Fdtxjr

Motivated by this question, My question pertains to closed form of the sum
$$sum_dfracphi(d)d$$

There are some formulae and expressions for $sum_nfracphi(n)n$, but what about when we replace $n$ by divisors of $n$ By seeing this question, I think the sum is multiplicative. Any hints? Thanks beforehand.

Here is not a closed form, but an alternative representation. Below, $(x,y)=rm gcd(x,y)$.

I claim that
$$sum_k=1^n (k,n)=nsum_dmid nfracphi(d)d.
$$
To see this, observe that if $dmid n$ is a divisor of $n$
$$
(k,n)=d iff left(frackd,fracndright)=1.
$$
In particular, the divisor, $d$, appears exactly $phi(n/d)$ times in the summation. Thus,
$$
sum_k=1^n (k,n)=sum_dmid nphi(fracnd)d.
$$
Now, letting $d'=n/d$, the sum can be rewritten as,
$$
sum_k=1^n (k,n)=nsum_d'mid nfracphi(d')d'.
$$
Hence, this object has the given operational meaning.

Now, for multiplicative property, take coprime $(m,n)$. Our goal is to show,
$$
left(sum_dmid mnfracphi(d)dright) = left(sum_dmid mfracphi(d)dright)left(sum_dmid nfracphi(d)dright).
$$
Check that, the left hand side sum has $phi(mn)=phi(m)phi(n)$ terms, so do the right hand side. Furthermore, each divisor $dmid mn$ can be uniquely decomposed into $d=d_1d_2$ where $d_1mid m$ and $d_2mid n$, since $(m,n)=1$. Hence, we deduce the given function is indeed multiplicative.

Remarks:
In fact, more is known about this function.

Theorem: For any positive integer $a$, there exists a positive integer $n$, such that,
$$
sum_dmid nfracphi(d)d=a.
$$

Try to prove this theorem, it is nice ;)

After a little observation, the sum is similar to evaluating $sum_ddphi(d)$. Since the function $sum_dfracphi(d)d$ is (weakly) multiplicative, therefore, it suffices to evaluate the sum for prime powers. We have, $$sum_p^kfracphi(d)d=1+fracp-1p+fracp^2(p-1)p^3+ldots+fracp^k(p-1)p^k+1$$
$$=1+kfracp-1p$$
Hence the form of the sum for any general $n=prod_ip_i^k_i$ is:
$$sum_dfracphi(d)d=prod_ileft(1+k_ifracp_i-1p_iright)$$
where $n$ is the product of $p_i$ distinct primes with multiplicity $k_i$