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By guessing, I obtained: $$\sum_{i=1}^{n}\sum_{d\mid i} (\lfloor \frac{i}{d+1} \rfloor + \lfloor \frac{i}{d-1} \rfloor)=\sum_{i=1}^{n}(\lfloor \frac{n}{i}\rfloor \lfloor \frac{n}{i+1}\rfloor + \lfloor \frac{n}{i(i+1)}\rfloor)$$But I did not figure out how to prove it. When $d=1$, just ignore this term $\lfloor \frac{i}{d-1} \rfloor$.

I know how to prove a similar equality, but I am not sure if the two could be generalized to a same form. The equality is: $$\sum_{i=1}^{n}\sum_{d\mid i} (\lfloor \frac{i}{d} \rfloor + \lfloor \frac{i}{d} \rfloor)=\sum_{i=1}^{n}(\lfloor \frac{n}{i}\rfloor \lfloor \frac{n}{i}\rfloor + \lfloor \frac{n}{i}\rfloor)$$

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  • $\begingroup$ The right hand sides seem to depend on $d$. Should there be a double sum? $\endgroup$ – Theo Bendit Oct 9 '18 at 3:06
  • $\begingroup$ @TheoBendit I am sorry, that was a typo $\endgroup$ – Mayoi Oct 9 '18 at 3:46
  • $\begingroup$ The right hand side is not defined when d=1, which it is for each i. $\endgroup$ – marty cohen Oct 9 '18 at 4:04
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Playing around.

$\begin{array}\\ s_f(n) &=\sum_{i=1}^n \sum_{d|i} f(i, d)\\ &=\sum_{d=1}^n \sum_{k=1}^{[\frac{n}{d}]} f(kd, d) \qquad\text{reversing the order of summation}\\ \end{array} $

If $f(i, d) =[\frac{i}{d}] $, then

$\begin{array}\\ s_f(n) &=\sum_{d=1}^n \sum_{k=1}^{[\frac{n}{d}]} [\frac{kd}{d}]\\ &=\sum_{d=1}^n \sum_{k=1}^{[\frac{n}{d}]} k\\ &=\sum_{d=1}^n \frac12 [\frac{n}{d}]([\frac{n}{d}]+1)\\ \end{array} $

which verifies your second equality.

I assume that this is the way you proved it.

If $f(i, d) =\lfloor \frac{i}{d+1} \rfloor + \lfloor \frac{i}{d-1} \rfloor $, then

$\begin{array}\\ s_f(n) &=\sum_{d=1}^n \sum_{k=1}^{[\frac{n}{d}]}\lfloor \frac{kd}{d+1} \rfloor + \lfloor \frac{kd}{d-1} \rfloor\\ \end{array} $

and I am not sure what you want to happen when $d=1$.

So I will leave it at this.

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  • $\begingroup$ Sorry about that, when d=1, just delete the that term (make it equal to 0) $\endgroup$ – Mayoi Oct 9 '18 at 4:18
  • $\begingroup$ I suspect this method does not work. $\sum_{k=1}^{\frac{n}{d}}f_2(kd,d)=\frac{1}{2}\lfloor \frac{n}{d}\rfloor(\lfloor \frac{n}{d}\rfloor+1)$ holds for each d in the second equality, but this is not true for the first one. $\endgroup$ – Mayoi Oct 9 '18 at 6:24

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