Analogue of the second Hardy-Littlewood conjecture for numbers of divisors?

Question

Let $f(n)$ denote the proposition "There exists some $k>1$ such that $$ \sum_{m=k}^{k+n-1}\tau(m) < \sum_{m=1}^n\tau(m) $$ where $\tau(m)$ is the number of the divisors of $m$." (This is like the so-called second Hardy-Littlewood conjecture on $\pi(x)+\pi(y)$ vs. $\pi(x+y)$ in testing the initial interval, which has guaranteed good behavior, vs. arbitrary intervals, which can perhaps be chosen cleverly.)

$f$ seems like a natural object, so I imagine this is not a novel question. What is known about $f(n)$ in terms of truth, falsity, or conjecture? The analogy with 2HL makes it look like $f(n)$ could be true for large $n$, but I can't find any.

I've convinced myself that $f(n)$ is false for $n<1000,$ but I don't have a proof. It's easy for small $n$ but it gets increasingly inconvenient for larger $n$.

Answer 1

This is false, as you suspect.

By definition, $$ \sum_{m=k}^{k+n-1} \tau(m)$$ is the sum of the number of divisors of integers $m$ in $[k, k+n)$. Equivalently, it is the sum over $d$ of the number of $m\in [k, k+n)$ such that $d$ divides $m$.

For each $d$, the number of $m\in [k, k+n)$ such that $d$ divides $m$ is at least $\lfloor \frac{n}{d} \rfloor $ since a half-open interval of length $\frac{n}{d}$ contains at least $\lfloor \frac{n}{d} \rfloor $ integers.

For the interval $[1,n)$, this inequality is an equality - there are exactly $\lfloor \frac{n}{d} \rfloor$ integers from $1$ to $n$ that are multiples of $d$.

Thus

$$ \sum_{m=k}^{k+n-1} \tau(m) \geq \sum_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor = \sum_{m=1}^n \tau(m).$$