A very strange sequence

There is a sequence defined as follows:

an=⌊n+12+n⌋a_n = \left \lfloor n + \dfrac {1}{2} + \sqrt{n} \right \rfloor

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

Note by Sharky Kesa
5 years, 4 months ago

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Comments

an=⌊n+12+n⌋=n+⌊12+n⌉ a_{n} = \left\lfloor n + \dfrac{1}{2} + \sqrt{n} \right\rfloor = n + \left\lfloor\dfrac{1}{2} + \sqrt{n} \right\rceil
Let n=k+γ \sqrt{n} = k + \gamma
k∈N,0≤γ<1 k \in N, 0 \le \gamma < 1
∴an=n+k+⌊12+γ⌋ \therefore a_{n} = n + k + \left\lfloor\dfrac{1}{2} + \gamma \right\rfloor
an=n+⌊n⌋+1,γ∈[12,1) a_{n} = n + \lfloor\sqrt{n}\rfloor+ 1 , \gamma \in \left[\dfrac{1}{2},1\right)
Or
an=n+⌊n⌋,γ∈[0,12) a_{n} = n + \lfloor\sqrt{n}\rfloor, \gamma \in \left[0,\dfrac{1}{2}\right)
It suffices to prove that if n+⌊n⌋=m2 n + \left \lfloor \sqrt{n} \right \rfloor = m^{2} for some integer m, then γ∈[12,1) \gamma \in \left[\dfrac{1}{2}, 1\right) .
∴an=m2+1 \therefore a_{n} = m^{2} + 1 and is not a perfect square.
Similarly if n+⌊n⌋+1=m2 n + \left \lfloor \sqrt{n} \right \rfloor + 1 = m^{2} then γ∈[0,12) \gamma \in \left[0,\dfrac{1}{2}\right)
∴an=m2−1 \therefore a_{n} = m^{2}-1 and is not a perfect square.

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

A Former Brilliant Member - 5 years, 4 months ago
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