$$ \sum_{m =1}^n {m\over\gcd(m,n)}$$
For example, for 1 it is
$${1\over\gcd(1,1)} =1;$$
for 5 it is $${1\over \gcd(1,5)}+{2\over \gcd(2,5)}+{3\over \gcd(3,5)}+{4\over \gcd(4,5)}+{5\over \gcd(5,5)}=\\ \frac11+\frac21+\frac31+\frac41+1 = \\ 1+2+3+4+1= \\ 11$$