Okay. Let's move on to exploring some variations on this function. The first of these will be the easiest to understand, and so it's probably a good place to start. It is, however, perhaps the most boring, so it might also be a bad place to start; subsequent ideas get more challenging and curiosity-provoking. Anyway, as a home base, here is the identity from the start.
Right.
And $f_1(n, 2)$ is
Okay.
Now, I want to alter it slightly.
For many of these experiments, my entire trick will be to introduce a helper function to the basic recursive definition and then see how the output of the function changes.
That sounds simple enough.
So let me introduce the first such helper function. Here it is, labeled arbitrarily for convenience as $a(x)$. (Here, $\fl{n}$ means the smallest whole number $\le n$)
What do you suppose it does?
Well... it looks like $a(n)$ must be 1 if $n$ is a perfect square and zero otherwise. So I guess it identifies perfect squares, basically.
Exactly. It looks like this.
Easy enough.
Right. Now let's use that helper function.
This will be pretty straightforward. I'll use the function $a(n)$ as a scaling factor in a recursive function similar to $f_k(n,j)$. This makes a new, related function, $g_k(n,j)$. I'll mark the changes in red:
Okay. But what does $g_1(n, 2)$ look like after that change?
See for yourself.
And what exactly do you think is going on here?
Well, this one seems stranger. It looks like the graph only increases at prime numbers raised to even powers. So that means...
That's right. So as a sum, we could write it as
$$g_1(n,2)= \sum_{p^{2 k} \le n} \frac{1}{k}$$What is its relationship to the original prime counting function?
Well, in terms of the original function $f_k(n, j)$, the new function is
$$g_1(n,2)= f_1(n\rr{^\frac{1}{2}}, 2)$$So, the insertion of an $a(x)$ scaling factor scales the domain by a square root factor.
Or in other words, the scaling factor $a(n)$ seems like it stretches the function out, then.
Right. If you want to watch this method of counting in action, here's another animation of it.
Is there something special about the power of 2, there?
Oh, no. This idea can be generalized. Suppose $m$ is a positive whole number. Then we can generalize $a(n)$ with a new parameter $m$ as
Do you have an example, to make this more concrete?
Sure. Suppose $m=3$. With this definition, $a_3(n) = 1$ if $n$ is a perfect cube and 0 otherwise.
And so what does that do?
Well, insert $a_3(n)$ as a scaling factor into the recursive prime counting function $g_1(n,2)$, and the following results:
Oh, okay. Now the function is stretched-out even more.
Correct. And given how infrequently primes cubed show up, this ought to be expected.
And so how does this generalization relate to the original prime counting function $f_1(n,2)$?
With this generalized version of $a_m(n)$, a little work shows that this new stretched function $g_1(n,2)$ relates to the original recursive function as
$$g_1(n,2)= f_1(n\rr{^\frac{1}{m}}, 2)$$Thus, changing the value of $m$ stretches out the domain of the prime counting function.
I guess this is clear enough. Can this be connected to the Riemann zeta function, too?
It can. With a touch more work, the above ideas connect easily to the Riemann zeta function like so.
First, introduce a scaling factor $j^{-s}$ to the recursive definition.
Okay.
Then apply the recursive definition to itself until it is entirely unrolled. With a suitable value of $s$, take the limit as $n$ goes to infinity. The following identity is the result of that process.
$$\log \zeta(2 \cdot s) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} (\zeta(2 \cdot s)-1)^k$$ $$\log \zeta(m \cdot s) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} (\zeta(m \cdot s)-1)^k$$This ends up being a basic extension ideas from our earlier conversation.
Okay. That seems straightforward enough.
It was... perhaps a little too straightforward. Anyway, that's one simple family of changes that can be made to $f_k(n,2)$. Let's move on to a more interesting one.