@MaxAlekseyev Sure, but we don't know how to do that efficiently as far as I know.

@PeterTaylor If I calculated correctly, Nate's answer implies that $ \binom{2n}{n}_q > \binom{2n}{n-a}_q \cdot q^{a^2} $, which means that the sum of the other coefficients is $ 2 \sum_{a=1}^n \binom{2n}{n-a}_q < 2 \binom{2n}{n}_q \sum_{a=1}^n q^{-a^2} < 2 \binom{2n}{n}_q \sum_{a=1}^\infty q^{-a^2} $. Since $ 2 \sum_{a=1}^\infty q^{-a^2} < 1 $ for any $ q \geqslant 3 $, the middle coefficient is larger than the sum of all the others when $ q \geqslant 3 $. For $ q = 2 $ this is perhaps not true, but taking into account the residues I am interested in, this case also holds.

Yes, I had in mind prime powers, although it should be true for all reals $ \geqslant 1 $ (by Nate's answer below).

Could you sketch a proof of this, or give a reference?

Thank you, this is very helpful!

@OfirGorodetsky Thanks for the references, I wasn't aware of the second one.

@LSpice To your first question - I don't know :) To your second question - yes, selecting some vector twice would mean that the condition of linear independence is violated. I'll edit my question to clarify.