Up to a(13) all the terms are of the form floor(sqrt(2^k1)) for some k.  Giovanni Resta, Jan 07 2019
From Chai Wah Wu, Jan 10 2019: (Start)
The terms are numbers of the form floor(sqrt(2^(2k1))) that are larger than 1/2 + sqrt(1/4 + 2^(2k1)  2^k) for some k > 0.
Theorem: if x is a term, then it is of the form floor(sqrt(2^(2k1))) for some k > 0. In addition, floor(sqrt(2^(2k1))) is a term if and only if it is larger than 1/2 + sqrt(1/4 + 2^(2k1)  2^k).
Proof: suppose x has k bits in its binary representation, i.e. 2^(k1) <= x < 2^k. Then x^2 has either 2k1 or 2k bits.
First we show that if x is a term of the sequence, then x^2 has 2k1 bits. Suppose x^2 has 2k bits, i.e. x^2 >= 2^(2k1). Then 2^2k  1  x^2 < 2^k  1  x. This is rearranged as x^2  x + 2^k  2^2k > 0. Solving this quadratic inequality leads to x > 2^k which contradicts the fact that x has k bits.
Thus x^2 < 2^(2k1) and 2^(2k1)  1  x^2 < 2^k  1  x. Solving this inequality and combining it with x < sqrt(2^(2k1)) shows that x must satisfy sqrt(2^(2k1)) > x > 1/2 + sqrt(1/4 + 2^(2k1)  2^k).
To complete the proof, we need to show that for each k, there is at most one integer satisfying this inequality. This is easily verified for k = 1. Assume that k > 1. Let a = sqrt(2^(2k1)) and b = 1/2 + sqrt(1/4 + 2^(2k1)  2^k).
Using the identity sqrt(x)  sqrt(y) = (xy)/(sqrt(x)+sqrt(y)) it follows that ab = 1/2 + (2^k  1/4)/(sqrt(2^(2k1))+sqrt(1/4 + 2^(2k1)  2^k) < 1/2 + 2^k/(sqrt(2^(2k1)) + sqrt(2^(2k1)2^k)).
Since k >= 2, sqrt(2^(2k1)2^k) = sqrt(2^(2k1)(12^(1k))) >= sqrt(0.5)*sqrt(2^(2k1)). This implies that ab < 1/2 + 2^k/((1+sqrt(0.5))*(sqrt(2^(2k1)))) = 2sqrt(2)/(sqrt(2)+1)  1/2 < 0.672. This implies that there is at most one integer between b and a.
The above discussion also provides another characterization of the sequence. floor(sqrt(2^(2k1))) is a term if and only if sqrt(2^(2k1))floor(sqrt(2^(2k1))) < ab where ab is as defined above.
The criterion can be simplified as:
floor(sqrt(2^(2k1))) is a term if and only if it is larger than 1/2 + sqrt(1/4 + 2^(2k1)  2^k). This concludes the proof.
Note that ab > 0.5 as k > oo, i.e. for large k, the fractional part of sqrt(2^(2k1)) should be less than about 0.5 in order for the integer part to be a term.
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