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The radical $\mathrm{rad}\; n$ of a positive integer $n$ is the product of $n$'s distinct prime factors. For example $\mathrm{rad}\; 12=2\cdot 3=6$.

Denote the sum of $n$'s distinct prime factors by $\mathrm{spf}\; n$.

Claim: Given any $k \ge 1$, for nearly half of positive integers $n$, $\mathrm{spf}\; n > \mathrm{spf}\; (n+k)$, but for only one-third of positive integers $n$, $\mathrm{rad}\; n > \mathrm{rad}\; (n+k)$.

Detailed explanation:

For any $k \ge 1$, there will be $n$ such that $\mathrm{rad}(n) > \mathrm{rad}(n+k)$ e.g. $n=15, k = 1$. We define

$$ a_{n,k} = \begin{cases} 1 & \frac{\mathrm{rad}(n)}{\mathrm{rad}(n+k)} > 1 \\ 0 &\text{ otherwise} \end{cases} $$

Question: What is limiting value

$$ \lim_{x \to \infty}\dfrac{1}{x}\sum_{n \le x}a_{n,k} $$

Instead of product when I used sum, the limiting value approached $0.5$ which means that it is equally likely the sum of distinct prime factors of a natural number is greater or less than that of the next natural number which makes intuitive sense.

Similarly I was expecting that product of the distinct prime factors of a number is equally likely to be greater than or less than that of the the next number but this was not the case. I was surprised to see that for $x = 1.5 \times 10^8$

$$ \dfrac{1}{x}\sum_{n \le x}a_{n,k} \approx 0.3386 $$ and was decreasing with $x$ hence the eventual limit is expected be be slightly less than this.

Note: The initial version of this question was with $k=1$ but the result doesn't change with $k$, hence I have updated the question.

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  • $\begingroup$ $p_n$ for primes $p(n)$ for partitions to be precise :) $\endgroup$ – Nilotpal Kanti Sinha Dec 6 '19 at 4:30
  • $\begingroup$ Well the Euler product is written $\prod_p \frac1{1-p^{-s}}$. $p$ is reserved for primes in this context, use $\Pi$ or anything. $\endgroup$ – reuns Dec 6 '19 at 4:35
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Each of these products will be $\frac{n}{p}$ and $\frac{n+1}{q}$ for some $p,q\in\mathbb{Z}^+$, where $p$ and $q$ are the "leftover" primes in the factorizations of $n$ and $n+1$ when one copy of each distinct prime has been removed.

If $p\neq q$, $\frac{n}{p}>\frac{n+1}{q}\iff p<q$. Which of $p$ and $q$ is larger will be effectively "random", in that the behavior will jump around a lot and one will be larger than the other with a limiting ratio of $0.5$. (A formal proof might prove difficult, but you can certainly gather some numerical evidence to support this - I think it makes as much intuitive sense as the sum case you mentioned.)

When $p=q$, the comparison will always go in favor of $n+1$. But since $n$ and $n+1$ are relatively prime, this only happens when $p=q=1$ - i.e., when $n$ and $n+1$ are each squarefree. So we want to find out what fraction of consecutive integer pairs are squarefree, and hence what limiting ratio to exclude from the even split mentioned above.

Fortunately, we can express this as an infinite product. For a number to be squarefree is just for it to not be divisible by $4, 9, 25, \ldots, p^2,\ldots$. The limiting fraction of $(n,n+1)$ pairs which do not contain a multiple of 4 is $2/4$; likewise, for neither to be a multiple of 9 is $7/9$, and so on.

However, congruence modulo the first $N$ primes is independent over large ranges by the Chinese remainder theorem. So our probability that a "random" pair are both squarefree is

$$\prod_p\frac{p^2-2}{p^2}$$

which looks to be around $0.322634...$, or a touch under one third. This means that your limiting value will be

$$\frac12\left(1-\prod_p\frac{p^2-2}{p^2}\right) \approx 0.33868295...$$

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If $n+1$ is squarefree then $p(n+1)=n+1\implies a_n=0$ thus $$\lim\sup_{n\to \infty} \dfrac{1}{x}\sum_{n \le x}a_n \le 1-\frac{6}{\pi^2}\approx 0.393$$

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  • $\begingroup$ You can quickly verify with some numerical calculations that 0.39 is much too high. Try n from 1 to 10,000. $\endgroup$ – RavenclawPrefect Dec 6 '19 at 4:20
  • $\begingroup$ Sure this is just a rigorous upper bound, your approach is a conjectural estimate. It seems hard to make it rigorous, you should show us a plot of how well does it fit. $\endgroup$ – reuns Dec 6 '19 at 4:23
  • $\begingroup$ For $x = 1.65 \times 10^8$, the value is $0.3386851$ so I guess there is scope to tighten the upper bound. $\endgroup$ – Nilotpal Kanti Sinha Dec 6 '19 at 4:28
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    $\begingroup$ If your value is correct the convergence should be in $O(x^{-1/2+\epsilon})$ (that's what predicts the random model for the primes) $\endgroup$ – reuns Dec 6 '19 at 4:29
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    $\begingroup$ Ah, I missed the ≤. Apologies. A graph of the ratios from n=1000 to 100,000 is at imgur.com/3LOWorl, with the blue line the constant estimated to around 9 digits. $\endgroup$ – RavenclawPrefect Dec 6 '19 at 4:41

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