can't find either of you guys' sequences...

 

3, 3, 5, 4, 4, 3, 5, 5, 4, __

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12, 24, 36, 48, 510, ...

612, 714, 816 ... the number is divided into two parts, and you add 1 to the first part and 2 to the second part.

 

-----

 

33, 61, 115, 130, ...

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Here it comes then:

There's two sequences here, kind of. You start out with 3, 5, 4, 3, 5, 4, 3, subtracting one from the previous number. However, this is an alternate universe where 3 somehow wraps back around to 5, so that 3 - 1 = 5. You then duplicate every other number, which gets you 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 3.

 

33, 61, 115, 130, ...

The only pattern I possibly see is that the last two digits of 115 added to it get 130... maybe that is hinting at some part of something more complex which explains the rest.

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33, 61, 115, 130, ...

The only pattern I possibly see is that the last two digits of 115 added to it get 130... maybe that is hinting at some part of something more complex which explains the rest.

 

No, it's not mathematical, sorry.

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It's a sequence of all really long and tedious levels.

 

Hmm, I guess they are, but that's not what I was looking for.

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I analysed both the passwords and the chip counters, nothing there, could be related to their respective T-bold times though but I haven't checked that up.

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Alright here is a completely random one that I don't know if anyone will get right, because the idea is dumb.

 

All I will say - it has something to do with grid coordinates of CC1 (you'll want an editor)

 

3, 10, 16, 22, ___

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All the unanswered sequences so far (I'll try to keep this list updated as much as possible):
 

a, e, i, u, __

Μ, A, Γ, N, Υ, __

 

50, 9, 9, 9, ___

 

3, 10, 16, 22, ___

 

:facepalm:  :star:  1star

 

610, 570, 510, 480, 440, 400, 140, ?

 

1, 6, 13, 22, 25, 30, 51, 61, 67, 80, 94, 136, __, __

Edited by random 8

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fidhatpbmkwsrt?mi: q

 

Finally! (Y)

 

0,0,4,34,240,1680,11764,... - Need a hint? I'll give you seven. (and maybe a calculator)

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Here's one (shouldn't be hard):

 

5, 15, 11-15, 31-15, 13-21-15...

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Here's one (shouldn't be hard):

 

5, 15, 11-15, 31-15, 13-21-15...

 

11-13-12-21-15

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5, 15, 11-15, 31-15, 13-21-15...

 

Perhaps you could give a slight clue or context? I'm completely lost on that one.

 

0,0,4,34,240,1680,11764,... - Need a hint? I'll give you seven. (and maybe a calculator)

 

I thought that last hint was pretty powerful. Oh well....

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Incorrect. :)

 

I was going to say the exact same thing as Tyler. Kinda shocked it's incorrect...O_o

 

In any case: 610, 570, 510, 480, 440, 400, 140, ?

 

Edit:

 

33, 61, 115, 130, ...

 

134 (, 147, 148)

Edited by James

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Here's one (shouldn't be hard):

 

5, 15, 11-15, 31-15, 13-21-15...

 

Now i see where Tyler was going...

 

31-13-12-15?

 

-----

 

For this next one, the answer is probably already in front of you:

 

17, 1, 26, 23, 19, 24, 5, 4, 3, 18, 6, 22, ?

Edited by random 8

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For this next one, the answer is probably already in front of you:

 

17, 1, 26, 23, 19, 24, 5, 4, 3, 18, 6, 22, ?

 

Seems that this is what it is:

https://oeis.org/A124517

So the next term is 20.

(Yeah, I guess that's probably cheating.)

 

Here's one:

 

1, 6, 13, 22, 25, 30, 51, 61, 67, 80, 94, 136, __, __

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Correct.
 
Weird...i tried searching the sequence on Google and that site didn't show up.
 
EDIT:

EXTENSIONS

Corrected by Eric M. Schmidt, May 19 2013


lol

Edited by random 8

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