Let $\mu(x)$ be the mobius function and $\Lambda(x)$ be the vonmangolt function. $$ f(n) = \sum_{d|n} \mu(d-1)\Lambda(d - 1)$$. Is there a way to simplify or estimate this function.
$f(3) = -\log(2)$
$f(4) = -log(3)$
$f(6) = -\log(10)$
$f(8) = -\log(21)$
$f(9) = -\log(2)$
$f(12) = -\log(66)$
$f(14) = -\log(13)$
$f(15) = -\log(2)$
$f(16) = -\log(21)$
Barry's response helped me simplify it a bit more. If n is odd, then |f(n)| is nonzero if 3 divides n. If n is even and if |f(n)| is nonzero then n is congruent 0 or $\pm 2 $modulus 3.
This is more comment than answer, but it might be of help.
Since $\Lambda(m)=0$ unless $m$ is a prime power, say $p^k$, and $\mu(p^k)=0$ if $k\gt1$, the only contributions to the sum come from divisors $d$ that are $1$ more than a prime, in which case $\mu(d-1)\Lambda(d-1)=-\log p$, so we can rewrite (simplify?) the function as
$$f(n)=-\sum_{(p+1)\mid n}\log p$$
If we ignore the minus sign and exponentiate, we get an integer sequence
$$1,1,2,3,1,10,1,21,2,1,1,66,1,13,2,21,\ldots$$
(where $f(1)=f(2)=0$ to get the sequence started). This sequence is not (yet) in the OEIS. It's easy to see that
$$ f(n)= \cases{0\text{ if }n\equiv\pm1\mod6\\\\ \log2\text{ if }n\equiv3\mod6} $$
since $p+1$ cannot divide an odd number $n$ if $p$ is odd. It might make sense therefore to focus on the (exponentiated) entries for even $n$:
$$1,3,10,21,1,66,13,21,\ldots$$
but this doesn't (yet) appear in the OEIS either.
That's about all I have, except to note that the sum is sometimes fairly large. For example,
$$f(48)=-(\log2+\log3+\log5+\log7+\log11+\log23+\log47)=-\log2497110$$
There might be an interesting connection with Sophie Germain primes.