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Let $ n $ denote a square free positive composite integer, $ \omega(n) $ its number of prime factors, $ P_{i}(n) $ its $ i $ -th prime factor.

Can we determine an asymptotics for the number of $ n $ below $ x $ such that $ P_{\omega(n)}-P_{1}(n)<\left(\dfrac{n}{\omega(n)}\right)^{1/\omega(n)} $?

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  • $\begingroup$ $i$th biggest prime factor, or $i$th smallest prime factor? $\endgroup$
    – Gerry Myerson
    Commented May 22, 2017 at 12:55
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    $\begingroup$ $ i $-th prime factor in increasing order. For example, if $ n=42=2.3.7 $, $ P_{1}(n)=2 $ , $ P_{2}(n)=3 $ , $ P_{3}(n)=7 $ . $\endgroup$
    – Sylvain JULIEN
    Commented May 22, 2017 at 13:15
  • $\begingroup$ You can divide out by the fractional power of n (which I call w) to get (b-c) is less than 1, which means the largest prime factor P is always less than twice the geometric mean of the factors (w). If P is near 2w, then the number of factors must be immensely large, and all near and most less than w. If P is near w, again most of the other factors must also be near w, and there can't be many far away from w. It almost becomes a counting problem, where I would be surprised if the number of solutions was greater than w. Gerhard "The A And The W" Paseman, 2017.05.22. $\endgroup$
    – Gerhard Paseman
    Commented May 22, 2017 at 15:42

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Please ignore the previous version of this answer.

Motivated by Lucia's comment, we use smooth numbers to show that the number in question is $o(x)$. First note that for $100\%$ of all integers one has $P_1(n)\leq \log \log \log n$, as a typical application of the naive Eratosthenes' sieve. Also by Hardy-Ramanujan one has for $100\%$ of all integers $n\leq x$ that $$ n^{1/\omega(n)}\leq x^{2/\log \log x}.$$ Therefore $$ p|n,n\leq x\Rightarrow p\leq x^{2/\log \log x}+\log \log \log x \leq x^{3/\log \log x}.$$ Now by estimates $(1,7),(1.8)$ here http://www.dms.umontreal.ca/~andrew/PDF/msrire.pdf one has that for every $A>0$ the number of integers $n\leq x$ in the question is at most $$\ll x \left(\frac{3\mathrm{e}}{(\log \log x)(\log \log \log x)}\right)^{(\log \log x)/3}\ll_A \frac{x}{(\log x)^A}.$$

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    $\begingroup$ This seems wrong. It is $\log \log P_i(n)$ that is close to $i$ often. Indeed even the $i$-th prime itself is around $i\log i$. In fact, the number of integers satisfying the relation in the question is $o(x)$. $\endgroup$
    – Lucia
    Commented May 24, 2017 at 18:33
  • $\begingroup$ Yes indeed, sorry, i will try to fix that. $\endgroup$
    – Dr. Pi
    Commented May 24, 2017 at 22:41

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