[BruceZ]

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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.

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

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

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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