$\alpha _n=e^{2 \pi i/n}$

$$f(x_1,x_2,x_3,\ldots,x_n)=(x_1+\alpha _n x_2+ \alpha _n ^2 x_3+\cdots+\alpha _n ^{n-1} x_n)^n$$

Maximum how many different results can have with all permutation of inputs? I have read in Jim Brown's paper on page 5. http://www.math.caltech.edu/~jimlb/abel.pdf

Lagrange showed that If n=3 then $f(x_1,x_2,x_3)$ Maximum can have 2 different results with all permutations of $(x_1,x_2,x_3)$

If n=4 then $f(x_1,x_2,x_3,x_4)$ Maximum can have 3 different results with all permutations of $(x_1,x_2,x_3,x_4)$

If n=5 then $f(x_1,x_2,x_3,x_4,x_5)$ Maximum can have 6 different results with all permutations of $(x_1,x_2,x_3,x_4,x_5)$

Is there any general formula for n and which method is used to find the general formula?

Thanks for answers

$\alpha _n=e^{2 \pi i/n}$

$$f(x_1,x_2,x_3,\ldots,x_n)=(x_1+\alpha _n x_2+ \alpha _n ^2 x_3+\cdots+\alpha _n ^{n-1} x_n)^n$$

Maximum how many different results can have with all permutation of inputs? I have read in Jim Brown's paper on page 5. http://www.math.caltech.edu/~jimlb/abel.pdf

Lagrange showed that If n=3 then $f(x_1,x_2,x_3)$ Maximum can have 2 different results with all permutations of $(x_1,x_2,x_3)$

If n=4 then $f(x_1,x_2,x_3,x_4)$ Maximum can have 3 different results with all permutations of $(x_1,x_2,x_3,x_4)$

If n=5 then $f(x_1,x_2,x_3,x_4,x_5)$ Maximum can have 6 different results with all permutations of $(x_1,x_2,x_3,x_4,x_5)$

Is there any general formula for n and which method is used to find the general formula?

Thanks for answers