jasperE3 wrote: If $n\ge2$ is an integer, prove the equality $$\lfloor\log_2n\rfloor+\lfloor\log_3n\rfloor+\ldots+\lfloor\log_nn\rfloor=\left\lfloor\sqrt n\right\rfloor+\left\lfloor\sqrt[3]n\right\rfloor+\ldots+\left\lfloor\sqrt[n]n\right\rfloor.$$ Let integer $c(m)$ be the number of integers $k\in\{2,3,...,n\}$ such that $\left\lfloor\sqrt[k]n\right\rfloor=m$ We have $S=\sum_{k=2}^n\left\lfloor\sqrt[k]n\right\rfloor=\sum_{m=1}^{+\infty}mc(m)$ $\left\lfloor\sqrt[k]n\right\rfloor=m$ $\iff$ $m+1>n^{\frac 1k}\ge m$ $\iff$ $\frac{\log m+1}{\log n}>\frac 1k\ge\frac{\log m}{\log n}$ If $m=1$, this is $k>\frac{\log n}{\log 2}$ and so $c(1)=n-\left\lfloor\frac{\log n}{\log 2}\right\rfloor$ (remember $k\le n$) If $m>1$, this is $\frac{\log n}{\log m}\ge k>\frac{\log n}{\log m+1}$ $\iff$ $\left\lfloor\frac{\log n}{\log m}\right\rfloor\ge k>\left\lfloor\frac{\log n}{\log m+1}\right\rfloor$ And so (remember $k\ge 1$ : $\forall m<n$ : $c(m)=\left\lfloor\frac{\log n}{\log m}\right\rfloor-\left\lfloor\frac{\log n}{\log m+1}\right\rfloor$ $\forall m\ge n$ : $c(m)=0$ And so $S=c(1)+\sum_{m=2}^{n-1}mc(m)$ $=n-\left\lfloor\frac{\log n}{\log 2}\right\rfloor$ $+\sum_{m=2}^{n-1}m\left(\left\lfloor\frac{\log n}{\log m}\right\rfloor-\left\lfloor\frac{\log n}{\log m+1}\right\rfloor\right)$ $=n+\sum_{m=2}^{n-1}\left\lfloor\frac{\log n}{\log m}\right\rfloor-(n-1)\left\lfloor\frac{\log n}{\log n}\right\rfloor$ $=1+\sum_{m=2}^{n-1}\left\lfloor\frac{\log n}{\log m}\right\rfloor$ $=\sum_{m=2}^{n}\left\lfloor\frac{\log n}{\log m}\right\rfloor$ Q.E.D.

Beautiful problem

Ok so this one actually has a really sick solution.

https://www.youtube.com/watch?v=kpW2V0hD0Ro&t=27s

oty wrote: see here