Low-prime intervals

Question

• $\mathbb P\ $ -- the set of all natural prime numbers.

• $\mathbb N=\{1\,\ 2\,\ \ldots\}\ $ -- the set of all natural numbers.

• $\forall_{d\, x\,\in\,\mathbb R_{_{>0}}}$, $E(d\ x) := \frac{\log(d)}{\log(x)} $ -- we will say that $\ E(d\ x)\ $ is the exponent of divisor $\,d\,$ with respect to $\,x\ $ (e.g. $E(d\,\ d^2) = \frac 12$).

Let $\ 1 < a < b\ $ and $\,\ a\,\ b\in\mathbb N.\ $ Define:

$$ M(a\,\ b)\ :=\ \max\left(\, E(p\,\ b):\ p\in\mathbb P\ \ \text{and}\ \ p\, | \prod_{x=a}^b x \ \right) $$

The computational challenge:$\quad$ given real $\ f\in(0;1),\ $ find as long as possible intervals $\ [a;b]\ $ such that $\ M(a\ b)\le f.\ $ (A variation: $\ \ldots M(a\ b)< f.)$

This implies a table of the record length results $\ L_f\ $ (and of $\ Ł_f$). One may also consider record results related to the positions of intervals $\ [a;b].$

One may also reverse the notion of records. We may fix the length $\ N:=b-a\ $ of the intervals $\ [a\,\ b]$, then we will be concerned with the respective minimal interval divisor exponents $\ F_N$.

The mathematical challenge: Given $\ f\in(0;1),\ $ are there only finitely many intervals $\ [a;b]\ $ as described above?

In either case, the question arises about the distribution of such intervals.

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EXAMPLES

$$ M(8\,\ 9)\ =\ \frac 12\qquad\text{but}\qquad M(8\,\ 10)\ =\ \frac{\log(5)}{\log(10)}\ = \ 0.6989700\ldots $$ Next, $$ M(63\,\ 64)\ =\ \frac{\log(7)}{\log(64)}\ =\ 0.467892\ldots\qquad\text{but}\qquad M(63\,\ 65)\ =\ \frac{\log(13)}{\log(65)}\ =\ 0.61673995\ldots $$

Next, $$ M(98\,\ 100)\ =\ \frac{\log(11)}{\log(100)}\ =\ 0.5206963\ldots $$

-- not too bad but Dave Benson's examples are much-much better (see the comments below).

Answer 1

Let me attempt to turn my comment into an answer. The paper

Balog, Antal; Wooley, Trevor D., On strings of consecutive integers with no large prime factors, J. Aust. Math. Soc., Ser. A 64, No. 2, 266-276 (1998). ZBL0942.11041.

contains (a stronger version of) the following result: For each positive integer $k$ and each $\epsilon > 0$, there are infinitely many $n$ where none of $n+1, \dots, n+k$ has a prime factor exceeding $n^\epsilon$.

Here's the seed of the idea. For 3 consecutive numbers, one can look at $x^m-1$, $x^m$, $x^m+1$, where $m$ is odd and has many small prime factors. Certainly $x^m$ has only small prime factors relative to its size. And $x^m-1$ and $x^m+1$ factor into cyclotomic polynomials of degree at most $\phi(m)$. So all the prime factors should be of size at most about $x^{\phi(m)}=(x^m)^{\phi(m)/m}$, and $\phi(m)/m$ can be made as small as one wants (by pumping $m$ full of small odd primes). Ultimately this comes down to the Euler's result that $\sum 1/p$ diverges.

It might seem we are stuck, because (e.g.) $x^m-2$ will not have such a nice factorization. But we are free to choose $x$. Suppose we try $x=2^a$. Then $x^m-2 = 2(2^{am-1}-1)$. If we pick $a$ appropriately, we can make $am-1$ also chock full of small primes (we can't get any of the primes appearing in $m$, but we can still put in enough to make sure the reciprocal sum of the primes dividing $am-1$ is large). Then $2^{am-1}-1$ will have only small prime factors relative to its size, for reasons like the last paragraph. And one can play similar games to extend the string to any length, but extracting a quantitative result (as in the Balog--Wooley paper) gets a bit messy.