Sum over k of Sum over j of Floor of n + jb^k over b^k+1
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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.
Corollary
Let $n, b \in \Z$ such that $n < 0$ and $b \ge 2$.
Then:
- $\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }= n + 1$
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:
\(\ds \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }\) | \(=\) | \(\ds \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{0 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b} - \floor {\dfrac n {b^{k + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - 1} \paren {b - 1} } 2 + \dfrac {\paren {1 - 1} } 2 + 1 \floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } }\) | Summation over k of Floor of mk+x over n | |||||||||||
\(\ds \) | \(=\) | \(\ds \floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } }\) |
Thus:
\(\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }\) | \(=\) | \(\ds \sum_{k \mathop \ge 0} \paren {\floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } }\) | Definition of Telescoping Series | |||||||||||
\(\ds \) | \(=\) | \(\ds n\) |
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 $44$