# Need help

1. Jan 22, 2008

### cc123

who can fill the brace:
-3,0,26,252,()
thanks a lot!

2. Jan 22, 2008

### HallsofIvy

?? In what way? is there any reason that brace could not contain any integer? What are the conditions you are not telling us about?

3. Jan 22, 2008

### rohanprabhu

most probably it's some pointless IQ test where we go on filling in patterns and numbers and all that stuff..

4. Jan 22, 2008

### CRGreathouse

Whew. I get
$$\sigma(4n\cdot2^{2^{2^{3-1}}})-4=2,097,144$$
There's no rounding needed since it's even.

5. Jan 22, 2008

### rohanprabhu

d00d seriously.. what did u just do??

6. Jan 22, 2008

### HallsofIvy

Oh, well, of course!

7. Jan 22, 2008

### dodo

Can you change the first to -6, instead of -3?

8. Jan 23, 2008

### cc123

in any way i think.there are many answers if you can tell the reason.
to Dodo, if u change the first to -6,what is your answer?thank u!

9. Jan 23, 2008

### dodo

With a -6... is it ambiguous. It would be the sequence
$$2^2-10, \ \ \ \ \ \ 2^3-8, \ \ \ \ \ \ 2^5-6, \ \ \ \ \ \ 2^8-4$$​
so the next is $$2^n-2$$, and n depends on how you interpret the sequence of exponents: if it is Fibonacci, then n=11 for a value of 2046, if it is a quadratic, n=12 and you get 4094.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook