MHB Find $x$ for $\sum_{k=1}^{x}\lfloor{\sqrt[4]{k}}\rfloor=2x$

[anemone]

Find all integers $x$ such that $\left\lfloor{\sqrt[4]{1}}\right\rfloor+\left\lfloor{\sqrt[4]{2}}\right\rfloor+\left\lfloor{\sqrt[4]{3}}\right\rfloor+\cdots+\left\lfloor{\sqrt[4]{x}}\right\rfloor=2x$.

 

[chisigma]

Find all integers $x$ such that $\left\lfloor{\sqrt[4]{1}}\right\rfloor+\left\lfloor{\sqrt[4]{2}}\right\rfloor+\left\lfloor{\sqrt[4]{3}}\right\rfloor+\cdots+\left\lfloor{\sqrt[4]{x}}\right\rfloor=2x$.

[sp]Writing the sequence...

$\displaystyle S_{n} = \sum_{k=1}^{n} a_{k}\ (1)$

... where...

$\displaystyle a_{k} = 1\ \text {if}\ 1 \le k \le 15,\ = 2\ \text{if}\ 16 \le k \le 80,\ = 3\ \text{if}\ 81 \le k \le 255, ...$

... we have to search the value of n for which $\displaystyle S_{n}= 2\ n$. It is easy to see that $\displaystyle S_{15} = 15$ and $\displaystyle S_{80}= 15 + 2\ 65 = 145$. For n> 80 $\displaystyle S_{n}$ increases by 3 at each step so that the value of n for which is $\displaystyle S_{n}= 2\ n$ will be 80 + 160 - 145 = 95. Clearly that is the only solution to the problem...[/sp]

Kind regards

$\chi$ $\sigma$