21 (number)

Note that a necessary condition for ${\displaystyle n}$ is that for any ${\displaystyle a}$ coprime to ${\displaystyle n}$, ${\displaystyle a}$ and ${\displaystyle n-a}$ must satisfy the condition above, therefore at least one of ${\displaystyle a}$ and ${\displaystyle n-a}$ must only have factor 2 and 5.

Let ${\displaystyle A(n)}$ denote the quantity of the numbers smaller than ${\displaystyle n}$ that only have factor 2 and 5 and that are coprime to ${\displaystyle n}$, we instantly have ${\displaystyle {\frac {\varphi (n)}{2}}<A(n)}$.

We can easily see that for sufficiently large ${\displaystyle n}$, ${\displaystyle A(n)\sim {\frac {\log _{2}(n)\log _{5}(n)}{2}}={\frac {\ln ^{2}(n)}{2\ln(2)\ln(5)}}}$, but ${\displaystyle \varphi (n)\sim {\frac {n}{e^{\gamma }\;\ln \ln n}}}$, ${\displaystyle A(n)=o(\varphi (n))}$ as ${\displaystyle n}$ goes to infinity, thus ${\displaystyle {\frac {\varphi (n)}{2}}<A(n)}$ fails to hold for sufficiently large ${\displaystyle n}$.

In fact, For every ${\displaystyle n>2}$, we have

${\displaystyle A<1+\log _{2}(n)+{\frac {3\log _{5}(n)}{2}}+{\frac {\log _{2}(n)\log _{5}(n)}{2}}}$

and

${\displaystyle \varphi (n)>{\frac {n}{e^{\gamma }\;\log \log n+{\frac {3}{\log \log n}}}}}$

so ${\displaystyle {\frac {\varphi (n)}{2}}<}$ fails to hold when ${\displaystyle n>273}$ (actually, when ${\displaystyle n>33}$).

Just check a few numbers to see that '= 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 21.