1

For each $n \in \mathbb{N}$, let $f(n)$ map $n$ to the product of the primes that divide $n$. So for $n=112$, $n=2^4 \cdot 7^1$, $f(n)= 2 \cdot 7 = 14$. For $n=1000 = 2^3 \cdot 3^3$, $f(1000)=6$. Continuing in this manner, I arrive at the following plot:
           Base Primes
Essentially: I would appreciate an explanation of this plot, "in the large": I can see why it contains near-lines at slopes $1,2,3,4,\ldots,$ etc., but, still, somehow the observational regularity was a surprise to me, due, no doubt, to my numerical naivety—especially the way it degenerates near the $x$-axis to such a regular stipulated pattern.

I'd appreciate being educated—Thanks!

Joseph O'Rourke
  • 30,327

1 Answers1

3

I cannot tell how much you know. If a number $n$ is squarefree then $f(n) = n.$ This is the most frequent case, as $6/\pi^2$ of natural numbers are squarefree. Next most frequent are 4 times an odd squarefree number, in which case $f(n) = n/2,$ a line of slope $1/2,$ as I think you had figured out. These numbers are not as numerous, a count of $f(n) = n/2$ should show frequency below $6/\pi^2.$ All your lines will be slope $1/k$ for some natural $k,$ but it is not just a printer effect that larger $k$ shows a less dense line. Anyway, worth calculating the actual density of the set $f(n) = n/k.$

Meanwhile, note that a computer printer does not actually join up dots in a line into a printed line, that would be nice but is not realistic. There are optical effects in your graph that suggest we are seeing step functions. If so, there are artificial patterns not supported mathematically.

Alright, i am seeing an interesting variation on frequency that I did not expect. Let us define $$ g(k) = \frac{\pi^2}{6} \cdot \mbox{frequency of} \; \left\{ f(n) = n/k \right\}. $$ Therefore $$ g(1) = 1. $$ What I am finding is, for a prime $p,$ $$ g(p) = \frac{1}{ p \,(p+1)}, $$ $$ g(p^2) = \frac{1}{ p^2 \,(p+1)}, $$ $$ g(p^m) = \frac{1}{ p^m \,(p+1)}. $$

Furthermore $g$ is multiplicative, so when $\gcd(m,n) = 1,$ then $$g(mn) = g(m) g(n). $$

Note that it is necessary that the frequency of all possible events be $1,$ so $$ \sum_{k=1}^\infty \; g(k) = \frac{\pi^2}{6}.$$ I will need to think about the sum, it ought not to be difficult to recover this from known material on $\zeta(2).$

EDIT: got it. see EULER. For any specific prime, we get $$ G(p) = g(1) + g(p) + g(p^2) + g(p^3) + \cdots = \left( 1 + \frac{1}{p(p+1)} + \frac{1}{p^2(p+1)} + + \frac{1}{p^3(p+1)} + \cdots \right) $$ or $$ G(p) = 1 + \left( \frac{1}{p+1} \right) \left( \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \cdots \right) $$ or $$ G(p) = \frac{p^2}{p^2 - 1} = \frac{1}{1 - \frac{1}{p^2}}. $$ Euler's Product Formula tells us that $$ \prod_p \; G(p) = \zeta(2) = \frac{\pi^2 }{6}. $$ The usual bit about unique factorization and multiplicative functions is $$ \sum_{k=1}^\infty \; g(k) = \prod_p \; G(p) = \zeta(2) = \frac{\pi^2}{6}.$$

Will Jagy
  • 139,541