Maths extravaganza

You call me nerd?? Anyways I have an idea for yez another silly Space Junk game. Create a number sequence and the next person has to come out with a formula matching the outcome. Then he creates number sequence of his own and so on...
Example:
8, 15, 22, 29, 36, 43... ===> 7n+1
For more complicated formulas you might want to extend the sequence a littlebit to avoid getting too simple formulas. Always give answers for n=1, 2, 3, 4, 5... (basically positive integers)

Ill start out easy:
5.141, 7.141, 9.141, 11.141, 13.141, etc......



edit: You can also include questions like fill in the missing numbers in the sequence: 1, 2, 3, ..., ..., 6, 7 - If you want

 

Up by 2.0?

Here's another easy one

0 1 1 2 3 .... 8 13 21 .... 55 89 144 233 377 .....

 

Actually it was 2n+3.141 (or pi, if you will...). Up by 2.0 is no formula

Oh and they are Fibonacci numbers... couldnt remember the damn name xD You find N by adding the 2 Ns before it. Gimme a second...

Find formula for this one:
1, 8, 27, 64, 125, 216, 343, 512, 729

 

Fiiiine. n+2 sure as hell is a formula.

 

i hate maths >.>

 

TheWorker thats N^3

How about:

0, 42, 336, 2394, 16800, 117642...

 

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

continue

 

11131221133112132113212221
3113112221232112111312211312113211

Not sure if second line is right >>

 

1 2 3 4 5 6 ? 7 ? ? 10....

 

@ LordKerwyn: 7*(previous)+42
(or, more strictly, sum(k=0, n-1)(7^k * 42) for n > 0, and 0 for n == 0)
next is 823536.

@ Fenix: Is that first '?' supposed to be there? If it is, I don't know. Else, the second two are 8, 9 and your formula is n.

Mine:
1, 11, 104, 1009, 10016, 100025, 1000036,...

I like these pattern problems.

 

123,1454, 3543,4654,5678,6789,7909,8983,98234,31234, 2346456,8793820985, 198734568746, 295474778289, 988876648396197878938, 678765456434567687786756435677545686

I don't expect you to figure this out

 

Sodium is it (n^2)+(10^n) beggining with n=0 ?

Nice answer to my solution Sodium. I actually intended it to be (7^n)-7 but yours definetly works as well.

1, 7, 56, 448, 3584, 28672, 229376, 1835008...

 

1000000tan(180/1000000) = ?

answer in 2 letters

 

F U?


I get the feeling that this is 11's and LK's new favorite thread

 

Im a general nerd and a mathmatical freak of nature so I like threads like this that call on both of those aspects...

 

Oh god... I think that I am on the right track to solve this but my 1st year of highschool cripples me.

I keep getting - 1, 7, 8, 49, 56, 64, 343, 392, 448, 512, 2401, 2744, 3136, 3584, 4096, 16807, 19208, 21952, 25088, 28672, 32768, 117649, 134456, 153664, 175616, 200704, 229376, 262144, 823543, 941192, 1075648, 1229312, 1404928, 1605632, 1835008, 2097152.
I can see the gaps getting wider by 1 each time and that the number trend to bundle into little groups which also grow... some kind of triangle or what... but thats about it.

I hope that it at least helped somebody else.

 

I hope superscript is made available soon and that you drop the habit of using ^ instead of that, because it makes formulas harder to read.

I'm more of an observer here because I don't really have the brains to solve these.

 

Well, I haven't figured Wick742's out... yet...

LordKerwyn's is (I can best describe it in words) start with 1, multiply by 7, then for all subsequent steps multiply by 8. That would make the next term 14680064... unless I missed something.

ursawarrior16's isn't really a sequence of numbers, and I think he wants the answer pi. Unfortunately, if you calculate it (using degree mode), you get
3.14159265360012313836... and not
3.14159265358979323846... While they are close, they are not the same. lim(x->inf)(x*tan(180/x)) = pi.

Now, mine:
3, 7, 13, 19, 29, 37, 43, 53, 61, 71,...

 

Nice sodium the next number would be 79 correct?

Also, there is an actual N equation that governs my set its not just numbers randomly multiplied together, but your next number is correct.

 

Okay, I came up with a formula just absurd enough to work for that one, LordKerwyn (sorry about spelling your username incorrectly before!) Get ready...

7^[1 - ((sin(π*n))^2 + (sin(π*n))/(π*n))] * 8^[ |n-1| - (sin(π*n))/(π*n)]

I am sure there is a more reasonable solution, but once I got this idea I couldn't help myself.