the definition of $Y$) occurring at $P-1 ,P$ (by **(P3)**) and of height 1 (by **(P0)**). Since $f$ has only one drop, it can also have only one increase (since its values can only be 0 or 1). The location of that unique increase can then be determined to be at $P/2-1,P/2$ (due to **(P0)**), and this gives the whole statement.

**Example.** We provide another $\mu$ than $(6,6,6,6,6)$ for $\vec{\gamma} = (30,1), \vec{\delta} = (2,3,5)$ (see Section \[sec.nonintegral\]). In that case, the values taken by $f(x;(30,1),(2,3,5))$ are $$

[c|p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}p[0.0025]{}]{} x 30&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20&21&22&23&24&25&26&27&28&29&$\cdots$\
f&0& 1& 1& 1& 1& 1& 0& 1& 1& 1& 0& 1& 0& 1& 1& 0& 0& 1& 0& 1& 0& 0& 0& 1& 0& 0& 0& 0& 0& 1&$\cdots$

$$ We see that $P = 30$, $Y= \left\{ 5,9,11,14,17,19,23,29\right\}$ and $A_0 = \{0, 6, 10, 12, 15, 16, 18, 20, 21,$ $22,$ $24,$ $25,$ $26, 27, 28\}$, with $A_1$ the complement to $A_0$. One can check that $A_0 + A_0 = \{0,\cdots,28\}$ (remember this is in $\mathbb{Z}/30\mathbb{Z}$!), with $29$ missing. Because $Y$ is not a singleton, any element of $Y$ different from 29 leads to a $\mu$. Take $9\in Y$. Then, we solve (non-constructively and $\mod 30$) $9 