How can I compute the sum of $ {m\over\gcd(m,n)}$? $$ \sum_{m =1}^n {m\over\gcd(m,n)}$$
For example,
for 1 it is 
$${1\over\gcd(1,1)} =1;$$
for 5 it is
$${1\over \gcd(1,5)}+{2\over \gcd(2,5)}+{3\over \gcd(3,5)}+{4\over \gcd(4,5)}+{5\over \gcd(5,5)}=\\
\frac11+\frac21+\frac31+\frac41+1 = \\
1+2+3+4+1= \\ 11$$

If $\ n=p\ $ is prime, we can say
$$\sum_{m=1}^p\frac{m}{gcd(m,n)}=1+\frac{2}{1}+...+\frac{p}{p}=1+2+...+(p-1)+1=\frac{(p-1)p}{2}+1$$
But in general, in spite of the fact that the number $\,22\,$ appears several times in the first few non-prime numbers, there doesn't seem to be an easy way out. For example:
$$\sum_{m=1}^{12}\frac{m}{gcd(m,n)}=1+\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+\frac{5}{1}+\frac{6}{6}+\frac{7}{1}+\frac{8}{4}+\frac{9}{3}+\frac{10}{2}+\frac{11}{1}+\frac{12}{12}=39$$
