# Logarithms of ratios of squarefree numbers

Let $M \geq 1$, and $N=2^M$. Let $a_1, \dots, a_N$ be the set of all the numbers that you get when forming all square-free products of the first $M$ primes.

For example, for $M=2$ and $N=4$ you get $\{a_1, a_2, a_3, a_4\}=\{1,2,3,6\}$.

The question is: Find a good upper bound for the sum $$\sum_{1 \leq m \neq n \leq N} \frac{1}{|\log a_m/a_n|}.$$ I want to find an upper bound for this sum which is polynomial in $N$. (From the maximal size of $a_n$ on can quite easily deduce an upper bound of size roughly $\exp(\log N \log \log N)$, which does not help me. It really has to be polynomial in $N$.)

The problem turns up in the context of proving an upper bound for $$\int_0^T \sum_{1 \leq m,n \leq N} \cos(t (\log a_m/a_n))~dt,$$ where $T$ is of size $N^\beta$ for some (arbitrary, but fixed) exponent $\beta$. Evidently, the terms for which $n=m$ contribute $TN$. If for a single summand we have $n \neq m$, then the integral is $$\frac{\sin (T (\log a_m/a_n))}{\log a_m/a_n} \leq \frac{1}{|\log a_m/a_n|}.$$ The question is, whether the $TN$ from the terms with $n=m$ is the dominant contribution, or whether the other terms can be significantly larger. Thus a small number of pairs $m,n$ for which $a_m/a_n$ is very close to 1 is no problem, since the integral is always bounded by $T$. That means, $cN$ terms for some constant $c$ may be discarded from the sum.

The problem is somewhat related to Baker's theorems on linear forms of logarithms, etc. (which is not strong enough to get something here). However, the setting here is quite specific, so maybe someone has a good idea.

• I get 0 for the sum as written. What do you get? – The Masked Avenger Feb 12 '14 at 3:04
• Evidently the denominator should be $\left| \log a_m/a_n \right|$, not $\log \left| a_m/a_n \right|$ (NB $a_m/a_n$ is already positive). – Noam D. Elkies Feb 12 '14 at 3:16
• Or perhaps m is strictly less than n? In another MO question mathoverflow.net/questions/86502/estimate-about-primes/… the number of small ratios greater than 1 seems to be superpolynomial in N, so I suspect disappointment is in store for the poster. – The Masked Avenger Feb 12 '14 at 3:27
• Oops. Looks like I switched M and N. The poster may get his desired bound after all, as the largest summand is less than NM and there are less than N^2 terms. I think the largest summand is exponential in M, however. – The Masked Avenger Feb 12 '14 at 4:22
• To fix the scale here: let $B$ be the largest of the $a_j$, so that $B \approx e^{M \log M} = M^M$ is the product of the first $M$ primes. The type of nearness that would rule out a polynomial upper bound would be something like: two numbers around $B^{1/3}$, each squarefree and $(\log B)$-friable, whose difference is about $B^{1/6}$. The numbers $\frac16$ and $\frac13$ aren't important - they can be replaced by any $0<\alpha<\beta\le\frac12$. – Greg Martin Feb 12 '14 at 10:06