Euclidean division and parity of the quotient I need this result for a solution of a problem to work. I believe it is true
but I'm having a hard time proving it.
Given two numbers $n,m \in \mathbb{N}$, $n\gg m$ exists $a\in \mathbb{N}$ with $a\in o(n)$ such that $\lfloor \frac{n}{m+a} \rfloor$ is odd.
$m$ is fixed. The proof can work for $n>n_0$ for some $n_0$ arbitrarily
large that can deppend on $m$.
$a$ must be small, I believe that can be taken $a\leq m$.
Thanks in advance.
 A: $a$ at least can't necessarily be $\leq m$, if $n$ is allowed to be large enough; consider $n=2\cdot m\cdot(m+1)\cdots(2m)$, when all of the floor functions are exact divisions that yield even results.
On the other hand, letting $k=m+a$, then $\frac{n}{k}-\frac{n}{k+1} = \frac{n}{k(k+1)}$, and so as soon as $k\geq\sqrt{n}$ then the successive terms of the sequence $a_k=\frac{n}{k}$ differ by less than one (but more than $\frac12$), which means that one of $\lfloor\frac{n}{k}\rfloor$, $\lfloor\frac{n}{k+1}\rfloor$, or $\lfloor\frac{n}{k+2}\rfloor$ must be odd.  Thus, $a$ is at most $O(\sqrt{n})$.  I suspect even better bounds could be derived, but hopefully this is good enough for your needs; the counterexample above shows that $a$ can be as bad as roughly $\Omega(\log n)$.
A: For any $\epsilon>0$ and $n>n_0(\epsilon)$ one may take $a<\max(2m, n^\epsilon)$. To see this consider the sequence $n/k\bmod 2$ where $K\leq k\leq 2K$. If $K>n^\epsilon$, it is asymptotically equidistributed (by a standard application of Weyl-van der Corput and Erdoes-Turan). In particular for about half of the values $k$ we have that $\lfloor n/k\rfloor$ is even, and for about half of them this quantity is odd. Hence take $K=\max(n^\epsilon/2, m)$, and our claim follows. The bound $2m$ can be improved by explicitly computing the discrepancy of the sequence, this would yield $(1+o(1))m$, provided that $n<m^c$. The $o$ then depends on $c$. The $\epsilon$ can also be made explicit. If this is necessary, one should try to copy Weyl's bound for $\zeta(1+it)$, or if this is not sufficient, the work by Vinogradoff. Wooley's efficient congruencing would certainly be better, but probably really difficult.
