In attempt to deepen my understanding of Dedekind sums, I've proven the following identity $$ \sum_{i = 0}^{t} \sum_{j = 0}^{b(t-i)} \left \lbrace c \left( t - i - \frac{j}{b} \right) \right \rbrace = \frac{t(t+1)}{4}(b - \gcd(b,c)), $$ where $b,c,t \in \mathbb{N}$ and $\lbrace \cdot \rbrace$ denotes the fractional part function.

I'd like to compute the following more general double sums $$ \sum_{i = 0}^{a t} \sum_{j = 0}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace \qquad \text{and} \qquad \sum_{i = 1}^{a t} \sum_{j = 1}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace \qquad a, b, c, t \in \mathbb{N}, $$ where $\lfloor \cdot \rfloor$ denotes the floor function. The sums clearly vanish if $a$ and $b$ are divisors of $c$ (by the vanishing of the fractional part) and (I expect) grow as $O(t^{2})$ otherwise. Are such sums in the literature, even for some special cases of pairwise coprime coefficients $a, b$ and $c$? Any help and/or hints are certainly appreciated!

Harder Problem(s): If the above is trivial, then suppose $a, b, c$ and $t$ are real and compute the following sums $$ \sum_{i = 0}^{\lfloor a t \rfloor} \left \lbrace b \left( t - \frac{i}{a} \right) \right \rbrace \qquad \text{and} \qquad \sum_{i = 0}^{\lfloor a t \rfloor} \sum_{j = 0}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace $$ as well as $$ \sum_{i = 1}^{\lfloor a t \rfloor} \left \lbrace b \left( t - \frac{i}{a} \right) \right \rbrace \qquad \text{and} \qquad \sum_{i = 1}^{\lfloor a t \rfloor} \sum_{j = 1}^{\lfloor b(t-i/a) \rfloor} \left \lbrace c \left( t - \frac{i}{a} - \frac{j}{b} \right) \right \rbrace. $$

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    $\begingroup$ I want to look at this, but I want to make sure I'm looking at the right function. Calling your double sum $f(a,b,c,t)$, can you confirm that the values of $f(2,3,5,t)$ for $t=1,\dots,9$ are $7/3,43/6,29/2,73/3,110/3,103/2,413/6,266/3,111$? $\endgroup$ Jul 7, 2011 at 5:20
  • $\begingroup$ @Gerry: Yes. The next one in the sequence is $\frac{815}{6}$. $\endgroup$
    – user02138
    Jul 7, 2011 at 5:36

1 Answer 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 $b = 1$.

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    $\begingroup$ Incredible work good sir /madam thankyou for sharing this with everyone $\endgroup$ Aug 22, 2018 at 16:10

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