Can it be shown that $\lceil n\cdot log_ab\rceil$ and $\lceil n\cdot log_\frac{b}{a}b\rceil$ are "complementary" ($1<a<b$ , $b$ is not a power of $a$) ?
By "complementary" I mean that they do not overlap and they "cover" all integers > 1 for (for integer $n>0$)
e.g. $\lceil n\cdot log_25\rceil$ for $n>0$ gives $\{3,5,7,10,12,14,17,19,21,24,26,28,31...\}$
and $\lceil n\cdot log_\frac{5}{2}5\rceil$ gives $\{2,4,6,8,9,11,13,15,16,18,20,22,23,25,27,29,30,32...\}$
I can show that for any $n$ there is an integer $i$ such that (Note: interchange $a$ and $\frac{b}{a}$ if $b>a^2$ in the below formula): $$\lceil i\cdot log_ab\rceil+1=\lceil n\cdot log_\frac{b}{a}b\rceil=\lceil (i+1)\cdot log_ab\rceil-1$$
which means that a "$log_\frac{b}{a}b$" element is distinct from and surrounded by 2 "$log_ab$" elements, but that's not enough (I must show there is no other gap in consecutive "$log_ab$" elements) and I feel that I am missing a much simpler way to show this (or find a counter example).
Let $r=\log_ab$ and $s=\log_{b/a}b$. What is $1/r+1/s$? $$\frac1r+\frac1s=\log_ba+\log_b\frac ba=\log_bb=1$$ So the two sequences are complementary Beatty sequences, and partition the integers from $2$ onwards into two sets.