Separate the natural numbers into two sequences , namely $F(n)$ & $G(n)$ such that $F(n)$ contains the squares and $G(n)$ contains the non squares, i.e.,
$F(n)=1,4,9,16,25,36,49,....$
& $G(n)=2,3,5,6,7,8,10,........$
Surely $F(n)=n^{2}$.Now is there any formula to calculate $G(n)$?If there is ,then what is it?
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Sequences of interest
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- Phlembac Adib Hasan
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Re: Sequences of interest
I don't understand.What did you mean by 'formula'?
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Re: Sequences of interest
\[ G(n) = \left\lceil n-\frac{1}{2}+\sqrt{n-\frac{1}{2}}\right\rceil \]
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- nafistiham
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Re: Sequences of interest
There's the PM button.I hope now you can see it.SANZEED wrote:
Note:I don't find any option to send private message.can anyone help me to find it please?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Sequences of interest
Sorry,But my answer is as follows:
$n$ plus the ceiling of the sum of half and the square root of$n$
Is it the same as yours?I am confused
$n$ plus the ceiling of the sum of half and the square root of$n$
Is it the same as yours?I am confused
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: Sequences of interest
Yes! Let, $[x]$ denotes $x$ rounded to closest integer. Then it is easy to find, $[x] = \lfloor x + \frac{1}{2} \rfloor$
And, another obvious thing is, $\lceil x \rceil = \lfloor x \rfloor +1$ iff $x \not\in \mathbb{Z}$
It can also be easily found that, $\left[\sqrt{n}\right]=\left[\sqrt{n-\frac{1}{2}}\right]$ for all $n\in \mathbb{N}$
So, the solution is actually $G(n)= n +\left[\sqrt{n}\right] $
And, another obvious thing is, $\lceil x \rceil = \lfloor x \rfloor +1$ iff $x \not\in \mathbb{Z}$
It can also be easily found that, $\left[\sqrt{n}\right]=\left[\sqrt{n-\frac{1}{2}}\right]$ for all $n\in \mathbb{N}$
So, the solution is actually $G(n)= n +\left[\sqrt{n}\right] $
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