For all you math puzzle nuts... - Page 2

Homer · #11 10-04-2005, 05:28 PM

Damn, I hate you Bruce. [img]/images/graemlins/smile.gif[/img]

I was getting close, but couldn't figure out how to turn the odd n terms to even without changing the even n terms. Basically, this part is where I was stuck -> [1 - (-1)^n]/2

-- Homer

cookie · #12 10-04-2005, 05:28 PM

why isnt it like:
N: 1 2 3 4 5 6 7 8 9
...0 1 1 2 2 3 3 4 4

???

TBag · #13 10-04-2005, 05:36 PM

[ QUOTE ]
why isnt it like:
N: 1 2 3 4 5 6 7 8 9
...0 1 1 2 2 3 3 4 4

???

[/ QUOTE ]

The book doesn't specify, but I'm pretty sure it's supposed to be like that.

My answer is in white below:
<font color="white"> n/2 - [1 - (-1)^n ]/4
n >= 1 </font>

BruceZ · #14 10-04-2005, 05:48 PM

[ QUOTE ]
[ QUOTE ]
why isnt it like:
N: 1 2 3 4 5 6 7 8 9
...0 1 1 2 2 3 3 4 4

???

[/ QUOTE ]

The book doesn't specify, but I'm pretty sure it's supposed to be like that.

My answer is in white below:
<font color="white"> n/2 - [1 - (-1)^n ]/4
n >= 1 </font>

[/ QUOTE ]

That's the same as my second solution if you break mine into two terms.

It's arbitrary whether you start counting from 0 or 1. Math people usually start from 1, while computer science people index arrays from 0.

Cosimo · #15 10-04-2005, 06:08 PM

grunching.

-1^n : 1 -1 1 -1 1 -1
2n 0 2 4 6 8 10
+ 1 1 5 5 9 9

so,

(2n+(-1^n)-1)/4

produces 0 1 1 2 2 ... etc for n=1..infinity

#16 10-04-2005, 06:11 PM

[ QUOTE ]
I came across this problem while doing my Calc hw today. Took me awhile to get it, I thought it was a pretty neat problem.

Find a formula for the nth term of this sequence.
0, 1, 1, 2, 2, 3, 3, 4, ... (each positive integer repeated).

The whole time I was doing this I was thinking about how much easier it would be to just write a C code for it, damn calculus and it's lack of ability to round down to the nearest factor of 2. Enjoy.

Edit - fixed sequence typo, and edited to add no using mod or anything such as that. Only standard math terms, ie. + , - , x , divide, and powers.

Edited again to say that you can only use one function that will apply to your whole sequence, no dividing odds and evens between two seperate functions.

[/ QUOTE ]

What math is this?