For all you math puzzle nuts...

TBag · #1 10-04-2005, 04:48 PM

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.

Homer · #2 10-04-2005, 04:55 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, 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.

[/ QUOTE ]

I assume it should be 0, 1, 1, 2, 2, 3, 3...

(n - n mod 2)/2

Homer · #3 10-04-2005, 05:01 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.

[/ QUOTE ]

Now you say that. Bastard! [img]/images/graemlins/smile.gif[/img]

Isn't mod a standard math term, though.

Okay, I'll try again.

TBag · #4 10-04-2005, 05:02 PM

Haha, when I said I was thinking about how easy it would be to write a C code for it instead of doing standard math procedures, I thought that would've gotten the point across [img]/images/graemlins/grin.gif[/img]

BruceZ · #5 10-04-2005, 05:10 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.

[/ QUOTE ]

f(n) = n/2 for n even
f(n) = (n+1)/2 for n odd

cookie · #6 10-04-2005, 05:13 PM

[ QUOTE ]
[ 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.

[/ QUOTE ]

f(n) = n/2 for n even
f(n) = (n+1)/2 for n odd

[/ QUOTE ]
f(n) = (n+1)/2 for n odd should be (n-1)/2, right?

TBag · #7 10-04-2005, 05:14 PM

Only one function is allowed. I'll post my solution (although it's a sloppy solution and I think a better one exists) before I go to class in 30 mins if no one else comes up an answer.

cookie · #8 10-04-2005, 05:21 PM

[ QUOTE ]
Only one function is allowed. I'll post my solution (although it's a sloppy solution and I think a better one exists) before I go to class in 30 mins if no one else comes up an answer.

[/ QUOTE ]
Yeah also thought the double function solution seemed to easy...

Homer · #9 10-04-2005, 05:22 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, 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.

[/ QUOTE ]

{n + (-1)^n)}/2

EDIT - [censored], that's wrong.

BruceZ · #10 10-04-2005, 05:26 PM

[ QUOTE ]
[ QUOTE ]
[ 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.

[/ QUOTE ]

f(n) = n/2 for n even
f(n) = (n+1)/2 for n odd

[/ QUOTE ]
f(n) = (n+1)/2 for n odd should be (n-1)/2, right?

[/ QUOTE ]

No, f(0) = 0, f(1) = 1, unless you want to start from f(1), and then it would be (n-1)/2.

One equation? OK, f(n) = {n + [1 - (-1)^n]/2}/2

Or if you start from f(1) = 0, then f(n) = {n - [1 - (-1)^n]/2}/2