Alternating Power Fibonacci Sequence

Python 2, 30 bytes

f=lambda n:n<1or f(n/4)-f(n/2)

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One-indexed.

The sequence felt like a puzzle, something that Dennis generated by having a short way to express it. The power-of-two repetitions suggest recursing by bit-shifting (floor-dividing by 2). The alternating-sign Fibonacci recursion f(n)=f(n-2)-f(n-1) can be adapted to bitshift in place of decrementing. The base case works out nicely because everything funnels to n=0.

Jelly, 5 bytes

BLCÆḞ

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How?

Extending the Fibonacci series back into negative indexes such that the relation f(i) = f(i-2) + f(i-1) still holds:

  i   ...   -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   4   5 ...
f(i)  ...  -21  13  -8   5  -3   2  -1   1   0   1   1   2   3   5   8 ...

Going back from i=0 the numbers are those we need to repeat 2^n-1 times, and Jelly's Fibonacci built-in, ÆḞ, will calculate these.

We can find the -i (a positive number) that we need by taking the bit-length of n and subtracting 1.

Since we want i (a negative number) we can instead perform 1-bitLength and Jelly has an atom for 1-x, C, the complement monad.

BLCÆḞ - Main link: n               e.g.  500
B     - convert n to a binary list      [1,1,1,1,1,0,1,0,0]
 L    - get the length                   9
  C   - complement                      -8
   ÆḞ - Fibonacci                       -21

Mathematica, 43 36 24 bytes

Fibonacci@-Floor@Log2@#&

7 bytes saved thanks to @GregMartin, and a further 12 thanks to @JungHwanMin.