= Sixth Fourier moment as a three-sum collision count
For $A$ in a finite <abelian group> and normalized <Fourier transform> $\widehat{1_A}$, the normalized number of sextuples satisfying
$$
x_1+x_2+x_3=x_4+x_5+x_6
$$
is
$$
\sum_{\gamma\in\widehat G}|\widehat{1_A}(\gamma)|^6.
$$
This follows by inserting <orthogonality of complex exponentials> for the displayed equation: the three variables on one side contribute $\widehat{1_A}(\gamma)^3$ and those on the other side contribute its <complex conjugate>.
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