Sum over k of Sum over j of Floor of n + jb^k over b^k+1

Theorem
Let $n, b \in \Z$ such that $n \ge 0$ and $b \ge 2$.

Then:


 * $\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = n$

where $\floor {\, \cdot \,}$ denotes the floor function.

Proof
We have that $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$ is in the form $\floor {\dfrac {m k + x} n}$ so that:

Thus:

Hence the result.