Gevorg Hmayakyan proposed last year (17 September 2017) on StackExchange the following relationship concerning the mobius function. Gevorg stated that if $$∏^{n−1}_{i=1}(1−x^i)=∑_{k=0}^{n(n−1)/2}a_{n,k}x^{k} $$ then $$μ(n)=a_{n,1}+a_{n,n+1}+a_{n,2n+1}+a_{n,3n+1}+...$$
Does anyone know how prove this?
During related computer computation on this function, I also noted that $$\phi (n)=a_{n,0}+a_{n,n}+a_{n,2n}+a_{n,3n}+...$$
but why this is the case I do not know? Any insights?
PS. I had noticed that the above generating function is equivalent to a row of Euler Truncated Product Triangle (see the paper by Alex Mennen at http://www.alex.mennen.org/mahoniantri.pdf). Thus Using Mennens notation then they become $$μ(n)=∑_{k=0}^{n(n−1)/2}P(n−2,kn+1)$$ $$\phi(n)=∑_{k=0}^{n(n−1)/2}P(n−2,kn)$$
For $\phi(n)$, series multisection means that $$\sum_{k}a_{n,kn}x^{kn}=\frac1n \sum_{j=0}^n f_n(\zeta^j x)$$ where $f_n(x)=\prod_{i=1}^{n-1}(1-x^i)$ and $\zeta=\exp(2\pi i/n)$. Therefore $$\sum_{k}a_{n,kn}=\frac1n \sum_{j=0}^n f_n(\zeta^j).$$ If $\gcd(j,n)>1$ then $f_n(\zeta^j)=0$. For $\gcd(j,n)=1$ then $f_n(\zeta^j)=f_n(\zeta)=\prod_{j=1}^{n-1}(1-\zeta_j)$ and so $$\sum_{k}a_{n,kn}=\frac{\phi(n)}n f_n(\zeta).$$ But $$f_n(\zeta)=\lim_{x\to1}\frac{(x-1)(x-\zeta)(x-\zeta^2)\cdots(x-\zeta^{n-1})}{x-1}=\lim_{x\to1}\frac{x^n-1}{x-1}=n.$$ We conclude $$\sum_{k}a_{n,kn}=\phi(n).$$
ADDED IN EDIT
The $\mu(n)$ case follows similar lines. In this case $$\sum_{k}a_{n,kn+1}=\frac1n \sum_{j=0}^n\zeta^{-j} f_n(\zeta^j).$$ This reduces to $\sum_{j:gcd(n,j)=1}\zeta^{-j}$. This is the sum of all primitive $n$-th roots of unity, which is $\mu(n)$.