Fractional Part Double Summations

Solution 1:

Let $a, b$ and $c$ be positive integers with no common factor, i.e., $\gcd(a,b,c) = 1$. Define $a^{\prime} = \gcd(b,c)$, $b^{\prime} = \gcd(c,a)$, $c^{\prime} = \gcd(a,b)$ and $d = a^{\prime} b^{\prime} c^{\prime}$.

The following identity holds: \begin{align} \sum_{i = 0}^{at} \sum_{j = 0}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace = \tfrac{ab -d }{4} t^{2} + \gamma t \end{align} where $t \in \mathbb{Z}_{\geq 0}$ and \begin{align} \gamma & = a^{\prime} \ \mathfrak{s}(\tfrac{bc}{d},\tfrac{aa^{\prime}}{d}) + b^{\prime} \ \mathfrak{s}(\tfrac{ca}{d},\tfrac{bb^{\prime}}{d}) + c^{\prime} \ \mathfrak{s}(\tfrac{ab}{d}, \tfrac{cc^{\prime}}{d}) + \tfrac{1}{4}(a - a^{\prime} + b - b^{\prime} + c^{\prime} ) - \tfrac{a^{2} b^{2} + c^{2} (c^{\prime})^{2} + d^{2}}{12 a b c}. \end{align} Here, $\mathfrak{s}(p,q) = \tfrac{1}{4q} \sum_{k = 1}^{q-1} \cot(\tfrac{\pi k}{q}) \cot( \tfrac{\pi k p}{q} )$ is the standard Dedekind sum. This more general identity simplifies to the identity above (in the post) as well as $$ \sum_{i = 0}^{at} \sum_{j = 0}^{\lfloor t - i/a \rfloor} \left \lbrace c \left( t- \frac{i}{a} - j \right) \right \rbrace = \frac{t(t+1)}{4} (a - \gcd(a,c)), $$ after an application of the Dedekind Sum Reciprocity Law and taking $b = 1$.