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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
10 months, 2 weeks 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. Vighnesh Shenoy · 10 months, 2 weeks ago

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