1. Quick question: What kind of formula yields something like odd, odd, even, even, odd, odd, even, even, etc. ?

If it starts with even or odd doesn't matter, as long as it goes in pairs. That's all. (if you're wondering, this has to do with finding a general formula for higher derivatives of a function) Oh, and no this is not homework

2.

3. You are opposed to piecewise functions?

4. Whatever works (why?)

5. Well then just define it piecewise. For example define your function to be 1 for all integers congruent to 0 or 1 (mod 4), and be 2 for all integers congruent to 2 or 3 (mod 4).

6. Blargh! My ignorance is killing me... Never mind, it can't be piecewise-defined. But thanks a lot anyway.

Shortly after I posted, I had an idea and figured out the problem. But I'm sure you had an answer in mind, so I didn't want to shut you off.

And now, I realized it has to be a single formulation due to the context of the problem. Sorry

7. In case anyone's wondering, here's the explicit formula I found:

This particular one goes 1, 1, 2, 2, 3, 3, 4, 4 ...

8. Starting with n=3 that is! But yeah that works much better.

9. Yeah. When I referred to an "alternating-ish sequence", I was referring to the signs.

The order of derivatives went - , - , + , + , - , - , + , + ... So I thought instead of finding a rule for alternating pairs of signs I could find a simpler rule for alternating pairs of evens/odds and raise -1 to that power. And much courtesy to Wolfram Alpha for its sequence tool.

10. Originally Posted by halorealm
Quick question: What kind of formula yields something like odd, odd, even, even, odd, odd, even, even, etc. ?

Recognize the sequence?

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