For $a$ and $n$, there is formula to calculate
$a + 2a + 3a + \cdots + na = \frac{n(n+1)}{2} a.$
Is there formula:
$\lfloor a\rfloor + \lfloor 2a\rfloor + \lfloor 3a\rfloor + \cdots + \lfloor na\rfloor $
For $a$ and $n$, there is formula to calculate
$a + 2a + 3a + \cdots + na = \frac{n(n+1)}{2} a.$
Is there formula:
$\lfloor a\rfloor + \lfloor 2a\rfloor + \lfloor 3a\rfloor + \cdots + \lfloor na\rfloor $
There is a closed form solution to the sum $\sum_{0\le k. Using this fact, the terms of the sum can be reordered in a way that leads to a radical simplification: $ \sum_{0\le k
There seems no exact formula for the sum, considering that the sum depends highly sensitively on the fraction part of $a$. But we can give an asymptotic formula:
Case 1. If $a$ is rational, write $a = p/q$ where $p$ and $q$ are positive coprime integers. Then $kp \ \mathrm{mod} \ q$ attains every value in $\{0, 1, \cdots, q-1\}$ exactly once whenever $k$ runs through $q$ successive integers. Thus if we write $n = mq + r$,
$ \begin{align*} \sum_{k=1}^{n} (ka - \lfloor ka \rfloor) &= \sum_{k=1}^{mq} (ka - \lfloor ka \rfloor) + \sum_{k=1}^{r} (ka - \lfloor ka \rfloor) \\ &= \frac{m(q-1)}{2} + O(1) = \frac{n(q-1)}{2q} + O(1). \end{align*}$
This gives
$ \sum_{k=1}^{n} \lfloor ka \rfloor = \frac{1}{2}n\left(n+\frac{1}{q}\right) + O(1). $
Case 2. If $a$ is irrational, then the fractional parts $\langle ka \rangle := ka - \lfloor ka \rfloor$ is equidistributed on $[0, 1]$ by Weyl's criterion. Thus $ \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \langle ka \rangle = \int_{0}^{1} x \, dx = \frac{1}{2} \quad \Longrightarrow \quad \sum_{k=1}^{n} \langle ka \rangle = \frac{n}{2} + o(n) $ and we have $ \sum_{k=1}^{n} \lfloor ka \rfloor = \frac{n^2}{2} + o(n). $