General form for this integer-factor-shifting bijection?

Question

Let $f(k,s)$ be a function that takes all the factors in an integer $k$ and increases their prime indices by $s$.

e.g.

$$f(28,2)=f(2^2 \cdot 7,2)=f(p_1^2 p_4,2)=p_3^2 p_6 =5^2 \cdot 13=325.$$

I've determined that for any positive integer $n$,

$$\left\vert\{1 \leq k \leq n : f(k,1)>2n\}\right\vert=\left\vert\{n+1 \leq k \leq 2n : f(k,1)\leq 2n\}\right\vert.$$

I understand why this happens; $x\iff f(x,1)$ is a bijection in $\mathbb{N}$, so symmetry requires it.

Through experimentation, it also seems that

$$\left\vert\{1 \leq k \leq n : f(k,2)>3n\}\right\vert=\left\vert\{n+1 \leq k \leq 3n : f(k,2)\leq 3n\}\right\vert.$$

I'm less clear on why this works, particularly because there is no obvious candidate for $x$ in situations like $f(x,2)=2y$ that maintains a 1-to-1 mapping.

More to the point, I haven't been able to extend this pattern any further, and am wondering whether there is a good reason for that. If anyone can shed some insight on whether this general form can or cannot be extended for values of $s>2$, I'll consider my question answered.

Answer 1

First of all, the map $x \mapsto f(x, 1)$ isn't a bijection from $\mathbb N$ to $\mathbb N$; it's a bijection from $\mathbb N$ to the set of positive odd integers. Similarly, $x \mapsto f(x, s)$ is a bijection from $\mathbb N$ to the set of positive integers not divisible by any of $p_1 = 2, p_2 = 3, \dots, p_s$.

Your observation about $f(k, 1)$ can be rephrased as follows: if you add to each side the quantity $$ |\{1 \leq k \leq n: f(k, 1) \leq 2n\}|, $$ you get $$ n = |\{1 \leq k \leq 2n: f(k, 1) \leq 2n\}|. $$ This is easy to prove: There are exactly $n$ odd numbers below $2n$, each of them equals $f(k, 1)$ for exactly one choice of $k$, and this $k$ is less than $2n$ because $x \leq f(x, 1)$ for all $x$.

You can restate your claim about $f(k, 2)$ similarly: adding $$ |\{1 \leq k \leq n: f(k, 2) \leq 3n\}| $$ to each side gives $$ n = |\{1 \leq k \leq 3n: f(k, 2) \leq 3n\}|. $$ To prove this, it suffices to show that exactly $n$ of the numbers $1, \dots, 3n$ are divisible by neither 2 nor 3. You can check this by casework on the parity of $n$.

If you want to generalize this to study $f(k, s)$ for $s > 2$, the trick of adding a term to each side will still work; the only thing that will change is the problem of counting integers below $p_s n$ that are not divisible by a prime $\leq p_s$. For example, when $s = 3$, there are approximately $\frac 12 \cdot \frac 23 \cdot \frac 45 \cdot 5n = \frac{4n}{3}$ integers below $5n$ that are not divisible by 2, 3, or 5; but the exact number will fluctuate a little depending on congruence conditions on $n$.