[Math] What's next?

Hey guys. I'm back yet again with a different type of math game.

Here's the basics:

I will give you guys a series of numbers. You have to tell me what number/numbers is/are next... and why. Most creative person wins it. You can outdo someone's creativity even if we passed that question a LONG while back. "most creative" is entirely up to me and as such is not up for debate

Example series:

1) 1, 2, 3

Example responses:

1) 1, 2, 3

4, 5, 6; because you just add 1 each time.

1) 1, 2, 3

4, 6; it's numbers that divide into the number 6

The second response wins it. They both quoted the original problem to denote the sequence they're responding to.

A few days later even though we're on a new question....

1) 1, 2, 3

2, 1, 2, 3; it oscillates from 1 to 3

This new poster wins it and takes place of most creative.

Notes:
I'll update the second post with all of the questions and best answers.
The series do not necessarily have to go on forever (but can).
It could just end at the numbers you gave... or keep going from there, that won't make a difference in judging.


You each get ONE shot at responding to each series. No edits please :3

Questions and Answers

1) 1, 3, 5
IAmFear: 4; numbers describe the length of the previous number when written as a word.

2) 0, 6, 3
Stereo0 6 3 5 5 5 5 ...
-a) If the number produces a distinct integer when divided by 2, do that
-b) otherwise, count the number of segments the number represents on a 7 segment display clock, and switch to that number.

3) 2, 4, 8
Yokuyin: 15, 26, 42, 64
Dividing a 3-D cake into as many pieces as possible using n cuts.

4) 2, 6, 10

1, 3, 5, 25, 119, 721

seed number: 1
sequence: x_n = n! + (-1)^n

Also, creativity doesn't necessarily mean difficulty or finding some retardedly difficult formula that it happens to fit in to. There's tons of online ways to get those after all.

Just so that people don't get dissuaded by the more mathematical amongst SP.

also, it doesn't work for n=1, Kalo

You get:

1! + (-1)^1 = 0

1, 3, 5, 7, 11, 13, 17

They're all prime and odd numbers

1,3,5,7,9,11,13

You add by two.

also, it doesn't work for n=1, Kalo

You get:

1! + (-1)^1 = 0

Well, hence the seed number. You have sequences like Fibonacci that don't make sense for the first one(s).

Well, hence the seed number. You have sequences like Fibonacci that don't make sense for the first one(s).

Yeah~

I'm looking for explanations that work for everything I give~

If i'd have just given 3 then 5, sure.

Intruder.... 1 isn't prime.

So far Locked has this one xD


2) 0, 6, 3

1) 1, 3, 5

4

Each number describes the number of letters in the written form of the previous number

ONE - 3 Letters
THREE - 5 Letters
FIVE - 4 Letters

But then, FOUR has 4 letters, so...

Well now, that was sure interesting. IAmFear takes the #1 spot :O

2) 0, 6, 3

0, 3, 3/2, 0, 3/2, 3/4.

Terms 1, 2, 3 are halved to make terms 4, 5 and 6 respectively. Terms 4, 5, 6 are halved to create terms 7, 8, 9 respectively, and so on. Completely sidestepped any sequence making by letting the first term be a set containing the first 3 terms. lol

For a real sequence, try the 4th/5th/6th term = 9, 6, 12 obtained from an add 6/take away 3 sequence. From term 1, add 6 for term 2 then take away 3 for term 3. For term 4, add 6 for a result of 9. Term 5, take away 3 for 6, then add 6 for 12 etc.

2) 0, 6, 3

x_n = 12 - 12/((-2)^n), n = 0, 1, 2, 3...

Basically a converging series.

Next numbers would be 4.5, 3.75, etc.

#2's going to onion knight

Good thinking on using them as blocks of three. Though future use of this wont be as creative of course~!

Also, shidoshi, that -2 should be a positive 2. If you try it with negative 2 it fails every odd number.

3) 2, 4, 8,

2) 0, 6, 3

0 6 3 5 5 5 5 ...
1) If the number produces a distinct integer when divided by 2, do that
2) otherwise, count the number of segments the number represents on a 7 segment display, and switch to that number.

I like these rules because despite being over the set of single digit integers, there are 2 stable solutions - 5 maps to itself, and the sequence 2 1 2 1 repeats forever as well.
0: 6 3 5...
1: 2 1 [...]
2: 1 2 [...]
3: 5...
4: 2 1 [...]
5: 5...
6: 3 5...
7: 3 5... | 4 2 1 [...] - depends how you display sevens
8: 4 2 1 [...]
9: 5...

Out of 10 digits, 5 map to one solution, 5 map to the other (if your seven is like '7). awesome.

0 6 3 5 5 5 5 ...
1) If the number produces a distinct integer when divided by 2, do that
2) otherwise, count the number of segments the number represents on a 7 segment display, and switch to that number.

I like these rules because despite being over the set of single digit integers, there are 2 stable solutions - 5 maps to itself, and the sequence 2 1 2 1 repeats forever as well.
0: 6 3 5...
1: 2 1 [...]
2: 1 2 [...]
3: 5...
4: 2 1 [...]
5: 5...
6: 3 5...
7: 3 5... | 4 2 1 [...] - depends how you display sevens
8: 4 2 1 [...]
9: 5...

Out of 10 digits, 5 map to one solution, 5 map to the other (if your seven is like '7). awesome.

Don't get this part D:

Mind being a bit more.... specific?

Don't get this part D:

Mind being a bit more.... specific?

7 segments display is the typical digital clock representation: like here

o.o

3) 2, 4, 8,

Let's do the simple one first:

16, 32, 64: it doubles every step

Now the difficult one:

15, 26, 42, 64

Imagine having a three-dimensional, let's say cubical, cake. And you want to divide the cake in as many possible parts with only N cuts.

Let's start with the 2 dimensional case: a square pancake.

The first cut (N=1) divides the pancake in 2 parts.
The second cut (N=2) can divide the pancake in 3 or 4 parts, so the maximum number or parts is 4.

More generally: Imagine we have cut the pancake p-1 times, and we have f(p-1) parts. Now we add a pth cutline, to maximize the number of parts, line p should intersect all other p-1 lines, which causes line p to be divided in p segments. All those p segments divides a part by two, so the new number of parts is:

f(p) = f(p-1) + p


Now the three dimensional case

Imagine we have cut the 3D cake q-1 times, and we have g(q-1) parts. Now we add a qth cutplane, to maximize the number of parts, plane q should intersect all other q-1 planes. Now comes the different part: If we look at plane q, we see a 2 dimensional plane with q-1 cutlines (the intersection of two non-parallel planes is a line) with f(q-1) 2D parts: the solution for the 2 dimensional pancake. Each of them act as a dividing wall of their respective 3D parts, so the new number of 3D parts is:

g(q) = g(q-1) + f(q-1)

with g(0) = f(0) = 1, we get

2, 4, 8, 15, 26, 42, 64...