Sum over k of Sum over j of Floor of n + jb^k over b^k+1/Corollary
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Corollary to Sum over $k$ of Sum over $j$ of $\floor {\frac {n + j b^k} {b^{k + 1} } }$
Let $n, b \in \Z$ such that $n < 0$ and $b \ge 2$.
Then:
- $\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }= n + 1$
where $\floor {\, \cdot \,}$ denotes the floor function.
Proof
From the working of Sum over $k$ of Sum over $j$ of $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$:
- $\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } }$
But:
- $\forall k \in \Z_{> 0}: \dfrac n {b^{k + 1} } < 0$
and so:
- $\ds \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } } = n - \paren {-1}$
Hence the result.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $44$