I want to find out the best way to perform a concret summation. I know for instance that:

$$ \sum_{i=1}^{p}\left \lfloor \sqrt{i} \right \rfloor=\int_{1}^{p+1}\left \lfloor \sqrt{i} \right \rfloor di $$ and then $$ \int_{1}^{p+1}\left \lfloor \sqrt{i} \right \rfloor di=1/6\left (\left \lfloor \sqrt{p+1} \right \rfloor \right )\left ( 6p+5-2\left \lfloor \sqrt{p+1} \right \rfloor^{2}-3\left \lfloor \sqrt{p+1} \right \rfloor \right ) $$ I would like to know the best way to compute $$ \sum_{i=1}^{p}\left \lfloor \sqrt{ip} \right \rfloor $$ Is there an exact formula, like in the first case? If not, what would be the fastest method?

  • 1
    $\begingroup$ The answer is an integer. Why are you talking about decimal positions? $\endgroup$
    – WhatsUp
    Commented Jul 2, 2020 at 18:49
  • $\begingroup$ Right! Sorry, I was messing with another equation at the same time. I will edit the question. Thanks! $\endgroup$
    – Josi
    Commented Jul 2, 2020 at 18:53
  • $\begingroup$ No need to complicate this, think about it. How many times are you going to add $1$? ($3$ times), what about $2$? ($5$ times), what about $r$? ($2r+1$) times. Now try doing this by assuming certain restrictions on $p$, for example, $m^2 \leq p <(m+1)^2$ for some $m$. $\endgroup$
    – Nikunj
    Commented Jul 2, 2020 at 19:06


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