Sequences of interest

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SANZEED
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Sequences of interest

Unread post by SANZEED » Thu Jan 26, 2012 7:09 am

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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Phlembac Adib Hasan
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Re: Sequences of interest

Unread post by Phlembac Adib Hasan » Thu Jan 26, 2012 10:36 am

I don't understand.What did you mean by 'formula'?
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Corei13
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Re: Sequences of interest

Unread post by Corei13 » Thu Jan 26, 2012 3:17 pm

\[ G(n) = \left\lceil n-\frac{1}{2}+\sqrt{n-\frac{1}{2}}\right\rceil \]
:D
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Re: Sequences of interest

Unread post by nafistiham » Fri Jan 27, 2012 12:35 am

SANZEED wrote:
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SANZEED
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Re: Sequences of interest

Unread post by SANZEED » Tue Jan 31, 2012 10:46 pm

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 :?
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Corei13
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Re: Sequences of interest

Unread post by Corei13 » Tue Jan 31, 2012 11:02 pm

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] $
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