Is the relationship between these two sequences, identical but for signs, trivial?

Question

Sums of divisors of product of first $n$ primes has same sequence as products of totients of first $n$ primes, but with alternating signs. OEIS A005867 Mathematica code shown:

Table[Sum[ MoebiusMu[d] d, {d, Divisors[Product[Prime[j], {j, 1, n}]]}], {n, 1, 8}] Table[Product[EulerPhi[Prime[j]], {j, 1, n}], {n, 1, 8}]

{-1, 2, -8, 48, -480, 5760, -92160, 1658880}
{1, 2, 8, 48, 480, 5760, 92160, 1658880}

Edit an alternate for the first statement: $$\sum_{n=1}\ \sum_{(d_{1},\dots,d_{2^{n}})\ |\ \text{lcm}(p_{1},\dots,p_{n})}\mu(d_{m})d_{m}$$ Is this a trivial relationship?

Answer 1

$$\prod_{i=1}^n\phi(p_i)=\phi(\#_n),$$

where $\#_n:=\prod_{i=1}^np_i$, by multiplicity.

$$\sum_{d\ | \ \#_n}\mu(d)d$$

is your second sum.

Conjecture: for any $n$ that is a product of distinct, single power primes (e.g. $35=5\cdot 7$):

$$\sum_{d|n}\mu(d)d=(-1)^k\phi(n),$$

where $k$ denotes the number of $prime$ divisors of $n$, so for $n=35=5\cdot 7$, $k=2$.

Notice that $g(d)=d$ is multiplicative, so it suffices to establish the conjecture for $n=p$ prime:

$$\sum_{d|p}\mu(d)d=\mu(1)+\mu(p)p=1-p=-\phi(p).$$

Since $\phi(n)=\prod_{p|n} \phi(p)$ (remember $n$ has only single prime powers), so

$$\sum_{d|n}\mu(d)d=\phi(n)(-1)^k$$

So your observation has been proven.

Fun fact, for any integer $n$:

$\phi(n)=\sum_{d|n}\mu(n/d)d=\sum_{d|n}\mu(d)n/d,$

So you have kind of a mirror identity in your case. Neat!

Answer 2

Yes, this is trivial — at least for some definition of trivial. The sum is literally a term-by-term expansion of the product of $(p_i-1)$; it iterates all the members of the product, with the sign of a given term $\Pi_{i\in S}p_i$ determined by the parity of the number of distinct primes $|S|$ in the individual term - in other words, by $\mu(\Pi_{i\in S}p_i)$.