# A very strange sequence

There is a sequence defined as follows:

$a_n = \left \lfloor n + \dfrac {1}{2} + \sqrt{n} \right \rfloor$

for all positive $$n$$. Prove that this sequence goes through all non-square positive integers.

Note by Sharky Kesa
2 years, 4 months ago

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$$a_{n} = \left\lfloor n + \dfrac{1}{2} + \sqrt{n} \right\rfloor = n + \left\lfloor\dfrac{1}{2} + \sqrt{n} \right\rceil$$
Let $$\sqrt{n} = k + \gamma$$
$$k \in N, 0 \le \gamma < 1$$
$$\therefore a_{n} = n + k + \left\lfloor\dfrac{1}{2} + \gamma \right\rfloor$$
$$a_{n} = n + \lfloor\sqrt{n}\rfloor+ 1 , \gamma \in \left[\dfrac{1}{2},1\right)$$
Or
$$a_{n} = n + \lfloor\sqrt{n}\rfloor, \gamma \in \left[0,\dfrac{1}{2}\right)$$
It suffices to prove that if $$n + \left \lfloor \sqrt{n} \right \rfloor = m^{2}$$ for some integer m, then $$\gamma \in \left[\dfrac{1}{2}, 1\right)$$ .
$$\therefore a_{n} = m^{2} + 1$$ and is not a perfect square.
Similarly if $$n + \left \lfloor \sqrt{n} \right \rfloor + 1 = m^{2}$$ then $$\gamma \in \left[0,\dfrac{1}{2}\right)$$
$$\therefore a_{n} = m^{2}-1$$ and is not a perfect square.

I got stuck at proving the interval $$\gamma$$ belongs in.

- 2 years, 4 months ago