For two multiplicative arithmetic functions ๐,๐ the Dirichlet convolution is defined by $(๐โ๐)(๐)=โ_{๐ยท๐=n}=๐(๐)๐(๐). $
Let u(n) be the function that is 1 for all n and define ฮป(n) = $(-1)^{e_1 + e_2 +…+e_n }$ where n has prime factorization $n = {p_1}^{e_1}ยท…ยท{p_n}^{e_n} $
Find a closed form for ฮป*u(n).
Notes: so in the formula for calculating the Dirichlet convolution the u(b) term will be 1 for each term so it should just be the sum of ฮป(d) over all d which divide n. My thoughts are that the closed form depends on n's prime factorization and the sum of the coefficients on all the primes, but I'm not sure how to prove it.
Best Answer
Let $n= p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, we have \begin{eqnarray*} (g*f)(n) &=& \sum_{d \mid n} \lambda(d) =\sum_{i_{1}=0}^{e_1} \sum_{i_{2}=0}^{e_2} \cdots \sum_{i_{k}=0}^{e_k} (-1)^{i_1+i_2 +\cdots+i_k} \\ \end{eqnarray*} \begin{eqnarray*} &=& \sum_{i_1=0}^{e_1} (-1)^{i_1} \sum_{i_2=0}^{e_2} (-1)^{i_2} \cdots \sum_{i_k=0}^{e_k} (-1)^{i_k} \\ \end{eqnarray*} Now note that $\sum_{i_j=0}^{e_j} (-1)^{i_j}$ is zero if $e_j$ is odd and $1$ if it is even, so we require all the exponenets to be even, in other words $n$ must be an exact square.
So the convolution $1$ if $n$ is a square and $0$ otherwise.