$\left [ \frac{n}{3} \right ]+\left [ \frac{n+2}{6} \right ]+\left [ \frac{n+4}{6} \right ]=\left [ \frac{n}{2} \right ]+\left [ \frac{n+3}{6} \right ]$ (Wik...)
$\left [\frac{1}{2}+\sqrt{n+\frac{1}{2}}\; \right ]=\left [ \frac{1}{2}+\sqrt{n+\frac{1}{4}} \;\right ]$ (Ibid.)
$[\sqrt{n}+\sqrt{n+1}\;]=[\sqrt{4n+2}\;]$ (Ibid.) See also http://math.colgate.edu/~integers/w33/w33.pdf.
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