The beginning of the sequence of all algebraic numbers ordered as explained in the parent entry is as follows:

$0;\,\,-1,\,1;\,\,-2,\,-\frac{1}{2},\,-i,\,i,\,\frac{1}{2},\,2;\,\,-3,\, \frac{-1-\sqrt{5}}{2},\,-\sqrt{2},\, -\frac{1}{\sqrt{2}},\,\frac{1-\sqrt{5}}{2},\,\frac{-1-i\sqrt{3}}{2}, \,\frac{-1+i\sqrt{3}}{2},\,-\frac{1}{3},\,$

$-i\sqrt{2},\,-\frac{i}{\sqrt{2}},\,\frac{i}{\sqrt{2}},\, i\sqrt{2},\,\frac{1}{3},\,\frac{1-i\sqrt{3}}{2},\, \frac{1+i\sqrt{3}}{2},\,\frac{-1+\sqrt{5}}{2},\,\frac{1}{\sqrt{2}},\,\sqrt{2},\, \frac{1+\sqrt{5}}{2},\,3;\,\ldots$

The first number corresponds to the algebraic equation $x = 0$ , the two following numbers to the equations $x\pm 1 =0$ , the six following to the equations $x\pm 2 = 0$ , $2x\pm 1 = 0$ , $x^2+1 = 0$ , the twenty following to the equations $x\pm 3 = 0$ , $3x\pm 1 = 0$ , $x^2\pm x \pm 1 = 0$ , $x^2\pm 2 = 0$ , $2x^2\pm 1 = 0$ .

In practice, one cannot continue the sequence very far since the higher degree equations - quintic and so on - are non-solvable by radicals; instead we can list the equations satisfied by the numbers as far we want and tell how many roots they have. In principle, the number sequence does exist!