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 $\map \phi n \times \map \nu n = \map \sigma n$, where $\map \nu n$ is the number of divisors of $n$. $\map \phi {105} = 48$, $\map \nu {105} = 8$ and $\map \sigma {105} = 192$.


 * 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:
 * $\map \sigma n = \dfrac {\map \phi n \times \map \nu n} 2$.

The same applies to $35$.

The sequence such that $\map \phi n \times \map \nu n = \map \sigma n$ is in fact:
 * $1, 3, 14, 42$

See Integers whose Phi times Tau equal Sigma.