@Peter, I see. I had them install Latex on this machine. It is easy, and my writing is poor. Also, I get to preview things before putting on MSE, or the arXiv, or whatever.
@pourjour Similar notions of divisibility can be used to constrain the number of primes between $n$ and $2n$ using the central binomial coeifient, this was used by erdos in a proof of bertands postulate
@pourjour you start out by noting that $\binom{2n}{n} =\frac{(2n)!}{n!^2}$ is always an integer, and is divisible by all the primes between $n$ and $2n$, this allows one to obtain the estimate $\sum_{p\leq n}\ln(p)\leq n\ln(4)$
is there a reason I have never formally been taught what ZFC? I'm in college now and have taken the introductory number theory course but never formally learned the base axioms of math it seems
@Charlie, I did visit Copenhagen for a day on the way to Sweden. I walked out into the water, my idea was to kiss the statue of The Little Mermaid, which I did. But I took to long to get back down, all the tourists with cameras yelled at me.
@WillJagy The "only" thing I need to do is show that $$\sum_{d=1}^n f(d)\left\lfloor \frac n d\right\rfloor+\sum_{d\mid n+1}f(d)=\sum_{d=1}^{n+1}f(d)\left\lfloor \frac {n+1} d\right\rfloor$$ This depends on the relation between $n$ and $n+1$, and thus of their divisors.
@Ethan, that looks right. Peter posted it a few hours ago, it is in Landau's book without explanation. There is a proof by induction that I found. Your indicator function seems to be my exact idea.
$$\sum_{n\leq x} f(n)c_d(n)=\sum_{n\leq {x/d}}f(d)$$, because $c_d(n)$ is zero everywhere $n$ is not divisible by $d$ and there are only [x/d] multiples of $d$ less then or equal to $x$
In fact, I used $$\lfloor \frac {n+1}d\rfloor-\lfloor n/d\rfloor=1$$ in $$\sum_{d\mid n+1}f(d)$$ that is $$\sum_{d\mid n+1}f(d)=\sum_{d\mid n+1}f(d)\left(\lfloor \frac {n+1}d\rfloor-\lfloor n/d\rfloor\right)$$
@PeterTamaroff I think introducing the periodic indicator function is more natural it allows one to more readily see how to get asymptotics of arithmetic functions on polynomials for example, $\sum_{n\leq x} f(n^2+1)$
This means the function is zero everywhere except $d,2d,3d,...$
Because its periodic in $d$
Which means its zero everywhere except when $n$ is divisible by $d$
@PeterTamaroff you can estimate combinatoral sums of this sort even better in some cases by spliting each divisor sum across two different patches of divisors, a varient of this is the dirichlet hyperbola method
For example, $$\sum_{n\leq x} [\frac{x}{n}]=2\sum_{n\leq x^{1/2}}[\frac{x}{n}]-[\sqrt{x}]^2$$
@Peter, as I have tried to indicate, I do not really do these things in real time. I do not necessarily think on my feet. i would be quite happy if you took 24 hours and just played with it, eventually send me something. Ethan is enthusiastic and very, very fast, but I would rather you just do this little item and not get distracted.
@Charlie, there was a brief time when my favorite high scholl techer called me Billy, I believe he was trying to convince me to be more mainstream. He kind of had a point, you do not always understand the price you might be paying if you choose to be different, or simply are different.
Jonas, he is a good guy. Although once, in a game of American football, he ran into me and i chipped a tooth @Peter, that is fine. Email is good. I am not in a hurry.
