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

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The discussion focuses on finding integer values of \( x \) that satisfy the equation \( \sum_{k=1}^{x}\lfloor{\sqrt[4]{k}}\rfloor=2x \). The sequence \( a_k \) is defined based on ranges of \( k \), where \( a_k = 1 \) for \( 1 \leq k \leq 15 \), \( a_k = 2 \) for \( 16 \leq k \leq 80 \), and \( a_k = 3 \) for \( 81 \leq k \leq 255 \). Calculations show that \( S_{15} = 15 \) and \( S_{80} = 145 \), leading to the conclusion that for \( n > 80 \), \( S_n \) increases by 3 for each increment in \( n \). The only solution found is \( n = 95 \), which satisfies the equation.
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Find all integers $x$ such that $\left\lfloor{\sqrt[4]{1}}\right\rfloor+\left\lfloor{\sqrt[4]{2}}\right\rfloor+\left\lfloor{\sqrt[4]{3}}\right\rfloor+\cdots+\left\lfloor{\sqrt[4]{x}}\right\rfloor=2x$.
 
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anemone said:
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

$\chi$ $\sigma$
 
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