An important fact in analytic number theory, and especially the recent string of papers relating to the bounded gaps between primes, is the following: the generalized von Mangoldt function (defined below) vanishes when has more than distinct prime factors. I will give a proof of this fact here for those interested.
The generalized von Mangoldt function is defined as
where is the Mobius function. Unfortunately, like the usual von Mangoldt function (where ), the generalized von Mangoldt function is not multiplicative. We will first treat the case when is square-free. Write , where and the are distinct primes. Then we see that
Thus it suffices to check the following general polynomial identity
Here, the summands are all expressions of the form for subsets .
To simplify the calculations, let us consider the coefficient of a single monomial that appears after we expand the above polynomial expression. Since it follows that at most of the ‘s are non-zero, and without loss of generality, assume that is zero for . The only subsets we need to consider are those such that . Each eligible subset will contribute to the coefficient. There are many eligible subsets of size . Whence, the coefficient is equal to
by the binomial theorem and the fact that .
The general case is again done by verifying a general polynomial identity. For , where , write the linear form . Then the vanishing of the generalized von Mangoldt function can be deduced from
The key is to match the powers of in the above expression. For a fixed , consider all terms where the exact power of is . We would have
By strong induction, the term in the parentheses would vanish according to the polynomial identity proved above. Since this is true for every , we have the entire expression vanishing and we are done.
If anyone knows a shorter, cleaner proof I would be very happy to hear it.