I am hoping for explicit numerical estimates like the following sample (with made up numbers, though it might be true): for every $n \gt 10^6$ and every $b$ with $b^2 \lt n \lt b^3$, the number of somewhat smooth numbers in the interval $(n,n+b)$ is at least $b/4$ and at most $3b/7$. Here a number $m$ is somewhat smooth if each of its prime factors is at most the square root of $m$.

The literature on Dickman's function and recent surveys on smooth numbers by Moree and Granville say a lot for larger intervals than what is considered above, and in more generality, but the error terms are not explicit, and I have yet to construct a version with explicit terms from above based on my searches. The actual value approaches $b(1 - \log 2)$ as $n$ grows, but I do not know how fast. Also, somewhat smooth corresponds to $u=2$ in the literature, where $u=\log x / \log y$ is a common parameter in counting $y$-smooth positive integers less than $x$, but I may need to adapt somewhat smooth to $u$-smooth for $u$ some real between $2$ and $3$, so a good answer would address these other values of $u$. If smaller values of $b$ can be used, that would be great, but I think I may only need $n \lt b^3$, and doubt I will need $n \lt b^4$. Note that I am emphasizing explicit and reliable over asymptotic; I do not need tight estimates or small error terms especially if they are not explicit, and I do not want a set of exceptional $n$ or $b$ unless they are all less than a finite explicit and small bound.

Can someone point to or derive such explicit estimates of this type? Computations suggest that the sample above not only is true, but holds for $n$ much smaller than $10^6$.

Gerhard "Quite More Than Somewhat Unsure" Paseman, 2017.04.10.

  • $\begingroup$ Can't the lower bound be derived from the obvious sieve and an explicit Mertens estimate, of the kind surely to be found in Rosser–Schoenfeld? $\endgroup$ Commented Apr 12, 2017 at 0:37
  • $\begingroup$ Maybe it can. If you expand upon this idea a bit, I will follow it up. I am having troubles distinguishing obvious sieves from holes in the ground presently. Gerhard "Thank Goodness This Isn't Topology" Paseman, 2017.04.11. $\endgroup$ Commented Apr 12, 2017 at 0:53

1 Answer 1


Given $u$, for every prime $p$ let $S_p = \{m \in (n,n+b)\colon p\mid m,\, p^u > m \}$. Then a lower bound on the quantity you want is $(b-1) - \sum_p \#S_p$. (Indeed, this is the exact value when $u<2$.) The smallest $p$ you have to consider is $n^{1/u}$, and an upper bound for $\#S_p$ is $\lceil\frac{b-1}p\rceil \le \frac{b-2}p+1$. You're therefore led to need to estimate $\sum_{n^{1/u} < p < n+b} 1$ and $\sum_{n^{1/u} < p < n+b} 1/p$. For both of these, the complete sums (where the lower bound is $1$) should appear explicitly in the work of Rosser–Schoenfeld, so just subtract two of those from each other to get the interval versions above.

  • $\begingroup$ I'm guessing you mean $p^u \gt m$ in the definition of $S_p$. This looks good. I will do the calculations and post the results later. Gerhard "First, I Have To Sing" Paseman, 2017.04.11. $\endgroup$ Commented Apr 12, 2017 at 1:15
  • $\begingroup$ One question. For moderate sized $p$, how do I tell when $S_p$ is empty? Gerhard "Maybe Shanta Laishram Knows How?" Paseman, 2017.04.11. $\endgroup$ Commented Apr 12, 2017 at 1:25
  • $\begingroup$ Yeah, that's not easy. Of course computationally it is; but in a sum where many terms are expected to be $0$ rather than $1$, getting a bound better than $1$ for each term is often difficult. $\endgroup$ Commented Apr 12, 2017 at 1:28

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