The sequence $\{a_n\}^{∞}_{n=1} = \{2,3,5,6,7,8,10,...\}$ consists of all the positive integers that are not perfect squares.
Prove that $a_n= n+ [\sqrt{n} + \frac{1}{2}]$.
Well, I managed to prove that $ m^2 < n+ [\sqrt{n} + \frac{1}{2}] < (m+1)^2 $, where $[\sqrt{n} + \frac{1}{2}] = m$.
But is this enough to answer the question? Or do we also need to prove that $n+ [\sqrt{n} + \frac{1}{2}] $ can take all non-perfect square values. If yes, then how?
Any help is appreciated, thanks!
On
The following is just a first step toward the answer.
From $$ \eqalign{ & a_{\,n} = n + \left\lfloor {\sqrt n + {1 \over 2}} \right\rfloor \cr & a_{\,n + 1} - a_{\,n} = 1 + \left\lfloor {\sqrt {n + 1} + {1 \over 2}} \right\rfloor - \left\lfloor {\sqrt n + {1 \over 2}} \right\rfloor \cr} $$ since we have $$ \eqalign{ & \left\lfloor {x - y} \right\rfloor = - \left\lceil {y - x} \right\rceil = \cr & = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} - \left\{ y \right\}} \right\rfloor = \cr & = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor - \left[ {\left\{ x \right\} < \left\{ y \right\}} \right] \cr} $$ where $[P]$ denotes the Iverson bracket, then $$ \eqalign{ & a_{\,n + 1} - a_{\,n} = 1 + \left\lfloor {\sqrt {n + 1} + {1 \over 2}} \right\rfloor - \left\lfloor {\sqrt n + {1 \over 2}} \right\rfloor = \cr & = 1 + \left\lfloor {\sqrt {n + 1} - \sqrt n } \right\rfloor + \left[ {\left\{ {\sqrt {n + 1} + {1 \over 2}} \right\} < \left\{ {\sqrt n + {1 \over 2}} \right\}} \right] = \cr & = 1 + \left[ {\left\{ {\sqrt {n + 1} + {1 \over 2}} \right\} < \left\{ {\sqrt n + {1 \over 2}} \right\}} \right] \cr} $$
Now the Iverson bracket will be one, and thus we will have a jump in $a_{\,n}$, when $$ \left\{ {\sqrt {n + 1} + {1 \over 2}} \right\} < \left\{ {\sqrt n + {1 \over 2}} \right\} $$ which for $1 \le n$ , which means $sqrt{n+1} -\sqrt{n} < 1/2$, can occur only when $$ \sqrt n + {1 \over 2} < m < \sqrt {n + 1} + {1 \over 2} $$ with $m$ an integer, that is $$ \eqalign{ & \sqrt n < m - {1 \over 2} < \sqrt {n + 1} \quad \Rightarrow \cr & \Rightarrow \quad n < m^{\,2} + {1 \over 4} - m < n + 1\quad \Rightarrow \cr \quad \Rightarrow \cr & \Rightarrow \quad n - {1 \over 4} < m\left( {m - 1} \right) < n + {3 \over 4}\quad \Rightarrow \cr & \Rightarrow \quad n_ * = m\left( {m - 1} \right) \cr} $$
Thus $$ a_{\,n + 1} - a_{\,n} = 2\quad {\rm iff}\quad n = m\left( {m - 1} \right) $$
It remains then to demonstrate that $$ \forall q\;\exists m:\quad a_{\,n_{\, * } } + 1 = m\left( {m - 1} \right) + \left\lfloor {\sqrt {m\left( {m - 1} \right)} + {1 \over 2}} \right\rfloor + 1 = q^{\,2} $$
The increment $$a_{n+1}-a_n=1+\left\lfloor\sqrt{n+1}+\dfrac12\right\rfloor-\left\lfloor\sqrt{n}+\dfrac12\right\rfloor$$
is $1$, or $2$ when $\sqrt{n+1}$ rounds differently than $\sqrt n$.