Write out the Thue-Morse sequence

Pyth, 6 bytes

xMjR2Q

Try it online: Demonstration

Based on the solution from @ThomasKwa and a variation of @FryAmTheEggman.

It uses the following formular: the i-th digit in the Thue-Morse sequence is: xor(digits of i in base 2).

Explanation:

xMjR2Q   implicit: Q = input number
  jR2Q   convert each number in [0, 1, ..., Q-1] to its binary digits
xM       xor each binary list

CJam, 17 9 bytes

ri{2b:^}/

or

ri,2fb::^

Test it here.

Explanation

This uses the alternative definition given on Wikipedia, based on the parity of the number of 1s in the binary representation of the n.

ri   e# Read input and convert to integer n.
{    e# For each i from 0 to n-1...
  2b e#   Convert i to base 2.
  :^ e#   Fold XOR over the bits to compute the parity of the number of 1s.
}/

The alternative solution uses , to turn n explicitly into a range [0 ... n-1] before using infix operators to compute the binary representations and the XOR without needing a block.

Bonus Solutions

Here are some solutions based on other definitions. If there are two solutions, the shorter one will blow up memory very quickly (because the precomputation generates 2^n bits before truncating to n).

As an L-system with 0 --> 01 and 1 --> 10:

ri_2mL2,\{2,aA+f=s:~}*<
ri_2,\{2,aA+f=s:~}*<

By negating and appending the previous part:

ri_2mL2,\{_:!+}*<
ri_2,\{_:!+}*<

Using the recurrence relation given in the challenge, using divmod and XOR to distinguish between the two recursive definitions:

ri{Ta{2md\j^}j}/

(Although, of course, this recurrence relation is just a different way to express the Thue-Morse sequence as the parity of the 1-bits in the binary representation of n.)

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