For an integer $n$, denote by $P^+(n)$ the largest prime divisor of $n$. Then we have the following:

There exists some $c>0$, such that for all $x$ sufficiently large the number of integers $n\in[x, x+0.1\sqrt{x}]$ with $P^+(n)>x^{1/2+c}$ is $\geq c\sqrt{x}$.

The proof is quite standard: You essentially need a non-trivial bound for $$ \sum_{n\leq x^{1/2+c}}\Lambda(n)\left(\left[\frac{x+0.1\sqrt{x}}{n}\right]-\left[\frac{x}{n}\right]-\frac{0.1\sqrt{x}}{n}\right). $$ Now express $\Lambda$ by Vaughan's identity, approximate $[\cdot]$ by exponential sums, and bound the latter by Weyl-van der Corput.

However, while conceptually simple, the computations take at least two pages, which I would rather avoid since

a) journal space is precious

b) the computations are likely to distract the reader from the more algebraic main arguments,

c) the question is so natural that someone probably spent a lot of time on some quantitative version of it.

So my question is: Can anyone give me a reference for such a result? Note that for the application the value of $c$ is of no importance, while 0.1 should not be replaced by something much larger.

Thank you in advance!


1 Answer 1


See the very first paper on the subject by Ramachandra (A note on numbers with a large prime factor; JLMS 1969). Ramachandra's Theorem 1 states that there is a constant $\alpha <1/2$ such that for all large $x$ the interval $[x,x+x^{\alpha}]$ contains an integer with a prime factor larger than $x^{\frac 12+\frac{1}{13}}$. Subsequent works focus on the interval $[x,x+\sqrt{x}]$ and on improving the size of the largest prime factor. It should be noted that in the proof Ramachandra focuses for simplicity on the case $\alpha=\frac 12$, but he does indicate that only small changes to the argument are needed to get some $\alpha<\frac 12$.

  • 1
    $\begingroup$ Great question, great answer! $\endgroup$
    – GH from MO
    Aug 13, 2014 at 17:12

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