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I've come across the sum: $$\sum_{k=0}^n \lfloor\sqrt{k}\rfloor$$ According to Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I approached the formula: $$\sum_{k=0}^n \lfloor\sqrt{k}\rfloor = n\lfloor\sqrt{n}\rfloor-\frac{1}{3}\lfloor\sqrt{n}\rfloor^3-\frac{1}{2}\lfloor\sqrt{n}\rfloor^2+\frac{5}{6}\lfloor\sqrt{n}\rfloor$$ Now, programmaticaly I attempted this formula for first $100$ terms. The biggest possible $n$ for which this sum was a prime number should be $47$. Now, is there a precise mathematical approach to this? I can't seem to find a solution.

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    $\begingroup$ Why do you think that there should be "a greatest $n$" with this property? $\endgroup$ – Dietrich Burde Nov 27 '17 at 9:28
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Note that for $n\geq 1$, the sum $S_n$ is a positive integer which can be written as $$S_n:=\sum_{k=0}^n \lfloor\sqrt{k}\rfloor =\frac{1}{6}\cdot\lfloor\sqrt{n}\rfloor\cdot\left(6n-(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)\right).$$ where $f_1:=\lfloor\sqrt{n}\rfloor$ and $f_2:=\left(6n-(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)\right)$ are positive integers.

Therefore if $f_1:=\lfloor\sqrt{n}\rfloor>6$ (i.e. $n\geq 49$) and $$f_2=\left(6n-(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)\right)\geq \left(6n-(2n+3n-5)\right)=n+5>6$$ (i.e. $n\geq 2$) then $S_n=\frac{f_1\cdot f_2}{6}$ is not a prime (because $S_n$ is the product of two integer factors both greater than $1$).

Hence the largest $n$ such that $S_n$ is a prime, exists, it is less than $49$ and, according to your computations, we may conclude that it is equal to $47$ with $S_{47}=197$.

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  • $\begingroup$ Let me get the understanding of this. So if floor(sqrt(n)) is greater than six, that means i am able to factor something out of the bracket and that means it cannot be ever a prime. What i don't get is the line of the inequality: (6n-(2floor(sqrt(n))^+3floor(sqrt(n))-5))>= (6n-(2n+3n-5)) = n+5>6 $\endgroup$ – Michal Dvořák Nov 27 '17 at 15:49
  • $\begingroup$ Because $\lfloor\sqrt{n}\rfloor^2$ and $\lfloor\sqrt{n}\rfloor$ are less or equal to $n$. $\endgroup$ – Robert Z Nov 27 '17 at 15:56
  • $\begingroup$ Wait, I am still confused, would you please reveal it for me in more detail? $\endgroup$ – Michal Dvořák Nov 27 '17 at 16:11
  • $\begingroup$ $\lfloor\sqrt{n}\rfloor^2\leq n$, $\lfloor\sqrt{n}\rfloor\leq n$ imply $(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)\leq (2n+3n-5)=5n-5$. $\endgroup$ – Robert Z Nov 27 '17 at 16:45
  • $\begingroup$ Ah, I missed that $-$ sign there. Now it's clear $\endgroup$ – Michal Dvořák Nov 27 '17 at 16:49

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