Integers whose Divisor Sum equals Half Phi times Divisor Count/Mistake

Source Work

 * The Dictionary
 * $105$
 * $105$

Mistake

 * $105$ is the second number $n$ such that $\phi \left({n}\right) \times \nu \left({n}\right) = \sigma \left({n}\right)$, where $\nu \left({n}\right)$ is the number of divisors of $n$. $\phi \left({105}\right) = 48$, $\nu \left({105}\right) = 8$ and $\sigma \left({105}\right) = 192$.


 * The first such number is $35$.

Elementary arithmetic shows that in fact $48 \times 8 = 384$, not $192$, and the result in fact appears to be:
 * $\sigma \left({n}\right) = \dfrac {\phi \left({n}\right) \times \nu \left({n}\right)} 2$.

The same applies to $35$.

The sequence such that $\phi \left({n}\right) \times \nu \left({n}\right) = \sigma \left({n}\right)$ is in fact:
 * $1, 3, 14, 42$

See Integers whose Phi times Tau equal Sigma.