Exercise 4.3.14 ($S(n) = n^2 \phi(n)/3 + (-1)^{\omega(n)} \phi(n) s(n)/6$ for $n1$)

Question

In the notations of the two preceding problems, show that $S(n) = n^2 \phi(n)/3 + (-1)^{\omega(n)} \phi(n) s(n)/6$ for $n>1$.

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Hint. Use (4.1).

Richard Ganaye · Accepted Answer

\begin{proof} 
By Theorem 4.7, if $n>1$, 
\begin{align}
\sum_{d \mid n} \mu(d) = 0.
\end{align}
The equality (4.1) gives
\begin{align}
n \sum_{d \mid n} \frac{\mu(d)}{d} = \phi(n).
\end{align}
We proved in Problem 13 that
\begin{align}
 \sum_{d \mid n} d \mu(d) = (-1)^{\omega(n)} \phi(n) \frac{s(n)}{n}.
\end{align}
Starting with the result of Problem 12, using (1), (2) and (3) we obtain for $n>1$
\begin{align*}
S(n) &= n^2 \sum_{d \mid n} \frac{1}{6} \mu(d) \left( \frac{2n}{d} + 3 + \frac{d}{n} ight)\
&=\frac{n^3}{3}  \sum_{d \mid n} \frac{\mu(d)}{d} + \frac{n^2}{2}  \sum_{d \mid n}  \mu(d) + \frac{n}{6} \sum_{d \mid n} d \mu(d)\
&=\frac{n^2}{3} \phi(n)+ \frac{n}{6} (-1)^{\omega(n)} \phi(n) \frac{s(n)}{n}.
\end{align*}
In conclusion, for $n>1$,
$$S(n) = \frac{ n^2 \phi(n)}{3}+ (-1)^{\omega(n)} \frac{\phi(n) s(n)}{6},$$
that is
$$\sum_{j \in [\![1,n]\!],\, j \wedge n = 1} j^2 = \frac{1}{6} \phi(n) \left( 2n^2 + (-1)^{\omega(n)} \prod_{p \mid n} p ight).$$
\end{proof}

Exercise 4.3.14 ($S(n) = n^2 \phi(n)/3 + (-1)^{\omega(n)} \phi(n) s(n)/6$ for $n>1$)

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Exercise 4.3.14 ($S(n) = n^2 \phi(n)/3 + (-1)^{\omega(n)} \phi(n) s(n)/6$ for $n>1$)

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